1 00:00:00,000 --> 00:00:01,940 NARRATOR: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 Creative Commons License. 3 00:00:03,690 --> 00:00:06,640 Your support will help MIT OpenCourseWare continue to 4 00:00:06,640 --> 00:00:09,990 offer high quality educational resources for free. 5 00:00:09,990 --> 00:00:12,830 To make a donation or to view additional materials from 6 00:00:12,830 --> 00:00:16,760 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,760 --> 00:00:18,010 ocw.mit.edu. 8 00:00:36,900 --> 00:00:37,630 PROFESSOR: Hi. 9 00:00:37,630 --> 00:00:40,520 Welcome, once again, to another lecture on limits. 10 00:00:40,520 --> 00:00:44,880 Actually, from a certain point of view, today's lecture will 11 00:00:44,880 --> 00:00:47,290 be the same as the last lecture, only 12 00:00:47,290 --> 00:00:49,130 from a different viewpoint. 13 00:00:49,130 --> 00:00:51,850 Our lecture today is called 'Limits: a 14 00:00:51,850 --> 00:00:53,440 More Rigorous Approach'. 15 00:00:53,440 --> 00:00:57,690 And what our objective for the day is, aside from helping you 16 00:00:57,690 --> 00:01:01,660 gain experience and facility with using limit expressions 17 00:01:01,660 --> 00:01:05,630 and absolute values and the like, is to have you see how 18 00:01:05,630 --> 00:01:10,740 we can use the power of objective, well-defined 19 00:01:10,740 --> 00:01:15,220 mathematical definitions to find rather easy ways of 20 00:01:15,220 --> 00:01:18,210 solving certain types of problems. 21 00:01:18,210 --> 00:01:23,620 Now to this end, let's very briefly review our fundamental 22 00:01:23,620 --> 00:01:26,730 definition of last time. 23 00:01:26,730 --> 00:01:31,540 Namely, the limit of 'f of x' as 'x' approaches 'a' equals 24 00:01:31,540 --> 00:01:36,820 'l' means that for each epsilon greater than 0, we can 25 00:01:36,820 --> 00:01:40,770 find delta greater than 0, such that whenever the 26 00:01:40,770 --> 00:01:45,120 absolute value of 'x minus a' is less than delta but greater 27 00:01:45,120 --> 00:01:48,580 than 0, then the absolute value of ''f of x' minus 'l'' 28 00:01:48,580 --> 00:01:50,800 will be less than epsilon. 29 00:01:50,800 --> 00:01:56,270 To state that once again, but in more intuitive terms, for 30 00:01:56,270 --> 00:02:02,400 any tolerance limit epsilon at all, we can suitably find 31 00:02:02,400 --> 00:02:06,630 another tolerance limit, delta, such that whenever we 32 00:02:06,630 --> 00:02:12,130 make 'x' within delta of 'a', 'f of x' will automatically be 33 00:02:12,130 --> 00:02:14,800 within epsilon of 'l'. 34 00:02:14,800 --> 00:02:19,290 And again, the very, very important emphasis here, we do 35 00:02:19,290 --> 00:02:21,980 not allow 'x' to equal 'a'. 36 00:02:21,980 --> 00:02:25,680 Again, in terms of a diagram, if this is the curve 'y' 37 00:02:25,680 --> 00:02:28,650 equals 'f of x', this is 'x' equals 'a'. 38 00:02:28,650 --> 00:02:29,890 This is 'l'. 39 00:02:29,890 --> 00:02:34,710 If we call this 'l' plus epsilon, if we call this 'l' 40 00:02:34,710 --> 00:02:36,950 minus epsilon-- in other words, epsilon is this width 41 00:02:36,950 --> 00:02:38,230 over here-- 42 00:02:38,230 --> 00:02:44,520 then the way we find delta is to reflect back to the curve, 43 00:02:44,520 --> 00:02:47,100 emphasizing, again, in the neighborhood 44 00:02:47,100 --> 00:02:48,360 of the point 'a'-- 45 00:02:48,360 --> 00:02:51,230 I can't stress that point strongly enough, that if this 46 00:02:51,230 --> 00:02:55,510 curve were not 1:1, there are going to be other places where 47 00:02:55,510 --> 00:02:57,690 this line meets the curve. 48 00:02:57,690 --> 00:03:01,320 So if that happens, you must make sure that you pick the 49 00:03:01,320 --> 00:03:06,280 neighborhood of 'a', not some other point over here. 50 00:03:06,280 --> 00:03:08,530 We're interested in what happens near 'a'. 51 00:03:08,530 --> 00:03:11,660 But at any rate, again notice that the fact that these two 52 00:03:11,660 --> 00:03:15,670 intervals here were equal does not guarantee that when you 53 00:03:15,670 --> 00:03:18,090 project down here, they will be equal. 54 00:03:18,090 --> 00:03:20,480 In fact, the only time that these two widths would be 55 00:03:20,480 --> 00:03:24,210 equal is if the curve happened to be a straight line. 56 00:03:24,210 --> 00:03:27,120 And what it means, again, is that the delta that we're 57 00:03:27,120 --> 00:03:30,170 talking about, for example, is the minimum 58 00:03:30,170 --> 00:03:32,140 of these two widths. 59 00:03:32,140 --> 00:03:38,310 In other words, as I've drawn this diagram, delta would be 60 00:03:38,310 --> 00:03:41,900 the distance from 'a' to this end point. 61 00:03:41,900 --> 00:03:46,560 And all we're saying is that once 'x' is in this open 62 00:03:46,560 --> 00:03:51,090 interval but not including 'a' itself, 'f of x' will be in 63 00:03:51,090 --> 00:03:52,960 this open interval. 64 00:03:52,960 --> 00:03:56,140 And again, notice, as we were emphasizing last time, once 65 00:03:56,140 --> 00:04:01,970 this delta happens to work, automatically any smaller 66 00:04:01,970 --> 00:04:03,700 delta will also work. 67 00:04:03,700 --> 00:04:06,640 In other words, if something is true for everything in 68 00:04:06,640 --> 00:04:10,250 here, it's certainly true for everything in some 69 00:04:10,250 --> 00:04:11,370 subinterval of this. 70 00:04:11,370 --> 00:04:13,440 What you must be careful about is not to 71 00:04:13,440 --> 00:04:14,520 reverse this process. 72 00:04:14,520 --> 00:04:18,100 Don't get outside and pick bigger deltas, then you might 73 00:04:18,100 --> 00:04:20,320 very well be in a little bit of trouble. 74 00:04:20,320 --> 00:04:23,200 At any rate, once we've reviewed what the basic 75 00:04:23,200 --> 00:04:27,900 definition is, it seems about the only way to show what 76 00:04:27,900 --> 00:04:33,290 mathematics is all about is to actually do a few theorems, 77 00:04:33,290 --> 00:04:37,050 that is, derive a few inescapable consequences of 78 00:04:37,050 --> 00:04:42,830 the definition that show how our theorems coincide with 79 00:04:42,830 --> 00:04:45,150 what we believe to be intuitively true 80 00:04:45,150 --> 00:04:46,470 in the first place. 81 00:04:46,470 --> 00:04:48,840 And for obvious reasons, we should start with what 82 00:04:48,840 --> 00:04:51,750 hopefully would be the simplest possible theorems and 83 00:04:51,750 --> 00:04:54,630 then proceed to tougher ones as we go along. 84 00:04:54,630 --> 00:04:58,130 So as my first one, I've chosen the following. 85 00:04:58,130 --> 00:05:02,170 The limit of 'c' as 'x' approaches 'a' is 'c', where 86 00:05:02,170 --> 00:05:04,080 'c' is any constant. 87 00:05:04,080 --> 00:05:09,520 Now again, that may look a little bit strange to you. 88 00:05:09,520 --> 00:05:10,560 Let's look at it this way. 89 00:05:10,560 --> 00:05:16,250 What I'm saying is let 'f of x' equal 'c', 90 00:05:16,250 --> 00:05:18,120 where 'c' is a constant. 91 00:05:18,120 --> 00:05:22,600 Then what we're saying is for this choice of 'f of x', the 92 00:05:22,600 --> 00:05:27,410 limit of 'f of x' as 'x' approaches 'a' is 'c' itself. 93 00:05:27,410 --> 00:05:29,280 That's what this thing says. 94 00:05:29,280 --> 00:05:30,450 How do we prove this? 95 00:05:30,450 --> 00:05:33,680 Well, you see, we have a criteria given. 96 00:05:33,680 --> 00:05:36,190 Namely, what is our basic definition? 97 00:05:36,190 --> 00:05:39,750 Let's just juxtaposition our basic definition with this 98 00:05:39,750 --> 00:05:40,940 particular result. 99 00:05:40,940 --> 00:05:44,800 To prove that the limit here is 'c', notice that in this 100 00:05:44,800 --> 00:05:51,150 particular problem, if we come back and compare this with our 101 00:05:51,150 --> 00:05:54,180 basic definition, notice that we have the same basic 102 00:05:54,180 --> 00:06:00,630 definition as before, only the role of 'f of x' is played by 103 00:06:00,630 --> 00:06:03,390 'c' and 'l' is also played by 'c'. 104 00:06:03,390 --> 00:06:05,710 In other words, in this particular problem 'f of x' is 105 00:06:05,710 --> 00:06:09,230 'c' and 'l' is also 'c'. 106 00:06:09,230 --> 00:06:10,410 Now what must we do? 107 00:06:10,410 --> 00:06:14,430 We must show that for each epsilon greater than 0, or for 108 00:06:14,430 --> 00:06:16,910 an arbitrary epsilon greater than 0-- 109 00:06:16,910 --> 00:06:19,530 now what does it mean to say an arbitrary epsilon? 110 00:06:19,530 --> 00:06:22,590 In a way, think of it as being a game, a battle of wits 111 00:06:22,590 --> 00:06:25,320 between you and your worst enemy, and you're out to win 112 00:06:25,320 --> 00:06:27,160 and your worst enemy is out to beat you. 113 00:06:27,160 --> 00:06:31,180 So to make this as difficult a game as possible, you allow 114 00:06:31,180 --> 00:06:33,790 your worst enemy to choose the epsilon, provided only that 115 00:06:33,790 --> 00:06:36,570 it's positive. 116 00:06:36,570 --> 00:06:39,690 For whichever one he picks, you must be able to find the 117 00:06:39,690 --> 00:06:43,340 delta that matches that epsilon, such that what? 118 00:06:43,340 --> 00:06:47,750 Whenever 'x' is within delta of 'a' but not equal to 'a', 119 00:06:47,750 --> 00:06:51,170 'f of x' will be within epsilon of 'l'. 120 00:06:51,170 --> 00:06:52,900 So let's write that down over here. 121 00:06:52,900 --> 00:06:54,330 Let's write what that says. 122 00:06:54,330 --> 00:07:05,480 Given epsilon greater than 0, we must find a delta greater 123 00:07:05,480 --> 00:07:11,040 than 0, such that what? 124 00:07:11,040 --> 00:07:12,590 Well, by definition. 125 00:07:12,590 --> 00:07:19,330 Such that when the absolute value of 'x minus a' is less 126 00:07:19,330 --> 00:07:23,160 than delta but greater than 0, the absolute 127 00:07:23,160 --> 00:07:24,620 value of 'f of x'-- 128 00:07:24,620 --> 00:07:26,660 well, that of course, is 'c' in this case-- 129 00:07:26,660 --> 00:07:30,430 minus 'l', which is also 'c' in this case, the 'l' stands 130 00:07:30,430 --> 00:07:33,640 for the limit, has to be less than epsilon. 131 00:07:33,640 --> 00:07:37,740 And lo and behold, we find that this is a rather simple 132 00:07:37,740 --> 00:07:41,700 procedure because what is 'c' minus 'c'? 133 00:07:41,700 --> 00:07:45,090 'c' minus 'c' is 0, and automatically that will be 134 00:07:45,090 --> 00:07:47,710 smaller than any positive epsilon. 135 00:07:47,710 --> 00:07:50,520 In other words, what this says is even your worst enemy can't 136 00:07:50,520 --> 00:07:52,720 give you a tough time with this problem. 137 00:07:52,720 --> 00:07:56,340 Namely, no matter what epsilon he prescribes, no matter how 138 00:07:56,340 --> 00:08:01,020 sensitive, you say, well, for delta I'll pick anything. 139 00:08:01,020 --> 00:08:02,670 And it works. 140 00:08:02,670 --> 00:08:03,700 And why does that work? 141 00:08:03,700 --> 00:08:06,350 Well, here again we can emphasize 142 00:08:06,350 --> 00:08:07,920 the geometric approach. 143 00:08:07,920 --> 00:08:14,980 Namely, if we plot the graph 'f of x' equals 'c', observe 144 00:08:14,980 --> 00:08:18,180 that that plot equals a straight line, 'y' equals 'c'. 145 00:08:18,180 --> 00:08:21,150 Now take 'x' equals 'a' over here. 146 00:08:21,150 --> 00:08:23,290 What is 'f of a'? 147 00:08:23,290 --> 00:08:25,860 'f of a' is 'c'. 148 00:08:25,860 --> 00:08:28,860 Now pick an epsilon, which is this half-width over here, 149 00:08:28,860 --> 00:08:34,220 knock that off on either side of 'c'. 150 00:08:34,220 --> 00:08:36,780 And all you're saying in this particular case is no matter 151 00:08:36,780 --> 00:08:41,679 how far away 'x' is from 'a', 'f of x' will be within these 152 00:08:41,679 --> 00:08:43,860 tolerance limits of 'c'. 153 00:08:43,860 --> 00:08:45,050 And why is that? 154 00:08:45,050 --> 00:08:50,040 Because 'f' is defined in such a way that the output for any 155 00:08:50,040 --> 00:08:52,360 element in its domain is 'c' itself. 156 00:08:52,360 --> 00:08:55,930 In other words, every element maps up here. 157 00:08:55,930 --> 00:08:58,390 This is what you mean by saying that the graph is the 158 00:08:58,390 --> 00:09:00,160 straight line 'y' equals 'c'. 159 00:09:00,160 --> 00:09:02,110 By the way, another word of caution. 160 00:09:02,110 --> 00:09:09,930 We must always make sure that your solution does not depend 161 00:09:09,930 --> 00:09:10,930 on the diagram. 162 00:09:10,930 --> 00:09:13,160 You see, if a person were to look at this diagram very 163 00:09:13,160 --> 00:09:16,330 quickly, he would assume that 'c' had to be a positive 164 00:09:16,330 --> 00:09:19,470 constant here, because look at how I've drawn the line 'y' 165 00:09:19,470 --> 00:09:19,870 equals 'c'. 166 00:09:19,870 --> 00:09:21,430 It's above the x-axis. 167 00:09:21,430 --> 00:09:23,900 Well, 'c' could just as easily be a negative constant. 168 00:09:23,900 --> 00:09:25,990 And if I drew the diagram that way, 'c' would 169 00:09:25,990 --> 00:09:27,690 be below the x-axis. 170 00:09:27,690 --> 00:09:29,290 The important thing to check is this. 171 00:09:29,290 --> 00:09:32,750 When you draw a diagram, you can't have a certain number 172 00:09:32,750 --> 00:09:35,780 being positive and negative at the same time, so you choose 173 00:09:35,780 --> 00:09:37,650 it one way or the other. 174 00:09:37,650 --> 00:09:41,960 Always make sure when you do this that your formal proof, 175 00:09:41,960 --> 00:09:45,150 your analytic proof, goes through word for word if you 176 00:09:45,150 --> 00:09:47,800 reverse the signature of the sign of the number that you're 177 00:09:47,800 --> 00:09:48,920 working with. 178 00:09:48,920 --> 00:09:53,010 Make sure that your answer does not depend on the picture 179 00:09:53,010 --> 00:09:54,240 that you've drawn. 180 00:09:54,240 --> 00:09:57,040 Make sure that your answer follows, inescapably, from 181 00:09:57,040 --> 00:09:58,410 your basic definition. 182 00:09:58,410 --> 00:10:00,000 And notice that this is what we did here. 183 00:10:00,000 --> 00:10:01,200 We showed what? 184 00:10:01,200 --> 00:10:04,210 That no matter what epsilon we were given, we could find a 185 00:10:04,210 --> 00:10:06,420 delta-- in fact, in this case, any delta-- 186 00:10:06,420 --> 00:10:09,950 such that when 'x' was within delta of a as long as 'x' 187 00:10:09,950 --> 00:10:10,790 wasn't equal to 'a'. 188 00:10:10,790 --> 00:10:12,940 Well in this case, even if 'x' equaled 'a', 189 00:10:12,940 --> 00:10:14,160 there was no harm done. 190 00:10:14,160 --> 00:10:17,780 But the point is we want to keep away from a 0 over 0 form 191 00:10:17,780 --> 00:10:19,760 so we always impose this condition. 192 00:10:19,760 --> 00:10:23,940 In this case, once 'x' was within delta of 'a', 'f of x' 193 00:10:23,940 --> 00:10:27,410 was automatically within epsilon of 'c', as long as 194 00:10:27,410 --> 00:10:29,920 epsilon was positive, because 'f of x' was 195 00:10:29,920 --> 00:10:31,570 already equal to 'c'. 196 00:10:31,570 --> 00:10:34,010 The difference was already 0. 197 00:10:34,010 --> 00:10:36,620 Now again, notice what happens here. 198 00:10:36,620 --> 00:10:41,850 The more pragmatic student will say, why did you use this 199 00:10:41,850 --> 00:10:46,660 long math when it was obvious from either the diagram or 200 00:10:46,660 --> 00:10:49,820 from intuition that this is the correct answer? 201 00:10:49,820 --> 00:10:52,380 And the reason, is as we have already seen and as we will 202 00:10:52,380 --> 00:10:55,640 see many, many times during our course, the intuitive 203 00:10:55,640 --> 00:10:58,270 answer and the correct answer will not 204 00:10:58,270 --> 00:11:00,200 necessarily be the same. 205 00:11:00,200 --> 00:11:03,650 The point is we always want to make sure that when we prove 206 00:11:03,650 --> 00:11:09,110 something, the proof follows from the assumed definitions 207 00:11:09,110 --> 00:11:11,770 in a logically rigorous way. 208 00:11:11,770 --> 00:11:15,080 If it happens once you prove this that you can intuitively 209 00:11:15,080 --> 00:11:18,340 recognize the same result, that's like a double reward 210 00:11:18,340 --> 00:11:21,240 because now you won't have to memorize what the result was, 211 00:11:21,240 --> 00:11:22,980 you'll just use this thing automatically. 212 00:11:22,980 --> 00:11:25,910 But the beauty is that for somebody who doesn't have the 213 00:11:25,910 --> 00:11:29,290 same intuitive insights that you have, if he says it's not 214 00:11:29,290 --> 00:11:32,240 self-evident to me, explain to me what's happening. 215 00:11:32,240 --> 00:11:35,110 Then, you see, once he's accepted the basic definition 216 00:11:35,110 --> 00:11:38,210 and assuming that he knows the rules of mathematics, he has 217 00:11:38,210 --> 00:11:40,770 to come up with the same answer that you did. 218 00:11:40,770 --> 00:11:43,850 And this is what we mean by an objective criterion for doing 219 00:11:43,850 --> 00:11:45,560 mathematics, okay. 220 00:11:45,560 --> 00:11:47,550 Well, this was kind of an easy one. 221 00:11:47,550 --> 00:11:50,425 Let's do another kind of easy one that's harder. 222 00:11:50,425 --> 00:11:54,370 Let's make some gradual transitions, here. 223 00:11:54,370 --> 00:11:57,230 Let's pick another theorem. 224 00:11:57,230 --> 00:11:59,520 Here's another one that sounds pretty self-evident. 225 00:11:59,520 --> 00:12:04,410 The limit as 'x' approaches 'a' is 'a'. 226 00:12:04,410 --> 00:12:07,310 In fact, what that seems to say in a self-evident way is 227 00:12:07,310 --> 00:12:11,620 that as 'x' gets arbitrarily close to 'a', 'x' gets 228 00:12:11,620 --> 00:12:14,450 arbitrarily close to equal to 'a'. 229 00:12:14,450 --> 00:12:17,490 And if anything is a truism, I guess that's it. 230 00:12:17,490 --> 00:12:20,890 So we certainly suspect that this is a true statement. 231 00:12:20,890 --> 00:12:23,530 All I'm trying to get you used to is not to make a mountain 232 00:12:23,530 --> 00:12:25,780 out of a mole hill, not to have to think that mathematics 233 00:12:25,780 --> 00:12:30,050 becomes a severe thing where we try to find hard ways of 234 00:12:30,050 --> 00:12:33,375 doing easy things, but rather that we can find an 235 00:12:33,375 --> 00:12:37,660 unambiguous, logically constructed language from 236 00:12:37,660 --> 00:12:41,020 which all of our results can be proven without recourse to 237 00:12:41,020 --> 00:12:44,260 intuition once our basic definitions are chosen. 238 00:12:44,260 --> 00:12:46,010 So let's see how this would work. 239 00:12:46,010 --> 00:12:48,820 Let's again go back to our basic definition. 240 00:12:48,820 --> 00:12:51,850 You see, again, what we're saying here is that this is 241 00:12:51,850 --> 00:12:55,780 just a special case now where 'f of x' equals 'x'. 242 00:12:55,780 --> 00:12:57,740 You see the function is 'f of x'. 243 00:12:57,740 --> 00:13:00,720 In this case, it's 'x', and this is a special case where 244 00:13:00,720 --> 00:13:02,550 'a' is 'l'. 245 00:13:02,550 --> 00:13:04,650 Now what are you saying here? 246 00:13:04,650 --> 00:13:08,540 You're saying given epsilon greater than 0, we must find a 247 00:13:08,540 --> 00:13:11,380 delta greater than 0, such that what? 248 00:13:11,380 --> 00:13:14,510 Such that whenever 'x' is within delta of 'a' but not 249 00:13:14,510 --> 00:13:20,900 equal to 'a', then 'x' will be within epsilon of 'a'. 250 00:13:20,900 --> 00:13:24,340 And now you just look at this thing, I hope, and you say 251 00:13:24,340 --> 00:13:26,640 well, there's sort of a similarity over here. 252 00:13:29,690 --> 00:13:34,850 How close should 'x' be chosen to 'a' if you want 'x' to be 253 00:13:34,850 --> 00:13:36,920 within epsilon of 'a'? 254 00:13:36,920 --> 00:13:38,680 And the answer, quite obviously in this case 255 00:13:38,680 --> 00:13:42,490 without, cause to define again, is to simply say what 256 00:13:42,490 --> 00:13:43,460 in this case? 257 00:13:43,460 --> 00:13:48,520 For the given epsilon, choose, for example, 258 00:13:48,520 --> 00:13:50,940 delta to equal epsilon. 259 00:13:50,940 --> 00:13:54,590 Because certainly, if the absolute value of 'x minus a' 260 00:13:54,590 --> 00:13:57,810 is less than epsilon but greater than 0, then certainly 261 00:13:57,810 --> 00:14:02,920 the absolute value of 'x minus a' is less than epsilon. 262 00:14:02,920 --> 00:14:05,070 Don't be upset that the proof happened to be fairly 263 00:14:05,070 --> 00:14:06,280 easy in this case. 264 00:14:06,280 --> 00:14:08,890 More importantly, don't be upset that there was a more 265 00:14:08,890 --> 00:14:10,330 intuitive way of doing it. 266 00:14:10,330 --> 00:14:13,060 Remember what we want to do is to get these rigorous ways 267 00:14:13,060 --> 00:14:19,370 down pat, interpret them in terms of pictures wherever the 268 00:14:19,370 --> 00:14:20,960 pictures are available. 269 00:14:20,960 --> 00:14:23,560 And then, when we get to the case where pictures aren't 270 00:14:23,560 --> 00:14:28,110 available, to be able to extend the analytic concepts. 271 00:14:28,110 --> 00:14:30,780 Again, to see what happens here pictorially, you notice 272 00:14:30,780 --> 00:14:34,090 that if you start with the function 'f of x' equals 'x,' 273 00:14:34,090 --> 00:14:37,220 its graph is a straight line 'y' equals 'x'. 274 00:14:37,220 --> 00:14:41,620 And I'll risk some freehand drawing here. 275 00:14:41,620 --> 00:14:45,600 So in other words, we suspect that this is 'a'. 276 00:14:45,600 --> 00:14:48,550 It better be, if this says that all points on this line 277 00:14:48,550 --> 00:14:50,600 have the property that the y-coordinate is equal to the 278 00:14:50,600 --> 00:14:51,740 x-coordinate. 279 00:14:51,740 --> 00:14:52,770 Now what do we want to do? 280 00:14:52,770 --> 00:14:58,610 We pick an epsilon and knock off a plus 281 00:14:58,610 --> 00:15:01,150 epsilon and a minus epsilon. 282 00:15:01,150 --> 00:15:05,150 And now what we want to find out is how close must x be to 283 00:15:05,150 --> 00:15:10,440 a on the x-axis in order that 'f of x'-- namely 'y' in this 284 00:15:10,440 --> 00:15:11,730 case, or 'x' itself-- 285 00:15:11,730 --> 00:15:15,660 be within this prescribed tolerance limits of 'a'? 286 00:15:15,660 --> 00:15:17,070 You see what happens here? 287 00:15:17,070 --> 00:15:20,230 This is that one case where it happens that what? 288 00:15:20,230 --> 00:15:27,320 If you draw these lines over and project down, but by 289 00:15:27,320 --> 00:15:28,800 proportional parts-- 290 00:15:28,800 --> 00:15:30,140 and this is crucial here. 291 00:15:30,140 --> 00:15:33,520 Even if this weren't the 45 degree line, the fact that 292 00:15:33,520 --> 00:15:36,580 these two pieces here are equal would guarantee that 293 00:15:36,580 --> 00:15:39,440 these two pieces here are equal, in spite of how I've 294 00:15:39,440 --> 00:15:40,410 drawn this. 295 00:15:40,410 --> 00:15:44,280 The fact that this is the 45 degree line not only says that 296 00:15:44,280 --> 00:15:46,920 these two pieces here are equal but it says what? 297 00:15:46,920 --> 00:15:52,330 That each of these pieces is equal to each of these pieces. 298 00:15:52,330 --> 00:15:54,930 I wish I had drawn this better for you, but in fact the worse 299 00:15:54,930 --> 00:15:57,590 I draw it, the more you have to rely on being able to 300 00:15:57,590 --> 00:16:00,040 visualize abstractly what's happening here. 301 00:16:00,040 --> 00:16:03,290 What I'm saying is that this point here would be labeled 'a 302 00:16:03,290 --> 00:16:06,410 plus epsilon', this point here would be labeled 'a minus 303 00:16:06,410 --> 00:16:09,540 epsilon', and this is the pictorial version of what it 304 00:16:09,540 --> 00:16:13,660 means to say you could have chosen delta to equal epsilon, 305 00:16:13,660 --> 00:16:16,180 in this particular case. 306 00:16:16,180 --> 00:16:20,040 Well, that's enough of the easier ones, so let's pick one 307 00:16:20,040 --> 00:16:22,510 that gets slightly tougher. 308 00:16:22,510 --> 00:16:27,700 And this one gets tougher in one sense, but insidiously-- 309 00:16:27,700 --> 00:16:29,250 meaning it's real sneaky-- 310 00:16:29,250 --> 00:16:30,950 simple in another sense. 311 00:16:30,950 --> 00:16:34,280 In other words, it turns out that one can reason falsely 312 00:16:34,280 --> 00:16:37,690 and get the right answer just by a quirk. 313 00:16:37,690 --> 00:16:40,020 You see, what I want to prove next is a very important 314 00:16:40,020 --> 00:16:43,450 theorem that says that the limit of a sum is equal to the 315 00:16:43,450 --> 00:16:45,740 sum of the limits. 316 00:16:45,740 --> 00:16:49,690 Written out more formally, if I have two functions 'f of x' 317 00:16:49,690 --> 00:16:54,240 and 'g of x' and formed the sum ''f of x' plus 'g of x'' 318 00:16:54,240 --> 00:16:56,610 and I want to take the limit of that sum as 'x' approaches 319 00:16:56,610 --> 00:17:00,660 'a', what this theorem says is you can find the limit of each 320 00:17:00,660 --> 00:17:03,320 of the functions separately first, and 321 00:17:03,320 --> 00:17:05,500 then add the two results. 322 00:17:05,500 --> 00:17:07,980 Now at first glance, you might be tempted to say, well, what 323 00:17:07,980 --> 00:17:10,140 else would you expect to happen? 324 00:17:10,140 --> 00:17:11,859 The answer is, I don't know, again, what else you would 325 00:17:11,859 --> 00:17:12,609 expect to happen. 326 00:17:12,609 --> 00:17:13,970 But this is a luxury. 327 00:17:13,970 --> 00:17:17,010 You see, evidently what's happened here is that we've 328 00:17:17,010 --> 00:17:19,819 reversed the order of operations. 329 00:17:19,819 --> 00:17:20,950 You see, this says what? 330 00:17:20,950 --> 00:17:24,329 First add these two, and then take the limit. 331 00:17:24,329 --> 00:17:26,440 What are we doing down here? 332 00:17:26,440 --> 00:17:30,130 First we're taking the limits and then we're adding. 333 00:17:30,130 --> 00:17:33,380 Now, is it self-evident that just by changing the order of 334 00:17:33,380 --> 00:17:35,740 operations you don't change anything? 335 00:17:35,740 --> 00:17:39,290 We've seen many examples already in the short time that 336 00:17:39,290 --> 00:17:42,660 this course has been in existence where changing the 337 00:17:42,660 --> 00:17:46,720 order, changing the voice inflection, what have you, 338 00:17:46,720 --> 00:17:48,930 changes the answer. 339 00:17:48,930 --> 00:17:50,830 And in fact, we're going to see more drastic 340 00:17:50,830 --> 00:17:52,510 examples later on. 341 00:17:52,510 --> 00:17:54,470 I guess this is one of the tragedies of 342 00:17:54,470 --> 00:17:56,710 a course like this. 343 00:17:56,710 --> 00:17:58,740 I guess it's typical of problems every place. 344 00:17:58,740 --> 00:18:01,320 If the place that the person is going to get into trouble 345 00:18:01,320 --> 00:18:05,010 comes far beyond the time at which you're lecturing to him, 346 00:18:05,010 --> 00:18:08,030 it's kind of empty to threaten him with the trouble he's 347 00:18:08,030 --> 00:18:09,220 going to get into. 348 00:18:09,220 --> 00:18:11,560 So I'm not going to threaten you with the trouble you're 349 00:18:11,560 --> 00:18:13,510 going to get into until we get into it. 350 00:18:13,510 --> 00:18:16,760 All I'm going to say is be careful when you say that it's 351 00:18:16,760 --> 00:18:19,960 self-evident that we can first add and then take the limit or 352 00:18:19,960 --> 00:18:22,620 whether we first take the limits and then add. 353 00:18:22,620 --> 00:18:25,230 In general, it does make a difference in which order you 354 00:18:25,230 --> 00:18:26,840 perform operations. 355 00:18:26,840 --> 00:18:28,760 Well, let's take this just a little bit more formally to 356 00:18:28,760 --> 00:18:30,770 see what this thing says. 357 00:18:30,770 --> 00:18:33,460 First of all, when we write something like this, we assume 358 00:18:33,460 --> 00:18:36,030 that the limit of 'f of x' as 'x' approaches 'a' exists, 359 00:18:36,030 --> 00:18:37,940 otherwise we wouldn't write this thing. 360 00:18:37,940 --> 00:18:39,770 So let's call that limit 'l1'. 361 00:18:39,770 --> 00:18:42,020 In other words, let the limit of 'f of x' as 'x' approaches 362 00:18:42,020 --> 00:18:43,570 'a' equal 'l1'. 363 00:18:43,570 --> 00:18:47,910 Let the limit of 'g of x' as 'x' approaches 'a' equal 'l2'. 364 00:18:47,910 --> 00:18:51,790 Now define a new function 'h of x' to be equal to ''f of x' 365 00:18:51,790 --> 00:18:53,070 plus 'g of x''. 366 00:18:53,070 --> 00:18:54,510 And by the way, this is something I didn't say 367 00:18:54,510 --> 00:18:57,050 strongly enough in one of our early lectures, and I want to 368 00:18:57,050 --> 00:19:00,950 make sure that it's very clear that this is understood. 369 00:19:00,950 --> 00:19:06,420 And that is, notice that when you define 'h' to be the sum 370 00:19:06,420 --> 00:19:11,000 of 'f' and 'g', you had better make sure that 'f' and 'g' 371 00:19:11,000 --> 00:19:13,810 have the same domain. 372 00:19:13,810 --> 00:19:17,540 You see, if some number 'x' belongs to the domain of 'f' 373 00:19:17,540 --> 00:19:21,590 but it doesn't belong to the domain of 'g', then how can 374 00:19:21,590 --> 00:19:24,280 you form ''f of x' plus 'g of x''? 'g' 375 00:19:24,280 --> 00:19:26,320 doesn't operate on 'x'. 376 00:19:26,320 --> 00:19:29,250 By the way, this is not quite as serious a problem as it 377 00:19:29,250 --> 00:19:33,600 seems if you understand the language of our new 378 00:19:33,600 --> 00:19:35,670 mathematics and sets and the like. 379 00:19:35,670 --> 00:19:42,350 Namely, if the domain of 'f' happens to look like this and 380 00:19:42,350 --> 00:19:49,900 the domain of 'g' happens to look like this, what we do is 381 00:19:49,900 --> 00:19:51,730 we restrict the sum to the 382 00:19:51,730 --> 00:19:53,560 intersection of the two domains. 383 00:19:56,170 --> 00:20:00,150 In other words, referring back to our function 'h', we define 384 00:20:00,150 --> 00:20:01,980 the domain of 'h' to be the intersection 385 00:20:01,980 --> 00:20:03,290 of these two domains. 386 00:20:03,290 --> 00:20:06,580 And that way, for any 'x', which is in the domain of 'h', 387 00:20:06,580 --> 00:20:10,910 it automatically belongs to both the domain of 'f' and the 388 00:20:10,910 --> 00:20:11,950 domain of 'g'. 389 00:20:11,950 --> 00:20:14,730 And so this becomes well-defined. 390 00:20:14,730 --> 00:20:17,220 Another way of looking at this is what you're really saying 391 00:20:17,220 --> 00:20:21,820 is that 'f' and 'g' must include in their domain 392 00:20:21,820 --> 00:20:25,340 intervals surrounding 'x' equals 'a'. 393 00:20:25,340 --> 00:20:28,100 And since you're only interested in what's happening 394 00:20:28,100 --> 00:20:31,750 near 'x' equals 'a', you don't really care whether these have 395 00:20:31,750 --> 00:20:35,150 the same domains or not, provided they have what? 396 00:20:35,150 --> 00:20:38,520 An intersection that can serve as a common domain. 397 00:20:38,520 --> 00:20:41,600 But that, I think, is more clear from the context. 398 00:20:41,600 --> 00:20:44,280 It's a very, very important fine point. 399 00:20:44,280 --> 00:20:47,690 It's a tragedy to try to add two numbers, one of which 400 00:20:47,690 --> 00:20:48,890 doesn't exist. 401 00:20:48,890 --> 00:20:50,240 I don't know if it's a tragedy, 402 00:20:50,240 --> 00:20:52,760 it's certainly futile. 403 00:20:52,760 --> 00:20:55,710 At any rate, though, let's see what this thing then says. 404 00:20:55,710 --> 00:21:00,260 If we now define 'h' to be 'f plus g', what we want to prove 405 00:21:00,260 --> 00:21:04,250 is that the limit of 'h of x' as 'x' approaches 'a' equals 406 00:21:04,250 --> 00:21:06,700 'l1 plus l2'. 407 00:21:06,700 --> 00:21:08,310 Now here's the point again. 408 00:21:08,310 --> 00:21:11,420 What does this mean by definition? 409 00:21:11,420 --> 00:21:17,300 It means that given epsilon greater than 0, we must be 410 00:21:17,300 --> 00:21:21,170 able to find a delta such that when 'x' is within delta of 411 00:21:21,170 --> 00:21:25,540 'a' but not equal to 'a', 'h of x' is within 412 00:21:25,540 --> 00:21:28,380 epsilon of 'l1 plus l2'. 413 00:21:28,380 --> 00:21:31,000 That's probably kind of hard to keep track of, so I've 414 00:21:31,000 --> 00:21:32,860 taken the liberty of writing this down for 415 00:21:32,860 --> 00:21:34,740 you right over here. 416 00:21:34,740 --> 00:21:38,310 See, given epsilon greater than 0-- so that's given, we 417 00:21:38,310 --> 00:21:39,800 have no control over that-- 418 00:21:39,800 --> 00:21:45,180 what we must do is find delta greater than 0, such that 0 419 00:21:45,180 --> 00:21:48,280 less than the absolute value of 'x minus a' less than 420 00:21:48,280 --> 00:21:49,930 delta, implies-- 421 00:21:49,930 --> 00:21:51,140 now, what is 'h of x'? 422 00:21:51,140 --> 00:21:53,860 It's ''f of x' plus 'g of x'', and our limit that we're 423 00:21:53,860 --> 00:21:56,640 looking for in this case is 'l1 plus l2'. 424 00:21:56,640 --> 00:22:00,640 So mathematically, we replace 'h of x' by ''f of x' plus 'g 425 00:22:00,640 --> 00:22:03,430 of x'', the limit is 'l1 plus l2'. 426 00:22:03,430 --> 00:22:06,935 So what must we show that the absolute value of the quantity 427 00:22:06,935 --> 00:22:10,830 of ''f of x' plus 'g of x'' minus the quantity 'l1 plus 428 00:22:10,830 --> 00:22:12,690 l2' is less than epsilon. 429 00:22:12,690 --> 00:22:15,560 And now we start to play detective again. 430 00:22:15,560 --> 00:22:17,980 This is the expression that we want to 431 00:22:17,980 --> 00:22:19,570 make less than epsilon. 432 00:22:19,570 --> 00:22:23,970 So what we do is we look at this particular expression and 433 00:22:23,970 --> 00:22:27,500 we try to see what kind of cute things we can do with it. 434 00:22:27,500 --> 00:22:29,220 Now, what do I mean by a cute thing? 435 00:22:29,220 --> 00:22:32,490 Well, we're assuming that the limit of 'f of x' as 'x' 436 00:22:32,490 --> 00:22:36,150 approaches 'a' equals 'l1'. 437 00:22:36,150 --> 00:22:39,230 Let me write this as an aside over here. 438 00:22:39,230 --> 00:22:42,730 Among other things, what this tells us is that we have a 439 00:22:42,730 --> 00:22:47,480 hold on expressions like this. 440 00:22:47,480 --> 00:22:50,080 In other words, the fact that the limit of 'f of x' as 'x' 441 00:22:50,080 --> 00:22:53,310 approaches 'a' is equal to 'l1' tells us that we can make 442 00:22:53,310 --> 00:22:55,970 this as small as we want. 443 00:22:55,970 --> 00:22:57,940 I'll use a subscript over here for epsilon 1 because it 444 00:22:57,940 --> 00:23:00,050 doesn't have to be the same epsilon that was given here. 445 00:23:00,050 --> 00:23:03,540 For any positive number, say epsilon 1, the point is I can 446 00:23:03,540 --> 00:23:08,980 make 'f of x' within epsilon 1 of 'l1' just by choosing 'x' 447 00:23:08,980 --> 00:23:11,330 sufficiently close to 'a' by definition 448 00:23:11,330 --> 00:23:12,610 of what limit means. 449 00:23:12,610 --> 00:23:15,390 So in other words, I like expressions of this form. 450 00:23:15,390 --> 00:23:20,140 And similarly, I like expressions of this form. 451 00:23:20,140 --> 00:23:22,530 And again, the reason is that since the limit of 'g of x' as 452 00:23:22,530 --> 00:23:26,260 'x' approaches 'a' is 'l sub 2', it gives me a hold on the 453 00:23:26,260 --> 00:23:29,100 difference between 'g of x' and 'l2'. 454 00:23:29,100 --> 00:23:33,420 So again, using the old adage that hindsight is better than 455 00:23:33,420 --> 00:23:36,860 foresight by a darn site, knowing exactly what it is I 456 00:23:36,860 --> 00:23:40,260 have to do, I come back here and try to doctor things up 457 00:23:40,260 --> 00:23:41,360 for myself. 458 00:23:41,360 --> 00:23:45,360 The first thing I observe is that this can be rewritten. 459 00:23:45,360 --> 00:23:46,800 There's no calculus in this, notice. 460 00:23:46,800 --> 00:23:49,090 Just plain ordinary algebra, arithmetic. 461 00:23:49,090 --> 00:23:53,830 This can be rewritten so that I can group the 'f sub 'f of 462 00:23:53,830 --> 00:23:58,160 x' and 'l1'' together and 'g of x' and 'l2' together. 463 00:23:58,160 --> 00:24:01,940 In other words, this is indeed nothing more than 1. 464 00:24:01,940 --> 00:24:04,700 The absolute value of the quantity of ''f of x' minus 465 00:24:04,700 --> 00:24:07,370 l1' plus the absolute value of the quantity 466 00:24:07,370 --> 00:24:09,290 ''g of x' minus l2'. 467 00:24:09,290 --> 00:24:12,980 Now, the point is since we already know that the absolute 468 00:24:12,980 --> 00:24:15,810 value of a sum is less than or equal to the sum of the 469 00:24:15,810 --> 00:24:18,760 absolute values, that tells me that-- 470 00:24:18,760 --> 00:24:21,480 treating this is one number and this is another number, 471 00:24:21,480 --> 00:24:26,831 the absolute value of a sum is less than or equal to the sum 472 00:24:26,831 --> 00:24:29,010 of the absolute values. 473 00:24:29,010 --> 00:24:32,700 What this now tells me is I can say that this, that I'm 474 00:24:32,700 --> 00:24:35,580 trying to get a hold on, is less than this. 475 00:24:35,580 --> 00:24:37,670 But look at this expression. 476 00:24:37,670 --> 00:24:40,560 This expression is the absolute value of 477 00:24:40,560 --> 00:24:42,240 ''f of x' minus l1'. 478 00:24:42,240 --> 00:24:44,680 And this expression is the absolute value of 479 00:24:44,680 --> 00:24:47,140 ''g of x' minus l2'. 480 00:24:47,140 --> 00:24:50,360 In other words, then, since I can make 'f of x' as 481 00:24:50,360 --> 00:24:54,300 arbitrarily nearly equal to 'l1' as I want and 'g of x' as 482 00:24:54,300 --> 00:24:57,290 close to 'l2' as I want just by choosing 'x' sufficiently 483 00:24:57,290 --> 00:25:02,150 close to 'a', why don't I choose 'x' close enough to 'a' 484 00:25:02,150 --> 00:25:05,140 so each of these will be less than epsilon over 2? 485 00:25:05,140 --> 00:25:07,700 Now again, this calls for a little aside. 486 00:25:07,700 --> 00:25:10,830 When one talks about epsilon, that is 'a', 487 00:25:10,830 --> 00:25:12,040 what shall we say? 488 00:25:12,040 --> 00:25:19,325 A generic name for any number which exceeds 0. 489 00:25:19,325 --> 00:25:22,590 Or I could have written that less mystically by just saying 490 00:25:22,590 --> 00:25:24,530 any positive number. 491 00:25:24,530 --> 00:25:27,020 Well, if epsilon is positive, what can you say 492 00:25:27,020 --> 00:25:28,860 about half of epsilon? 493 00:25:28,860 --> 00:25:30,620 It's also positive. 494 00:25:30,620 --> 00:25:32,700 In other words, if I had chosen a different epsilon, 495 00:25:32,700 --> 00:25:37,300 say 'epsilon sub 1', equal to the original epsilon divided 496 00:25:37,300 --> 00:25:41,310 by 2, then I'm guaranteed what? 497 00:25:41,310 --> 00:25:45,170 That I can get 'f of x' with an epsilon 2 of 'l1', 'g of x' 498 00:25:45,170 --> 00:25:46,860 with an epsilon 2 of 'l2'. 499 00:25:46,860 --> 00:25:49,560 And now adding these two together, if this term is less 500 00:25:49,560 --> 00:25:52,420 than epsilon over 2 and this term is less than epsilon over 501 00:25:52,420 --> 00:25:56,110 2, the whole sum is less than epsilon. 502 00:25:56,110 --> 00:25:58,770 And it seems now semi-intuitively-- 503 00:25:58,770 --> 00:26:00,460 what do I mean by semi-intuitively? 504 00:26:00,460 --> 00:26:03,100 Well, this is far from an intuitive job over here. 505 00:26:03,100 --> 00:26:05,150 It's quite mathematical. 506 00:26:05,150 --> 00:26:06,460 It's rigorous in that sense. 507 00:26:06,460 --> 00:26:08,790 It's intuitive in the sense that I'm not playing around 508 00:26:08,790 --> 00:26:09,620 with the deltas here. 509 00:26:09,620 --> 00:26:12,560 All I'm saying is look, I can make this as small is I want, 510 00:26:12,560 --> 00:26:15,230 I could make this as small as I want, therefore I can make 511 00:26:15,230 --> 00:26:16,810 the sum as small as I want. 512 00:26:16,810 --> 00:26:18,700 And the fancy way of saying that is I can make it less 513 00:26:18,700 --> 00:26:21,730 than any given epsilon, and therefore it appears that this 514 00:26:21,730 --> 00:26:24,390 will be the limit. 515 00:26:24,390 --> 00:26:27,960 Using our old adage again of being able, knowing what we 516 00:26:27,960 --> 00:26:31,460 want, to be able to doctor things up rigorously, once 517 00:26:31,460 --> 00:26:35,090 we've gone through this it's now relatively easy to clean 518 00:26:35,090 --> 00:26:36,310 up the details. 519 00:26:36,310 --> 00:26:39,350 In other words, for those of us who are 520 00:26:39,350 --> 00:26:43,130 mathematically-oriented enough to say, the way you've proven 521 00:26:43,130 --> 00:26:46,220 this last result is the same sloppiness that I was used to 522 00:26:46,220 --> 00:26:49,400 seeing in certain types of engineering proofs where 523 00:26:49,400 --> 00:26:51,860 people were more interested in the result than with the 524 00:26:51,860 --> 00:26:54,600 answer, I still don't see how you used the epsilons and the 525 00:26:54,600 --> 00:26:55,430 deltas here. 526 00:26:55,430 --> 00:26:58,790 Let me show you what a simple step it is to now go from the 527 00:26:58,790 --> 00:27:02,960 semi-rigorous approach to the completely rigorous approach. 528 00:27:02,960 --> 00:27:05,940 All we do is reword what we've done before. 529 00:27:05,940 --> 00:27:08,590 In fact, this is true in most mathematics. 530 00:27:08,590 --> 00:27:10,970 You take a geometry book and there's a theorem that says 531 00:27:10,970 --> 00:27:14,000 something like if 'a', 'b', 'c', and 'd' are true, then 532 00:27:14,000 --> 00:27:14,680 'e' is true. 533 00:27:14,680 --> 00:27:16,490 And you learn this proof quite mechanically. 534 00:27:16,490 --> 00:27:18,060 You sort of memorize it. 535 00:27:18,060 --> 00:27:20,820 Well you know, the man who proved that theorem didn't, in 536 00:27:20,820 --> 00:27:24,020 general, start out by saying, I wonder what happens if 'a', 537 00:27:24,020 --> 00:27:26,190 'b', 'c', and 'd' are true. 538 00:27:26,190 --> 00:27:28,890 In general, what he tries to do is to prove that some 539 00:27:28,890 --> 00:27:30,740 result, like 'e', is true. 540 00:27:30,740 --> 00:27:33,360 And as he's proving it, he hits pitfalls. 541 00:27:33,360 --> 00:27:35,650 And he says, you know, if I could only be sure 'a' was 542 00:27:35,650 --> 00:27:37,670 true, I could get over this pitfall. 543 00:27:37,670 --> 00:27:40,440 And if I could be sure 'b' was true, I could get over the 544 00:27:40,440 --> 00:27:42,310 second pitfall, et cetera. 545 00:27:42,310 --> 00:27:44,710 And when he makes enough assumptions to get over all 546 00:27:44,710 --> 00:27:48,940 the pitfalls and he has his answer, he then we writes down 547 00:27:48,940 --> 00:27:51,310 the answer in the reverse order from 548 00:27:51,310 --> 00:27:52,600 which he invented it. 549 00:27:52,600 --> 00:27:53,620 Namely, he says what? 550 00:27:53,620 --> 00:27:55,580 Suppose 'a', 'b', 'c', and 'd' are true. 551 00:27:55,580 --> 00:27:57,210 Let's prove that 'e' is true. 552 00:27:57,210 --> 00:28:00,620 And the student is then robbed of any attempt to see 553 00:28:00,620 --> 00:28:03,320 intuitively how this whole thing came about. 554 00:28:03,320 --> 00:28:05,450 As a case in point, let me show you what 555 00:28:05,450 --> 00:28:06,490 I'm driving at here. 556 00:28:06,490 --> 00:28:09,240 My first exposure to formal limit proofs was 557 00:28:09,240 --> 00:28:10,240 something like this. 558 00:28:10,240 --> 00:28:13,700 When somebody said prove the limit of a sum equals the sum 559 00:28:13,700 --> 00:28:16,590 of the limits, something like this would happen. 560 00:28:16,590 --> 00:28:19,410 The first statement in the book would say, given epsilon 561 00:28:19,410 --> 00:28:24,440 greater than 0, let epsilon 1 equal epsilon over 2. 562 00:28:24,440 --> 00:28:26,870 Now, I respected my teacher, I respected the 563 00:28:26,870 --> 00:28:27,630 author of the book. 564 00:28:27,630 --> 00:28:30,000 If he says let epsilon 1 equal epsilon over 565 00:28:30,000 --> 00:28:31,150 2, all right, fine. 566 00:28:31,150 --> 00:28:32,250 We can do that. 567 00:28:32,250 --> 00:28:34,730 And more to the point, it turned out that the problem 568 00:28:34,730 --> 00:28:37,220 worked if you did that. 569 00:28:37,220 --> 00:28:40,880 The part that bothered me is why did he say epsilon over 2? 570 00:28:40,880 --> 00:28:43,230 Why not 2 epsilon over 3? 571 00:28:43,230 --> 00:28:44,820 Or epsilon over 5? 572 00:28:44,820 --> 00:28:48,090 Or epsilon over 6,872? 573 00:28:48,090 --> 00:28:49,920 Why epsilon over 2? 574 00:28:49,920 --> 00:28:52,220 And the point was that he had cheated. 575 00:28:52,220 --> 00:28:55,720 He had already done the problem that we had over here, 576 00:28:55,720 --> 00:28:59,290 and knowing what he needed, then came back here and said 577 00:28:59,290 --> 00:29:02,120 let epsilon 1 equal epsilon over 2. 578 00:29:02,120 --> 00:29:03,670 And notice how this is going to mimic 579 00:29:03,670 --> 00:29:06,240 everything we said before. 580 00:29:06,240 --> 00:29:10,230 For this choice of epsilon 1, we can find a delta 1 greater 581 00:29:10,230 --> 00:29:14,910 than 0 such that if the absolute value of 'x minus a' 582 00:29:14,910 --> 00:29:18,200 is less than delta 1 but greater than 0, then the 583 00:29:18,200 --> 00:29:20,910 absolute value of ''f of x' minus l1' is less 584 00:29:20,910 --> 00:29:21,800 than epsilon 1. 585 00:29:21,800 --> 00:29:23,260 Why do we know that? 586 00:29:23,260 --> 00:29:28,720 That's the definition of what it means to say that the limit 587 00:29:28,720 --> 00:29:32,530 of 'f of x' as 'x' approaches 'a' equals 'l1'. 588 00:29:32,530 --> 00:29:35,820 In a similar way, he says we can find delta 2 greater than 589 00:29:35,820 --> 00:29:40,040 0, such that whenever the absolute value of 'x minus a' 590 00:29:40,040 --> 00:29:43,610 is greater than 0 but less than delta over 2, we can make 591 00:29:43,610 --> 00:29:49,020 the absolute value of ''g of x' minus 'l sub 2'' and be 592 00:29:49,020 --> 00:29:51,100 less than epsilon 1. 593 00:29:51,100 --> 00:29:53,610 And now comes the beautiful step. 594 00:29:53,610 --> 00:29:57,590 He says, now that these delta 1 and delta 2 exist 595 00:29:57,590 --> 00:30:01,210 separately, pick delta to be the minimum of these two. 596 00:30:01,210 --> 00:30:04,020 In other words, if I let delta equal the minimum of these 597 00:30:04,020 --> 00:30:06,230 two, what does that guarantee me? 598 00:30:06,230 --> 00:30:09,500 If delta is the minimum of these two, it guarantees me 599 00:30:09,500 --> 00:30:14,180 that both of these conditions are met at the same time. 600 00:30:14,180 --> 00:30:15,120 What does that tell me? 601 00:30:15,120 --> 00:30:20,160 It tells me that as soon as the absolute value of 'x minus 602 00:30:20,160 --> 00:30:23,430 a' is less than delta but greater than 0, automatically 603 00:30:23,430 --> 00:30:24,970 these two conditions hold. 604 00:30:24,970 --> 00:30:26,850 And that, in turn, tells me what? 605 00:30:26,850 --> 00:30:32,600 That the absolute value of ''f of x' minus l1' is less than 606 00:30:32,600 --> 00:30:37,380 epsilon 1 and the absolute value of ''g of x' minus l2' 607 00:30:37,380 --> 00:30:39,940 is less than epsilon 1. 608 00:30:39,940 --> 00:30:44,620 And now by adding unequals to equals, that tells me what? 609 00:30:44,620 --> 00:30:48,910 That this plus this is less than 2 epsilon 1. 610 00:30:48,910 --> 00:30:52,410 But see, using my hindsight, I picked epsilon to be what? 611 00:30:52,410 --> 00:30:53,980 Epsilon over 2. 612 00:30:53,980 --> 00:30:59,000 So 2 epsilon 1 is just another way of saying epsilon. 613 00:30:59,000 --> 00:31:02,100 In other words, what this now implies is that the absolute 614 00:31:02,100 --> 00:31:07,860 value of ''f of x' minus l1' plus the absolute value of ''g 615 00:31:07,860 --> 00:31:11,740 of x' minus l2' is less than 2 epsilon 1, 616 00:31:11,740 --> 00:31:14,140 which equals epsilon. 617 00:31:14,140 --> 00:31:19,110 But you see, this in turn is what? 618 00:31:19,110 --> 00:31:23,880 This is greater than the absolute value of ''f of x' 619 00:31:23,880 --> 00:31:28,105 minus l1' plus ''g of x' minus l2'. 620 00:31:31,380 --> 00:31:33,350 And so this was the thing we wanted to make 621 00:31:33,350 --> 00:31:35,010 smaller than epsilon. 622 00:31:35,010 --> 00:31:38,250 Since this is smaller than this and this is already 623 00:31:38,250 --> 00:31:40,910 smaller than epsilon, then this must be smaller than 624 00:31:40,910 --> 00:31:42,520 epsilon too. 625 00:31:42,520 --> 00:31:44,480 Again, notice, this is rigorous. 626 00:31:44,480 --> 00:31:47,740 But if you understand what's happening here piece by piece, 627 00:31:47,740 --> 00:31:51,010 you never really have to memorize a thing. 628 00:31:51,010 --> 00:31:54,080 Let's just take a look back here to reinforce what I'm 629 00:31:54,080 --> 00:31:55,050 saying here. 630 00:31:55,050 --> 00:31:55,860 Notice what we did. 631 00:31:55,860 --> 00:31:58,960 We started with the answer that we wanted to show, worked 632 00:31:58,960 --> 00:32:02,270 around to get ahold of things that we wanted. 633 00:32:02,270 --> 00:32:03,430 We were lucky enough-- 634 00:32:03,430 --> 00:32:05,320 and this is true in any game, for example. 635 00:32:05,320 --> 00:32:08,210 One can plot masterful strategy and still lose, 636 00:32:08,210 --> 00:32:10,170 there's no guarantee we're going to win with our 637 00:32:10,170 --> 00:32:11,730 masterful strategy. 638 00:32:11,730 --> 00:32:14,670 But we usually do in this course, usually. 639 00:32:14,670 --> 00:32:18,510 What we do is we masterfully come back to this, see what 640 00:32:18,510 --> 00:32:19,740 has to be done. 641 00:32:19,740 --> 00:32:22,770 Knowing what the right answer has got to be, we come back 642 00:32:22,770 --> 00:32:25,590 here and then formalize it. 643 00:32:25,590 --> 00:32:28,350 Essentially, what we've done is reverse the steps here. 644 00:32:28,350 --> 00:32:30,860 I guess there is one thing that bothers many people that 645 00:32:30,860 --> 00:32:32,830 I should make an aside about. 646 00:32:32,830 --> 00:32:36,580 Why do you need a different delta 1 and delta 2 for the 647 00:32:36,580 --> 00:32:38,020 same epsilon 1? 648 00:32:38,020 --> 00:32:40,980 See, if you're memorizing, there's a danger that you 649 00:32:40,980 --> 00:32:43,390 won't realize this and why this happens. 650 00:32:43,390 --> 00:32:48,670 Let me show you in terms of a picture what this means. 651 00:32:48,670 --> 00:32:51,710 Let this, for the sake of argument, be the curve 'y' 652 00:32:51,710 --> 00:32:53,480 equals 'f of x'. 653 00:32:53,480 --> 00:32:58,550 And let this be the curve 'y' equals 'g of x'. 654 00:32:58,550 --> 00:32:59,800 All we're saying is this. 655 00:33:02,750 --> 00:33:10,220 That when you prescribe an epsilon, the same epsilon that 656 00:33:10,220 --> 00:33:15,400 surrounds 'l2', if you have that surround 'l1'. 657 00:33:15,400 --> 00:33:16,910 Because the curves may have different slopes. 658 00:33:16,910 --> 00:33:18,360 Look what happens over here, even as badly 659 00:33:18,360 --> 00:33:19,150 as I've drawn this. 660 00:33:19,150 --> 00:33:22,760 Notice that in epsilon, neighborhood of 'l2', projects 661 00:33:22,760 --> 00:33:26,070 down into this size neighborhood around 'a'. 662 00:33:26,070 --> 00:33:28,380 On the other hand, an epsilon neighborhood of 'l1'. 663 00:33:31,200 --> 00:33:33,500 You get what you pay for, I guess. 664 00:33:33,500 --> 00:33:37,950 An epsilon, neighborhood of 'l1', projects into a much 665 00:33:37,950 --> 00:33:39,220 smaller region. 666 00:33:39,220 --> 00:33:40,810 And by the way, that's exactly what we meant. 667 00:33:40,810 --> 00:33:43,030 This is a delta 1, for example. 668 00:33:43,030 --> 00:33:46,470 This is delta 2. 669 00:33:46,470 --> 00:33:50,060 And when we said pick delta to be the minimum of delta 1 and 670 00:33:50,060 --> 00:33:53,730 delta 2, all we were saying was listen, if we guarantee 671 00:33:53,730 --> 00:33:57,770 that 'x' stays in here, then certainly both of these two 672 00:33:57,770 --> 00:34:00,730 things will be true at the same time. 673 00:34:00,730 --> 00:34:04,530 Well again, we have many exercises and reading material 674 00:34:04,530 --> 00:34:05,840 to reinforce these points. 675 00:34:05,840 --> 00:34:08,940 All I want to do with the lecture is to give you an 676 00:34:08,940 --> 00:34:12,030 overview as to what's happening so that you see 677 00:34:12,030 --> 00:34:12,650 these things. 678 00:34:12,650 --> 00:34:15,690 And I'm afraid I might cure you with details if I just 679 00:34:15,690 --> 00:34:18,670 keep hammering home these rigorous little points. 680 00:34:18,670 --> 00:34:20,570 As I say, I hope you get the main idea 681 00:34:20,570 --> 00:34:21,670 from what we're doing. 682 00:34:21,670 --> 00:34:25,360 Let me just, for the sake of argument, try to work with 683 00:34:25,360 --> 00:34:31,370 just one more idea and we'll see how this works out also. 684 00:34:31,370 --> 00:34:36,340 Let's, for example, try to play around with the idea that 685 00:34:36,340 --> 00:34:39,400 a companion to the limit of a sum equals the sum of the 686 00:34:39,400 --> 00:34:42,219 limits would be what? 687 00:34:42,219 --> 00:34:45,469 The limit of a product equals the product of the limits. 688 00:34:45,469 --> 00:34:49,050 In other words, as before, if the limit of 'f of x' as 'x' 689 00:34:49,050 --> 00:34:52,380 approaches 'a' is 'l1' and the limit of 'g of x' as 'x' 690 00:34:52,380 --> 00:34:55,659 approaches 'a' is 'l2', let's form a new function, which we 691 00:34:55,659 --> 00:34:58,170 could call 'k of x' which is equal to the 692 00:34:58,170 --> 00:34:59,810 product of 'f' and 'g'. 693 00:34:59,810 --> 00:35:03,800 Again, noticing that the 'f' and 'g' have to have a common 694 00:35:03,800 --> 00:35:05,780 domain here. 695 00:35:05,780 --> 00:35:08,330 You want to show that the limit of 'f of x' times 'g of 696 00:35:08,330 --> 00:35:12,940 x' is 'l1' times 'l2'. 697 00:35:12,940 --> 00:35:16,920 And this is the hard part of the course, this is the part 698 00:35:16,920 --> 00:35:19,110 of the course that I don't think anybody in the world can 699 00:35:19,110 --> 00:35:20,150 really teach. 700 00:35:20,150 --> 00:35:24,410 All one can do is try to expose the student to ideas 701 00:35:24,410 --> 00:35:27,400 and hope that the student has the knack of putting these 702 00:35:27,400 --> 00:35:30,550 things together to form his own repertoire. 703 00:35:30,550 --> 00:35:32,030 The idea is something like this. 704 00:35:32,030 --> 00:35:34,020 I'll show you alternative methods and the like. 705 00:35:34,020 --> 00:35:36,480 One way is we want to get ahold of 'f of 706 00:35:36,480 --> 00:35:38,710 x' times 'g of x'. 707 00:35:38,710 --> 00:35:40,380 Now what do we have a hold on? 708 00:35:40,380 --> 00:35:41,930 It's very clever what we do here. 709 00:35:41,930 --> 00:35:45,360 We have a hold of ''f of x' minus l1' and we have a hold 710 00:35:45,360 --> 00:35:48,540 of ''g of x' minus l2'. 711 00:35:48,540 --> 00:35:52,620 So as we so often do in mathematics, we simply add and 712 00:35:52,620 --> 00:35:53,870 subtract the same thing. 713 00:35:53,870 --> 00:35:56,570 We frequently add on zeroes in this cute way. 714 00:35:56,570 --> 00:35:58,360 We add and subtract the same thing. 715 00:35:58,360 --> 00:35:59,640 Notice what we did over here. 716 00:35:59,640 --> 00:36:04,520 We wrote 'f of x' as 'l1' plus ''f of x' minus l1'. 717 00:36:04,520 --> 00:36:06,180 Again, why did we do that? 718 00:36:06,180 --> 00:36:09,260 Because from our definition of the limit of 'f of x' as 'x' 719 00:36:09,260 --> 00:36:12,380 approaches 'a' equaling 'l1', we know that we can 720 00:36:12,380 --> 00:36:14,250 control this side. 721 00:36:14,250 --> 00:36:15,710 And the same thing is true over here. 722 00:36:15,710 --> 00:36:17,660 The fact that the limit of 'g of x' as 'x' approaches 'a' 723 00:36:17,660 --> 00:36:20,710 equals 'l2' means we have some control over this. 724 00:36:20,710 --> 00:36:23,710 Now let's just multiply everything out. 725 00:36:23,710 --> 00:36:27,975 This is what? 'f of x' times 'g of x'. 726 00:36:27,975 --> 00:36:30,650 I'm going to save myself some space and keep the board 727 00:36:30,650 --> 00:36:31,750 somewhat symmetric. 728 00:36:31,750 --> 00:36:34,360 When I multiply these terms out, I'm going to get what? 729 00:36:34,360 --> 00:36:37,770 An 'l1' times 'l2' term over here as one of my four terms? 730 00:36:37,770 --> 00:36:40,790 Let me already transpose that one so I kill two 731 00:36:40,790 --> 00:36:42,050 birds with one stone. 732 00:36:42,050 --> 00:36:44,440 One is I keep a little bit of symmetry in what 733 00:36:44,440 --> 00:36:45,510 I'm going to write. 734 00:36:45,510 --> 00:36:50,110 And secondly, notice that later on, this is what I want 735 00:36:50,110 --> 00:36:51,080 to get a hold on. 736 00:36:51,080 --> 00:36:53,350 In other words, notice that the proof of the limit of 'f 737 00:36:53,350 --> 00:36:57,320 of x' times 'g of x' as 'x' approaches 'a' is 'l1' times 738 00:36:57,320 --> 00:37:01,360 'l2', this is precisely the expression I must make small. 739 00:37:01,360 --> 00:37:04,170 Our show can be made as small is I wish just by picking 'x' 740 00:37:04,170 --> 00:37:05,550 sufficiently close to 'a'. 741 00:37:05,550 --> 00:37:08,050 And again, I will not go through all the details here, 742 00:37:08,050 --> 00:37:11,810 I will simply outline what we do here. 743 00:37:11,810 --> 00:37:14,430 Namely, let's multiply out the rest of this thing. 744 00:37:14,430 --> 00:37:15,570 We have what here? 745 00:37:15,570 --> 00:37:18,390 This times this, which is what? 746 00:37:18,390 --> 00:37:23,660 'l1' times ''g of x' minus l2'. 747 00:37:23,660 --> 00:37:32,160 Then we have this times this, which is 'l2' times 748 00:37:32,160 --> 00:37:35,880 ''f of x' minus l1'. 749 00:37:35,880 --> 00:37:37,530 And now we have what? 750 00:37:37,530 --> 00:37:39,770 This times this. 751 00:37:39,770 --> 00:37:47,140 That's ''f of x' minus l1' times ''g of x' minus l2'. 752 00:37:47,140 --> 00:37:49,970 Now, the point is this is the thing that we'd like to make 753 00:37:49,970 --> 00:37:52,140 very small in absolute value. 754 00:37:52,140 --> 00:37:54,900 Well again, the absolute value of this is equal to the 755 00:37:54,900 --> 00:37:57,770 absolute value of this, which is less than or equal to-- and 756 00:37:57,770 --> 00:38:00,560 I'm going to go through these details rather rapidly, and 757 00:38:00,560 --> 00:38:02,750 allow you to fill these in for yourself-- all 758 00:38:02,750 --> 00:38:03,930 I'm using is what? 759 00:38:03,930 --> 00:38:07,200 That the absolute value of a sum is less than or equal to 760 00:38:07,200 --> 00:38:08,740 the sum of the absolute values. 761 00:38:08,740 --> 00:38:11,010 The absolute value of a product is equal to the 762 00:38:11,010 --> 00:38:13,650 product of the absolute values, so this is going to be 763 00:38:13,650 --> 00:38:14,900 less than or equal to. 764 00:38:18,680 --> 00:38:19,930 Let's break these things up. 765 00:38:22,630 --> 00:38:25,050 And now, you see what the key idea is? 766 00:38:25,050 --> 00:38:31,720 'l1' and 'l2' are certain fixed numbers, fixed numbers. 767 00:38:31,720 --> 00:38:34,570 Notice that because 'g of x' can be made as close to 'l2' 768 00:38:34,570 --> 00:38:37,800 as I want and 'f of x' can be made as close to 'l1' as I 769 00:38:37,800 --> 00:38:41,130 want, notice that no matter what epsilon I'm given, I can 770 00:38:41,130 --> 00:38:45,660 certainly make this as small as I wish, just by picking 'x' 771 00:38:45,660 --> 00:38:47,150 close enough to 'a'. 772 00:38:47,150 --> 00:38:49,040 For example, for a given epsilon, how many 773 00:38:49,040 --> 00:38:50,220 terms do I have here? 774 00:38:50,220 --> 00:38:55,670 One, two, three. 775 00:38:55,670 --> 00:38:59,280 To make this whole sum less than epsilon, it's sufficient 776 00:38:59,280 --> 00:39:02,240 to make each of these three factors less than 777 00:39:02,240 --> 00:39:03,490 epsilon over 3. 778 00:39:08,310 --> 00:39:10,950 I can make these two less than epsilon over 3 pretty easy. 779 00:39:10,950 --> 00:39:13,880 How do I make this less than epsilon over 3? 780 00:39:13,880 --> 00:39:16,280 And the answer is, if you want this times this to be less 781 00:39:16,280 --> 00:39:18,820 than epsilon over 3 where these are positive numbers, 782 00:39:18,820 --> 00:39:22,070 make each of these less than the square root of 783 00:39:22,070 --> 00:39:23,690 epsilon over 3. 784 00:39:23,690 --> 00:39:27,200 Then when you multiply these two together, if this is less 785 00:39:27,200 --> 00:39:30,110 than this and this is less than this, this times this 786 00:39:30,110 --> 00:39:32,610 will be less than epsilon over 3. 787 00:39:32,610 --> 00:39:35,680 And now what you see what we do is very simple. 788 00:39:35,680 --> 00:39:39,760 To finish this proof off, all we do now is say, let epsilon 789 00:39:39,760 --> 00:39:41,430 greater than 0 be given. 790 00:39:41,430 --> 00:39:45,410 Choose epsilon 1 to equal epsilon over 3. 791 00:39:45,410 --> 00:39:48,480 Choose epsilon 2 to equal epsilon over 3. 792 00:39:48,480 --> 00:39:53,630 Choose epsilon sub 3 and epsilon sub 4 to each be the 793 00:39:53,630 --> 00:39:55,680 square root of epsilon over 3. 794 00:39:55,680 --> 00:39:58,870 Then we can find delta 1, delta 2, delta 795 00:39:58,870 --> 00:40:01,660 3, delta 4, et cetera. 796 00:40:01,660 --> 00:40:06,130 Clean up all of these, you see, and pick delta to be the 797 00:40:06,130 --> 00:40:08,470 minimum of the four deltas involved. 798 00:40:08,470 --> 00:40:11,800 Again, this is done much more explicitly both in the text 799 00:40:11,800 --> 00:40:13,090 and in the exercises. 800 00:40:13,090 --> 00:40:15,160 I just wanted you to get an overview here. 801 00:40:15,160 --> 00:40:17,920 And by the way, while we're speaking of this, there's 802 00:40:17,920 --> 00:40:20,390 always the danger that some of you may respect the 803 00:40:20,390 --> 00:40:21,670 professor too much. 804 00:40:21,670 --> 00:40:25,320 So that danger is one I don't mind living with. 805 00:40:25,320 --> 00:40:26,580 The problem is this. 806 00:40:26,580 --> 00:40:29,240 You may get the idea out of respect for me that I have 807 00:40:29,240 --> 00:40:31,280 invented a unique proof here. 808 00:40:31,280 --> 00:40:32,250 Let me tell you this. 809 00:40:32,250 --> 00:40:34,460 One, I did not invent this proof. 810 00:40:34,460 --> 00:40:37,070 Two, it is not unique. 811 00:40:37,070 --> 00:40:40,040 There are many different ways of proving the same result. 812 00:40:40,040 --> 00:40:44,690 For example, a person being told to work on this-- 813 00:40:44,690 --> 00:40:46,880 and I'm not going to carry the details out here-- 814 00:40:46,880 --> 00:40:50,230 but a person being told to work on something like this 815 00:40:50,230 --> 00:40:52,720 might have decided that instead of doing the clever 816 00:40:52,720 --> 00:40:55,650 thing that we did before, he was going to do the clever 817 00:40:55,650 --> 00:40:58,280 thing of adding and subtracting 818 00:40:58,280 --> 00:41:00,250 'l1' times 'g of x'. 819 00:41:00,250 --> 00:41:03,050 Because you see, if he did that, what would happen? 820 00:41:03,050 --> 00:41:06,450 He could now factor out a 'g of x' from here and rewrite 821 00:41:06,450 --> 00:41:10,450 this as 'g of x' times ''f of x' minus l1'. 822 00:41:13,680 --> 00:41:16,970 And these two terms could be combined together, we factor 823 00:41:16,970 --> 00:41:19,430 out an 'l1' times what? 824 00:41:19,430 --> 00:41:22,060 ''g of x' minus 'l2'. 825 00:41:22,060 --> 00:41:24,150 And even though the details would have been considerably 826 00:41:24,150 --> 00:41:26,420 different, the intuitive approach would have been-- 827 00:41:26,420 --> 00:41:28,060 well, look. 828 00:41:28,060 --> 00:41:33,730 This is pretty close to 'l2' when 'x' is near 'a'. 829 00:41:33,730 --> 00:41:35,820 This I can make as small as I want. 830 00:41:35,820 --> 00:41:38,580 Similarly, I can do the same things over here, and pretty 831 00:41:38,580 --> 00:41:42,010 soon you've got the idea that you can make this sum as small 832 00:41:42,010 --> 00:41:46,800 as you want just by choosing these sufficiently small. 833 00:41:46,800 --> 00:41:48,840 And there's no unique way of doing this. 834 00:41:48,840 --> 00:41:52,700 Now, here's a main point. 835 00:41:52,700 --> 00:41:56,900 Once all this work has been done, for a wide variety of 836 00:41:56,900 --> 00:42:01,730 problems we never again have to use an epsilon or a delta. 837 00:42:01,730 --> 00:42:04,010 Let me illustrate with one problem. 838 00:42:04,010 --> 00:42:06,760 Let's suppose we were given the problem limit of 'x 839 00:42:06,760 --> 00:42:10,010 squared plus 7x' as 'x approaches 3', and we wanted 840 00:42:10,010 --> 00:42:11,450 to find that limit. 841 00:42:11,450 --> 00:42:14,300 Our intuitive thing would be to do what? 842 00:42:14,300 --> 00:42:19,880 Let 'x' equal 3, in which case we get 9 plus 21 is 30. 843 00:42:19,880 --> 00:42:23,190 But we know by now that this instruction says that 'x' 844 00:42:23,190 --> 00:42:25,850 can't equal 3. 845 00:42:25,850 --> 00:42:26,620 This is the problem. 846 00:42:26,620 --> 00:42:28,960 We get what appears to be a nice answer that we believe 847 00:42:28,960 --> 00:42:31,090 in, but by an illegal method. 848 00:42:31,090 --> 00:42:34,410 Can we use our legal methods to gain the same result? 849 00:42:34,410 --> 00:42:36,000 The answer is yes. 850 00:42:36,000 --> 00:42:39,440 Because, you see, what is 'x squared plus 7x'? 851 00:42:39,440 --> 00:42:43,260 It's the sum of two functions, and we've just proven that the 852 00:42:43,260 --> 00:42:46,140 limit of the sum is the sum of the limits. 853 00:42:46,140 --> 00:42:48,170 For example, what I can say is this. 854 00:42:48,170 --> 00:42:49,340 I don't know if this is true. 855 00:42:49,340 --> 00:42:56,880 What I do know is true is that the limit of 'x squared plus 856 00:42:56,880 --> 00:42:59,930 7x' as 'x' approaches 3 is a limit of 'x squared' as 'x' 857 00:42:59,930 --> 00:43:03,960 approaches 3 plus the limit of '7x' as 'x' approaches 3. 858 00:43:03,960 --> 00:43:05,000 How do I know that? 859 00:43:05,000 --> 00:43:06,700 I've proven that the limit of a sum is 860 00:43:06,700 --> 00:43:07,900 the sum of the limits. 861 00:43:07,900 --> 00:43:09,100 Now what is this? 862 00:43:09,100 --> 00:43:11,330 This is really a product. 863 00:43:11,330 --> 00:43:14,210 This is really the limit of 'x' times 'x'. 864 00:43:14,210 --> 00:43:17,490 But we already know that the limit of a product is the 865 00:43:17,490 --> 00:43:18,580 product of the limits. 866 00:43:18,580 --> 00:43:20,130 See, we proved that theorem. 867 00:43:20,130 --> 00:43:23,240 Well, we almost proved it, certainly close enough so I 868 00:43:23,240 --> 00:43:27,000 think that we can say that we did. 869 00:43:27,000 --> 00:43:28,250 And this is a product also. 870 00:43:33,930 --> 00:43:37,765 So you see, we can get from here to here to here just by 871 00:43:37,765 --> 00:43:39,100 our theorem. 872 00:43:39,100 --> 00:43:42,420 Now didn't we also prove as one of our theorems earlier 873 00:43:42,420 --> 00:43:45,910 that the limit of 'x' as 'x' approaches 'a' is 'a' itself? 874 00:43:45,910 --> 00:43:46,870 Sure, we did. 875 00:43:46,870 --> 00:43:50,580 In particular then, the limit of 'x' as 'x' approaches 3 has 876 00:43:50,580 --> 00:43:54,390 already been proven to be 3. 877 00:43:54,390 --> 00:43:56,800 So this is 3, this is 3. 878 00:43:56,800 --> 00:43:59,670 What is the limit of 7 as 'x' approaches 3? 879 00:43:59,670 --> 00:44:03,810 Well, 7 is a constant, and we already proved that the limit 880 00:44:03,810 --> 00:44:05,690 of a constant as 'x' approaches 'a' 881 00:44:05,690 --> 00:44:07,080 is that same constant. 882 00:44:07,080 --> 00:44:08,740 So what is the constant here? 883 00:44:08,740 --> 00:44:09,830 It's 7. 884 00:44:09,830 --> 00:44:12,340 And what's the limit of 'x' as 'x' approaches 3? 885 00:44:12,340 --> 00:44:13,600 That's 3. 886 00:44:13,600 --> 00:44:15,810 So by using our theorems, we get from what? 887 00:44:15,810 --> 00:44:20,050 From here to here to here to here. 888 00:44:20,050 --> 00:44:22,645 And now hopefully from a theorem that comes from some 889 00:44:22,645 --> 00:44:26,480 place around the third grade, 3 times 3 is 9, 7 times 3 is 890 00:44:26,480 --> 00:44:29,220 21, the sum is 30. 891 00:44:29,220 --> 00:44:32,280 And now you see this is no longer a conjecture. 892 00:44:32,280 --> 00:44:35,660 This follows, inescapably, from the rules of our game, 893 00:44:35,660 --> 00:44:38,140 from the rules and our basic definition. 894 00:44:38,140 --> 00:44:43,470 By the way, you may have a tendency to feel when you see 895 00:44:43,470 --> 00:44:47,220 something like this that, why did we need the epsilons and 896 00:44:47,220 --> 00:44:48,210 deltas in the first place? 897 00:44:48,210 --> 00:44:49,450 Wasn't it a terrible waste? 898 00:44:49,450 --> 00:44:50,670 Well, two things. 899 00:44:50,670 --> 00:44:54,160 First of all, we couldn't prove our theorems without the 900 00:44:54,160 --> 00:44:55,430 epsilons and deltas. 901 00:44:55,430 --> 00:44:58,860 And secondly, and don't lose sight of this, in many real 902 00:44:58,860 --> 00:45:02,930 life situations you may very well be faced with the type of 903 00:45:02,930 --> 00:45:06,420 a problem that doesn't ask you to prove that the limit of 'x 904 00:45:06,420 --> 00:45:10,050 squared plus 7x' is 30 as 'x' approaches 3, but rather might 905 00:45:10,050 --> 00:45:15,710 say how close must 'x' be chosen to 3 if we want 'x 906 00:45:15,710 --> 00:45:21,100 squared plus 7x' to be less than 30.023. 907 00:45:21,100 --> 00:45:23,430 And then, you see, if that's the kind of a problem you 908 00:45:23,430 --> 00:45:26,030 have, these new theorems will not solve 909 00:45:26,030 --> 00:45:27,740 that problem for you. 910 00:45:27,740 --> 00:45:29,030 So we're not making a choice here. 911 00:45:29,030 --> 00:45:32,130 All we're saying is that the epsilons and deltas are the 912 00:45:32,130 --> 00:45:36,210 backbone of limits, but that fortunately through 913 00:45:36,210 --> 00:45:40,920 mathematical theorems, we can get simpler ways of getting 914 00:45:40,920 --> 00:45:41,980 important results. 915 00:45:41,980 --> 00:45:43,800 And that was our main purpose of today's 916 00:45:43,800 --> 00:45:44,970 lecture, these two things. 917 00:45:44,970 --> 00:45:47,870 That completes our lecture for today, and so on until next 918 00:45:47,870 --> 00:45:49,120 time, good bye. 919 00:45:51,440 --> 00:45:54,450 NARRATOR: Funding for the publication of this video was 920 00:45:54,450 --> 00:45:59,160 provided by the Gabriella and Paul Rosenbaum foundation. 921 00:45:59,160 --> 00:46:03,340 Help OCW continue to provide free and open access to MIT 922 00:46:03,340 --> 00:46:07,540 courses by making a donation at ocw.mit.edu/donate.