1 00:00:00,040 --> 00:00:01,940 GUEST SPEAKER: 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 --> 00:00:09,980 offer high quality educational resources for free. 5 00:00:09,980 --> 00:00:12,830 To make a donation, or 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 00:00:35,206 --> 00:00:35,720 HERBERT GROSS: Hi. 9 00:00:35,720 --> 00:00:40,630 Our lecture today involves the backbone of Calculus, namely 10 00:00:40,630 --> 00:00:43,440 the concept of limit, and perhaps no place in the 11 00:00:43,440 --> 00:00:47,200 history of mathematics is there a more subtle topic than 12 00:00:47,200 --> 00:00:48,950 this particular one. 13 00:00:48,950 --> 00:00:52,090 I'm reminded of the anecdote of the professor who was 14 00:00:52,090 --> 00:00:55,710 staring quite intently, philosophically, in his home, 15 00:00:55,710 --> 00:00:58,420 and his wife asked, what are you contemplating, and he 16 00:00:58,420 --> 00:01:01,090 said, that it's amazing how electricity works. 17 00:01:01,090 --> 00:01:03,350 And the wife said, well what's so amazing about that? 18 00:01:03,350 --> 00:01:05,820 All you have to do is flick the switch. 19 00:01:05,820 --> 00:01:08,890 And this somehow or other is exactly the predicament that 20 00:01:08,890 --> 00:01:10,740 the limit concept falls into. 21 00:01:10,740 --> 00:01:13,910 From the point of view of computation, it seems that 22 00:01:13,910 --> 00:01:18,370 what comes naturally 99% of the time gives us the right 23 00:01:18,370 --> 00:01:23,480 answer, but unfortunately, the 1% of the time turns up almost 24 00:01:23,480 --> 00:01:26,410 all the time in differential Calculus. 25 00:01:26,410 --> 00:01:29,380 For this reason, what we will do is introduce the concept of 26 00:01:29,380 --> 00:01:32,540 limit through the eyes of differential Calculus, and we 27 00:01:32,540 --> 00:01:37,370 shall call our lecture today 'Derivatives and Limits'. 28 00:01:37,370 --> 00:01:41,440 And what we shall do is go back to our old friend who 29 00:01:41,440 --> 00:01:44,540 makes an appearance in almost all of our lectures so far, 30 00:01:44,540 --> 00:01:47,080 the case of the freely falling object. 31 00:01:47,080 --> 00:01:48,820 A ball is dropped. 32 00:01:48,820 --> 00:01:53,590 It falls freely in a vacuum, and the distance, 's', that it 33 00:01:53,590 --> 00:01:57,380 falls in feet, at the end of 't' seconds, is given by 's' 34 00:01:57,380 --> 00:01:59,640 equals '16t squared'. 35 00:01:59,640 --> 00:02:03,820 The question that we would like to raise is, how fast is 36 00:02:03,820 --> 00:02:07,770 the ball falling when 't' equals one? 37 00:02:07,770 --> 00:02:10,940 In other words, notice that the ball is covering a 38 00:02:10,940 --> 00:02:12,730 different distance here as it falls. 39 00:02:12,730 --> 00:02:15,920 What we want to know is at the instant that the time is one. 40 00:02:15,920 --> 00:02:18,280 And by the way, don't confuse the speed with the 41 00:02:18,280 --> 00:02:19,060 displacement. 42 00:02:19,060 --> 00:02:22,617 We know that when the time is one, the object has fallen a 43 00:02:22,617 --> 00:02:25,530 distance of 16 feet. 44 00:02:25,530 --> 00:02:28,290 When the time is 1, the object has fallen 16 feet. 45 00:02:28,290 --> 00:02:31,560 What we want to know is how fast is it falling at that 46 00:02:31,560 --> 00:02:32,960 particular point. 47 00:02:32,960 --> 00:02:36,810 Now keep in mind that mathematics being a logical 48 00:02:36,810 --> 00:02:40,920 subject proceeds from the idea of trying to study the 49 00:02:40,920 --> 00:02:43,160 unfamiliar in terms of the familiar. 50 00:02:43,160 --> 00:02:46,860 What we are familiar with is the concept known as average 51 00:02:46,860 --> 00:02:48,030 rate of speed. 52 00:02:48,030 --> 00:02:52,560 So what we will do is duck the main question temporarily and 53 00:02:52,560 --> 00:02:54,500 proceed to a different question. 54 00:02:54,500 --> 00:02:58,640 What we shall ask is, what is the average speed of the ball 55 00:02:58,640 --> 00:03:03,330 during the time interval from t equals one to t equals two? 56 00:03:03,330 --> 00:03:06,230 And we all know how to solve this problem from our 57 00:03:06,230 --> 00:03:12,600 pre-calculus courses, namely, we compute where the ball is 58 00:03:12,600 --> 00:03:14,430 when 't' is two. 59 00:03:14,430 --> 00:03:17,960 We compute where the ball was when 't' is one. 60 00:03:17,960 --> 00:03:21,280 The difference between these two distances is the distance 61 00:03:21,280 --> 00:03:24,070 that the ball has fallen during this time interval, and 62 00:03:24,070 --> 00:03:26,740 we then divide by the time interval. 63 00:03:26,740 --> 00:03:30,400 And the distance divided by the time is precisely the 64 00:03:30,400 --> 00:03:32,890 definition of average rate of speed. 65 00:03:32,890 --> 00:03:35,820 In other words, during the time interval from 't' equals 66 00:03:35,820 --> 00:03:40,420 one to 't' equals two, the ball falls at an average speed 67 00:03:40,420 --> 00:03:42,690 of 48 feet per second. 68 00:03:42,690 --> 00:03:47,230 Again, in terms of our diagram, you see at 't' equals 69 00:03:47,230 --> 00:03:50,510 one, the object is over here. 70 00:03:50,510 --> 00:03:55,300 At 't' equals two, the object is over here, it's fallen 64 71 00:03:55,300 --> 00:03:59,270 feet, and so we compute the average speed by taking the 72 00:03:59,270 --> 00:04:03,150 total distance and dividing by the time that it took. 73 00:04:03,150 --> 00:04:07,470 Now again, what we have done is found the right answer, but 74 00:04:07,470 --> 00:04:08,690 to the wrong question. 75 00:04:08,690 --> 00:04:10,940 The question was not what was the average rate of speed from 76 00:04:10,940 --> 00:04:14,190 't' equals one to 't' equals two, it was, what is the speed 77 00:04:14,190 --> 00:04:16,360 at the instant 't' equals one? 78 00:04:16,360 --> 00:04:20,050 And we sense that between 't' equals one and 't' equals two, 79 00:04:20,050 --> 00:04:23,170 an awful lot can happen, and that therefore, our average 80 00:04:23,170 --> 00:04:26,450 speed need not be a good approximation for the 81 00:04:26,450 --> 00:04:27,780 instantaneous speed. 82 00:04:27,780 --> 00:04:31,630 What we do next, you see, is we say OK, let's do something 83 00:04:31,630 --> 00:04:32,850 a little bit differently. 84 00:04:32,850 --> 00:04:37,240 Let's now compute the average speed, but not from 't' equals 85 00:04:37,240 --> 00:04:40,710 one to 't' equals two, but rather from 't' equals one to 86 00:04:40,710 --> 00:04:42,970 't' equals 1.1. 87 00:04:42,970 --> 00:04:47,220 And computationally, we mimic the same procedure as before. 88 00:04:47,220 --> 00:04:50,910 We compute 's' when 't' is 1.1, we compute 89 00:04:50,910 --> 00:04:52,780 's' when 't' is one. 90 00:04:52,780 --> 00:04:54,510 By the way, the only difference is that it's a 91 00:04:54,510 --> 00:04:58,850 little bit tougher to square 1.1 and multiply that by 16 92 00:04:58,850 --> 00:05:01,930 than it was to square two and multiply that by 16. 93 00:05:01,930 --> 00:05:04,680 The arithmetic gets slightly messier if you want to call it 94 00:05:04,680 --> 00:05:06,900 that, but the concept stays the same. 95 00:05:06,900 --> 00:05:09,810 We find the total distance traveled during the time 96 00:05:09,810 --> 00:05:13,980 interval, we divide by the length of the time interval, 97 00:05:13,980 --> 00:05:17,560 and that quotient, by definition, is the average 98 00:05:17,560 --> 00:05:18,560 rate of speed. 99 00:05:18,560 --> 00:05:23,050 Again, in terms of our diagram, at t equals one, the 100 00:05:23,050 --> 00:05:25,640 object has fallen 16 feet. 101 00:05:25,640 --> 00:05:28,230 I'll distort this so that we can see what's happening here. 102 00:05:28,230 --> 00:05:34,210 At 't' equals 1.1, the object has fallen 19.36 feet, and now 103 00:05:34,210 --> 00:05:37,530 we compute the average speed from here to here in the usual 104 00:05:37,530 --> 00:05:39,050 high school way. 105 00:05:39,050 --> 00:05:44,030 Now again, the same question arises as before, namely, this 106 00:05:44,030 --> 00:05:45,390 is still an average speed. 107 00:05:45,390 --> 00:05:47,800 We wanted an instantaneous speed. 108 00:05:47,800 --> 00:05:51,350 And our come back is to say look at, we at least suspect 109 00:05:51,350 --> 00:05:56,940 intuitively that between one and 1.1, we will hopefully get 110 00:05:56,940 --> 00:06:00,430 a better idea as to what's happening at exactly one than 111 00:06:00,430 --> 00:06:03,320 when we did on the bigger interval from one to two. 112 00:06:03,320 --> 00:06:06,810 In other words, whereas we don't believe that 33.6 is the 113 00:06:06,810 --> 00:06:10,350 answer to our original question, we do believe that 114 00:06:10,350 --> 00:06:15,600 33.6 is probably a better approximation to the answer to 115 00:06:15,600 --> 00:06:19,560 our original question than the answer 48 feet per second that 116 00:06:19,560 --> 00:06:21,860 was obtained over the larger interval. 117 00:06:21,860 --> 00:06:25,300 What the mathematician does next is he says well, let's 118 00:06:25,300 --> 00:06:26,670 stop playing games. 119 00:06:26,670 --> 00:06:31,010 Let's not go between one and 1.1, or one and 1.01. 120 00:06:31,010 --> 00:06:37,530 Why don't we go between one and '1+h', where 'h' is any 121 00:06:37,530 --> 00:06:38,630 non-zero amount? 122 00:06:38,630 --> 00:06:41,860 In other words, let's find the average speed from 't' equals 123 00:06:41,860 --> 00:06:46,620 one to 't' equals '1+h', and then what we'll do is we will 124 00:06:46,620 --> 00:06:48,930 investigate to see what happens when h 125 00:06:48,930 --> 00:06:50,810 gets very, very small. 126 00:06:50,810 --> 00:06:54,510 Well again, we mimic exactly our procedure before. 127 00:06:54,510 --> 00:06:59,090 We feed in 't' equals '1+h' and compute 's', which of 128 00:06:59,090 --> 00:07:01,980 course is '16 times '1+h' squared'. 129 00:07:01,980 --> 00:07:04,890 Then we subtract off the distance that the particle had 130 00:07:04,890 --> 00:07:08,620 fallen when 't' is one, that's 16 feet, we divide by the 131 00:07:08,620 --> 00:07:09,960 length of the time interval. 132 00:07:09,960 --> 00:07:14,740 Well between one and '1+h', the time interval is 'h', we 133 00:07:14,740 --> 00:07:18,650 simplify algebraically, and we wind up with this particular 134 00:07:18,650 --> 00:07:23,680 expression, '32h plus 16h squared' over 'h'. 135 00:07:23,680 --> 00:07:28,210 And now we come to the crux of what limits, and derivatives, 136 00:07:28,210 --> 00:07:31,460 and instantaneous rate of speeds are all about. 137 00:07:31,460 --> 00:07:34,630 If you recall, it's quite tempting to look at this thing 138 00:07:34,630 --> 00:07:38,580 and say, ah-ha, I'll cancel an 'h' from both the numerator 139 00:07:38,580 --> 00:07:44,250 and the denominator, and that will leave me '32 plus 16h'. 140 00:07:44,250 --> 00:07:47,260 And notice though, the very important thing, that as we 141 00:07:47,260 --> 00:07:53,410 mentioned in previous lectures, since you cannot 142 00:07:53,410 --> 00:07:57,960 divide by 0, it is crucial at this stage that we emphasize 143 00:07:57,960 --> 00:08:00,090 that 'h' is not equal to 0. 144 00:08:00,090 --> 00:08:02,150 Now of course, this is not a difficult 145 00:08:02,150 --> 00:08:03,770 rationalization to make. 146 00:08:03,770 --> 00:08:06,510 Namely, in terms of our physical situation, we 147 00:08:06,510 --> 00:08:09,960 wouldn't want to compute an average rate of speed over a 148 00:08:09,960 --> 00:08:12,810 time interval during which no time transpired. 149 00:08:12,810 --> 00:08:16,630 And if we let 'h' be 0, obviously, from one to '1+h', 150 00:08:16,630 --> 00:08:18,000 no time transpires. 151 00:08:18,000 --> 00:08:20,290 But the key statement is, and we'll come back to exploit 152 00:08:20,290 --> 00:08:23,210 this not only for this lecture, but for the next one 153 00:08:23,210 --> 00:08:27,310 too, is to exploit the idea that in our entire discussion, 154 00:08:27,310 --> 00:08:31,170 it was crucial that h not be allowed to equal 0. 155 00:08:31,170 --> 00:08:36,559 In other words, the average speed between one and '1+h' is 156 00:08:36,559 --> 00:08:39,340 '32 plus 16h'. 157 00:08:39,340 --> 00:08:43,100 Now again, without trying to be rigorous, observe that our 158 00:08:43,100 --> 00:08:47,740 senses tell us that as 'h' gets us close to 0 as we want 159 00:08:47,740 --> 00:08:51,670 without ever being allowed to get there, '16h' gets us close 160 00:08:51,670 --> 00:08:56,260 to 0 as we want also, and hence, '32 plus 16h' 161 00:08:56,260 --> 00:09:01,490 apparently gets us close to 32 as we wish. 162 00:09:01,490 --> 00:09:09,720 In other words, it appears that the speed is 32 feet per 163 00:09:09,720 --> 00:09:12,180 second when t equals one. 164 00:09:12,180 --> 00:09:17,230 And to reuse the vernacular, what we're saying is we 165 00:09:17,230 --> 00:09:20,550 compute the average rate of speed-- 166 00:09:20,550 --> 00:09:24,770 see 's' of '1+h' minus 's' of '1/h'-- 167 00:09:24,770 --> 00:09:27,800 and see what happens to that as 'h' approaches 0. 168 00:09:27,800 --> 00:09:29,740 And in our particular case, notice that this 169 00:09:29,740 --> 00:09:30,700 simply means what? 170 00:09:30,700 --> 00:09:34,790 We came down to '32 plus 16h' over here, and then as 'h' got 171 00:09:34,790 --> 00:09:38,580 arbitrarily small, we became suspicious, and believed that 172 00:09:38,580 --> 00:09:41,230 as 'h' became arbitrarily small, '32 173 00:09:41,230 --> 00:09:44,270 plus 16h' became 32. 174 00:09:44,270 --> 00:09:46,970 And I'm saying that in great detail because this is not 175 00:09:46,970 --> 00:09:49,190 quite nearly as harmless as it may seem. 176 00:09:49,190 --> 00:09:52,050 There is a great deal of difficulty involved in just 177 00:09:52,050 --> 00:09:55,900 saying as 'h' gets close to 0, '32 plus 16h' 178 00:09:55,900 --> 00:09:57,100 gets close to 32. 179 00:09:57,100 --> 00:09:59,910 Well there's no difficulty in saying it. 180 00:09:59,910 --> 00:10:03,240 There's difficulty that comes up in trying to show what this 181 00:10:03,240 --> 00:10:06,660 means precisely, as we shall see in just a little while. 182 00:10:06,660 --> 00:10:10,280 By the way, it's rather easy to generalize this result. 183 00:10:10,280 --> 00:10:13,220 Namely, you may have noticed in our presentation there was 184 00:10:13,220 --> 00:10:16,730 nothing sacred about asking what was the speed when 't' 185 00:10:16,730 --> 00:10:17,550 equals one. 186 00:10:17,550 --> 00:10:21,500 One could just as easily have chosen any other time 't', say 187 00:10:21,500 --> 00:10:26,660 't sub one', and asked what was the speed between 't1' and 188 00:10:26,660 --> 00:10:27,830 't1' plus 'h'? 189 00:10:27,830 --> 00:10:31,130 We could have done exactly the same computation. 190 00:10:31,130 --> 00:10:34,670 In fact, notice down here, when we do this computation, 191 00:10:34,670 --> 00:10:40,670 the answer '32t one' plus '16h' becomes precisely '32 192 00:10:40,670 --> 00:10:44,500 plus 16h when 't' one happens to be one. 193 00:10:44,500 --> 00:10:48,600 In other words, this result here is the generalization of 194 00:10:48,600 --> 00:10:53,660 our previous result that had us work with the 195 00:10:53,660 --> 00:10:57,170 limit of '32 plus 16h'. 196 00:10:57,170 --> 00:11:02,010 At any rate, by this approach, it becomes clear then, if the 197 00:11:02,010 --> 00:11:05,090 equation of motion, the distance versus the time, is 198 00:11:05,090 --> 00:11:08,840 given by 's' equals '16t squared', then that time 't' 199 00:11:08,840 --> 00:11:12,670 equals 't one', the instantaneous speed appears to 200 00:11:12,670 --> 00:11:16,020 be 32 times 't' one. 201 00:11:16,020 --> 00:11:21,270 And in fact, since 't' one happens to be any arbitrary 202 00:11:21,270 --> 00:11:23,820 time that we choose, it is customary to drop the 203 00:11:23,820 --> 00:11:25,880 subscript, and simply say what? 204 00:11:25,880 --> 00:11:29,550 At any time, 't', our freely falling body in this case, 205 00:11:29,550 --> 00:11:33,130 will have a speed given by '32t'. 206 00:11:33,130 --> 00:11:37,400 And many of us will remember, as this is a revisited course, 207 00:11:37,400 --> 00:11:40,740 that this is precisely the recipe that turns out when one 208 00:11:40,740 --> 00:11:43,180 talks about bringing down the exponent and 209 00:11:43,180 --> 00:11:45,250 replacing it by one less. 210 00:11:45,250 --> 00:11:50,540 16 times two is 32, and 't' to a power one less than two is 211 00:11:50,540 --> 00:11:53,150 just t to the first power or 't'. 212 00:11:53,150 --> 00:11:56,200 But notice now, no recipes involved here. 213 00:11:56,200 --> 00:12:01,060 Just a question of intuitively defining instantaneous speed 214 00:12:01,060 --> 00:12:05,470 in terms of being a limit of average speeds. 215 00:12:05,470 --> 00:12:08,340 Now in order that we don't become hypocrites in what 216 00:12:08,340 --> 00:12:12,040 we're doing, recall also that we have spent an entire 217 00:12:12,040 --> 00:12:16,610 lecture talking about the value of analytic Geometry in 218 00:12:16,610 --> 00:12:18,070 the study of Calculus. 219 00:12:18,070 --> 00:12:21,470 The idea that a picture is worth 1,000 words. 220 00:12:21,470 --> 00:12:25,360 So what we would like to do now is to revisit our previous 221 00:12:25,360 --> 00:12:29,000 discussion in terms of a graph. 222 00:12:29,000 --> 00:12:33,950 You see, suppose we have a curve 'y' equals 'f of x', and 223 00:12:33,950 --> 00:12:36,440 I've drawn the curve over on this side over here. 224 00:12:36,440 --> 00:12:40,110 We'll make more reference to it later as we go along, but 225 00:12:40,110 --> 00:12:44,120 what does it mean if I now mechanically compute 'f of x1 226 00:12:44,120 --> 00:12:48,400 ' plus 'delta x' minus 'f of x1' over 'delta x'? 227 00:12:48,400 --> 00:12:53,240 And by the way, again, notice just a question of symbolism. 228 00:12:53,240 --> 00:12:56,510 'Delta x' happens to be the conventional symbol that one 229 00:12:56,510 --> 00:12:59,410 uses in the analytical Geometry approach where we 230 00:12:59,410 --> 00:13:01,600 were using h before, but this is simply 231 00:13:01,600 --> 00:13:03,490 a question of notation. 232 00:13:03,490 --> 00:13:12,410 Well the idea is this: observe that if we go to the graph, 'f 233 00:13:12,410 --> 00:13:17,060 of x1' plus 'delta x' is this particular height. 234 00:13:17,060 --> 00:13:21,790 On the other hand, this height is 'f of x1' . 235 00:13:21,790 --> 00:13:29,000 Consequently, our numerator is just this distance here, and 236 00:13:29,000 --> 00:13:31,960 our denominator, which is 'delta x', is just this 237 00:13:31,960 --> 00:13:33,460 distance here. 238 00:13:33,460 --> 00:13:37,690 And therefore, what we have done in our quotient, is we 239 00:13:37,690 --> 00:13:41,650 have found the slope of the straight line that joins 'p' 240 00:13:41,650 --> 00:13:45,870 to 'q.' And here then is why the question of straight line 241 00:13:45,870 --> 00:13:48,430 becomes so important in the study of 242 00:13:48,430 --> 00:13:49,900 differential Calculus. 243 00:13:49,900 --> 00:13:54,260 That what is average speed, in terms of analysis, becomes the 244 00:13:54,260 --> 00:13:57,660 slope of a straight line in terms of Geometry. 245 00:13:57,660 --> 00:14:01,030 And the idea, you see, is that this gives me the average 246 00:14:01,030 --> 00:14:06,460 speed of the straight line that joins the average rise of 247 00:14:06,460 --> 00:14:08,890 the straight line that joins 'p' to 'q'. 248 00:14:08,890 --> 00:14:12,800 And by the way, this is quite deceptive as any average is. 249 00:14:12,800 --> 00:14:14,990 It's like the man who sat with his feet in the oven and an 250 00:14:14,990 --> 00:14:17,100 ice pack on his head, and when somebody said how do you feel, 251 00:14:17,100 --> 00:14:18,850 he said, on the average, pretty good. 252 00:14:18,850 --> 00:14:21,380 You see, the same thing is happening over here. 253 00:14:21,380 --> 00:14:24,880 Notice if I had taken a completely different curve, a 254 00:14:24,880 --> 00:14:28,560 curve which looked nothing at all like my original curve, 255 00:14:28,560 --> 00:14:32,670 only that also passed through the points 'p' and 'q', notice 256 00:14:32,670 --> 00:14:35,190 that even though these two curves are quite different, 257 00:14:35,190 --> 00:14:38,760 the average rise from 'p' to 'q' is the 258 00:14:38,760 --> 00:14:40,220 same for both curves. 259 00:14:40,220 --> 00:14:43,260 In other words, when one deals with the average, one is not 260 00:14:43,260 --> 00:14:46,690 too concerned with how one got from the first 261 00:14:46,690 --> 00:14:47,870 point to the second. 262 00:14:47,870 --> 00:14:50,280 You see, this is the problem with average rate of space. 263 00:14:50,280 --> 00:14:53,500 In fact, the bigger a time interval we work over, the 264 00:14:53,500 --> 00:14:58,090 more danger there is that the average rate of speed will not 265 00:14:58,090 --> 00:15:01,250 coincide intuitively with what we would like to believe that 266 00:15:01,250 --> 00:15:03,450 the instantaneous rate of speed is. 267 00:15:03,450 --> 00:15:06,940 And you see, that's exactly what this idea here means. 268 00:15:06,940 --> 00:15:10,650 What we do to define the derivative, the instantaneous 269 00:15:10,650 --> 00:15:15,820 rise at 'x' equals 'x1' , is we take this quotient and 270 00:15:15,820 --> 00:15:20,050 investigate what happens to it as 'delta x' is allowed to get 271 00:15:20,050 --> 00:15:22,080 arbitrarily close to 0. 272 00:15:22,080 --> 00:15:24,210 But let's emphasize this again. 273 00:15:24,210 --> 00:15:28,160 'Delta x' is not allowed to equal 0. 274 00:15:28,160 --> 00:15:31,020 You see, it's over a smaller and smaller interval, and you 275 00:15:31,020 --> 00:15:33,120 see what's happening here geometrically? 276 00:15:33,120 --> 00:15:36,580 To let 'delta x' get smaller and smaller means, that for a 277 00:15:36,580 --> 00:15:41,810 given 'x1' , you are holding 'p' fixed, and allowing 'q' to 278 00:15:41,810 --> 00:15:44,560 move in closer and closer to 'p'. 279 00:15:44,560 --> 00:15:46,960 And what we are saying is, is that when 'q' assumes this 280 00:15:46,960 --> 00:15:51,290 position, let's call it 'q sub 1', the slope of the line that 281 00:15:51,290 --> 00:15:55,080 joints 'p' to 'q' one looks a lot more like the slope of the 282 00:15:55,080 --> 00:15:59,040 line that will be tangent to the curve at the point 'p'. 283 00:15:59,040 --> 00:16:03,140 And for this reason, what is instantaneous rate of speed, 284 00:16:03,140 --> 00:16:07,220 when we're talking about functions, become slopes of 285 00:16:07,220 --> 00:16:11,980 curves at a given point when we are talking about curves. 286 00:16:11,980 --> 00:16:15,270 You see that the tangent line to a curve is viewed as being 287 00:16:15,270 --> 00:16:19,395 the limiting position of a cord drawn from a point 'p' to 288 00:16:19,395 --> 00:16:21,150 a point 'q'. 289 00:16:21,150 --> 00:16:24,220 Now at any rate, this will be emphasized in the text. 290 00:16:24,220 --> 00:16:26,990 It will be emphasized in our exercises. 291 00:16:26,990 --> 00:16:31,180 The major issue now is what did we really do when it came 292 00:16:31,180 --> 00:16:34,620 time to compute this little gadget. 293 00:16:34,620 --> 00:16:38,050 In other words, when we talked about the limit of 'f of x' as 294 00:16:38,050 --> 00:16:41,980 'x' approached 'a', how did we compute this? 295 00:16:41,980 --> 00:16:46,220 The danger is, and this is why I have underlined a question 296 00:16:46,220 --> 00:16:48,980 mark here, the intuitive approach is to say something 297 00:16:48,980 --> 00:16:53,550 like, let's just replace x by a in here. 298 00:16:53,550 --> 00:16:56,360 In other words, let the limit of 'f of x', as 'x' approaches 299 00:16:56,360 --> 00:16:58,310 'a', be 'f of a'. 300 00:16:58,310 --> 00:17:03,110 In fact, this is what we did with our '32 plus 16h'. 301 00:17:03,110 --> 00:17:06,122 In a way, what we did was, with '32 plus 16h', do you 302 00:17:06,122 --> 00:17:06,970 remember what we did? 303 00:17:06,970 --> 00:17:10,260 We said, let's let 'h' get arbitrarily close to 0. 304 00:17:10,260 --> 00:17:12,450 And in our minds, we let 'h' equal 0. 305 00:17:12,450 --> 00:17:17,730 '16h' was then 0, and then '32 plus 16h' was then 32. 306 00:17:17,730 --> 00:17:20,140 But what we had done is somehow or other, if our 307 00:17:20,140 --> 00:17:24,530 intuition is correct, we had arrived at the correct answer 308 00:17:24,530 --> 00:17:26,300 but for wrong reasons. 309 00:17:26,300 --> 00:17:30,000 Because you see, in this expression, it is mandatory, 310 00:17:30,000 --> 00:17:32,620 as we've outlined this, is that 'x' never be 311 00:17:32,620 --> 00:17:36,030 allowed to equal 'a'. 312 00:17:36,030 --> 00:17:37,960 Well you say, who cares whether you 313 00:17:37,960 --> 00:17:38,850 allow it or you don't? 314 00:17:38,850 --> 00:17:40,750 Aren't we going to get the same answer? 315 00:17:40,750 --> 00:17:43,000 And just to give you a quickie, one that we talked 316 00:17:43,000 --> 00:17:45,910 about in our introductory lecture, consider the function 317 00:17:45,910 --> 00:17:48,690 'f of x' to be 'x squared minus nine' 318 00:17:48,690 --> 00:17:51,080 over 'x minus three'. 319 00:17:51,080 --> 00:17:53,970 And let's evaluate what this limit is as 320 00:17:53,970 --> 00:17:56,700 'x' approaches three. 321 00:17:56,700 --> 00:18:00,690 If we were to blindly assume that all we have to do is 322 00:18:00,690 --> 00:18:04,150 replace 'x' by 'a', that would mean what here? 323 00:18:04,150 --> 00:18:06,510 Replace 'x' by three. 324 00:18:06,510 --> 00:18:08,700 Notice that what we got is what? 325 00:18:08,700 --> 00:18:12,980 Three squared, which is nine, minus nine, (which is 0) over 326 00:18:12,980 --> 00:18:15,730 three minus three, which is also 0. 327 00:18:15,730 --> 00:18:19,010 In other words, if in the expression 'x squared minus 328 00:18:19,010 --> 00:18:23,770 nine' over 'x minus three' you replace 'x' by three, you do 329 00:18:23,770 --> 00:18:25,750 get 0 over 0. 330 00:18:25,750 --> 00:18:30,270 Unfortunately, or perhaps fortunately, somehow or other, 331 00:18:30,270 --> 00:18:33,310 this definition is supposed to incorporate the fact that you 332 00:18:33,310 --> 00:18:36,860 never allow 'x' to equal three. 333 00:18:36,860 --> 00:18:40,240 See, the way we got 0 over 0 was we violated what the 334 00:18:40,240 --> 00:18:41,960 intuitive meaning of limit was. 335 00:18:41,960 --> 00:18:46,630 We wound up with this 0 over 0 mess by allowing 'x' to equal 336 00:18:46,630 --> 00:18:50,130 the one value it is not allowed to equal here. 337 00:18:50,130 --> 00:18:54,250 To see what happened more anatomically in this 338 00:18:54,250 --> 00:18:57,750 particular problem, let's go back to our old friend of 339 00:18:57,750 --> 00:18:59,140 cancellation again. 340 00:18:59,140 --> 00:19:03,520 We observe that 'x squared minus nine' can be written as 341 00:19:03,520 --> 00:19:07,930 'x plus three' times 'x minus three', and then we divide by 342 00:19:07,930 --> 00:19:12,020 'x minus three', and here's where the main problem seems 343 00:19:12,020 --> 00:19:12,710 to come up. 344 00:19:12,710 --> 00:19:16,420 One automatically, somehow or other, gets into the habit of 345 00:19:16,420 --> 00:19:18,600 canceling these things out. 346 00:19:18,600 --> 00:19:21,160 But the point is, since we have already agreed that you 347 00:19:21,160 --> 00:19:25,490 cannot divide by 0, the idea then becomes that the only 348 00:19:25,490 --> 00:19:30,850 time this is permissible is when 'x' is unequal to three. 349 00:19:30,850 --> 00:19:36,990 In fact, to have been more precise, what we should have 350 00:19:36,990 --> 00:19:42,160 said was not that 'x squared minus nine' over 'x minus 351 00:19:42,160 --> 00:19:46,310 three' equals 'x plus three', but rather, 'x squared minus 352 00:19:46,310 --> 00:19:51,850 nine' over 'x minus three' equals 'x plus three', unless 353 00:19:51,850 --> 00:19:53,820 'x' equals three. 354 00:19:53,820 --> 00:19:57,470 And by the way, I should add here what? 355 00:19:57,470 --> 00:20:04,820 That if 'x' equals three, then 'x squared minus nine' over 'x 356 00:20:04,820 --> 00:20:08,290 minus three' is undefined. 357 00:20:08,290 --> 00:20:10,850 You see, that's what we call 0 over 0. 358 00:20:10,850 --> 00:20:14,000 Undefined, indeterminate. 359 00:20:14,000 --> 00:20:16,570 But here's what bails us out. 360 00:20:16,570 --> 00:20:20,410 As soon as we say the limit as 'x' approaches three, what are 361 00:20:20,410 --> 00:20:21,490 we assuming? 362 00:20:21,490 --> 00:20:27,070 We're assuming that 'x' can be any value at all except three, 363 00:20:27,070 --> 00:20:30,370 and the interesting point now becomes, that as long as 'x' 364 00:20:30,370 --> 00:20:34,120 is anything but three, the only time you could tell these 365 00:20:34,120 --> 00:20:39,000 two expressions apart was when 'x' happened to equal three. 366 00:20:39,000 --> 00:20:43,550 So if you now say we won't allow 'x' to equal three, then 367 00:20:43,550 --> 00:20:47,450 it is true, whereas the bracketed expression by itself 368 00:20:47,450 --> 00:20:49,200 is not equal to the parenthetical 369 00:20:49,200 --> 00:20:50,530 expression by itself. 370 00:20:50,530 --> 00:20:54,520 The limit as 'x' approaches three of these two expressions 371 00:20:54,520 --> 00:20:58,160 are equal, and in fact, one becomes tempted to say "what?" 372 00:20:58,160 --> 00:20:59,110 at this case. 373 00:20:59,110 --> 00:21:02,920 Oh, here it's quite easy to see, that as 'x' approaches 374 00:21:02,920 --> 00:21:08,520 three, 'x plus three' is equal to six. 375 00:21:08,520 --> 00:21:11,330 And let me put a question mark here also, because it 376 00:21:11,330 --> 00:21:15,120 certainly is true that if you replace 'x' by three, this 377 00:21:15,120 --> 00:21:17,310 will be six. 378 00:21:17,310 --> 00:21:20,600 But this says don't replace 'x' by three. 379 00:21:20,600 --> 00:21:24,030 Now before, when we replaced 'x' by three we got into 380 00:21:24,030 --> 00:21:25,860 trouble, so we said we can't do that. 381 00:21:25,860 --> 00:21:29,280 Now we replace 'x' by three, we get an answer that we like, 382 00:21:29,280 --> 00:21:32,210 and that is the danger that we'll say OK, we won't allow 383 00:21:32,210 --> 00:21:35,860 this to be done, unless we like the result. 384 00:21:35,860 --> 00:21:38,710 And of course this gives you a highly subjective subject 385 00:21:38,710 --> 00:21:39,640 matter here. 386 00:21:39,640 --> 00:21:42,400 You see, the point is that somehow or other we must come 387 00:21:42,400 --> 00:21:46,070 up with a more objective criteria for distinguishing 388 00:21:46,070 --> 00:21:47,700 what limits are. 389 00:21:47,700 --> 00:21:51,200 And by the way, for the Polyanna who has the intuitive 390 00:21:51,200 --> 00:21:55,370 feeling for saying things like well look at, this may give 391 00:21:55,370 --> 00:21:58,390 you trouble if you wind up with 0 over 0. 392 00:21:58,390 --> 00:22:02,240 But what is the likelihood that given 'f of x' at random, 393 00:22:02,240 --> 00:22:05,270 and 'a' at random, that the limit of 'f of x' as 'x' 394 00:22:05,270 --> 00:22:09,060 approaches 'a' is going to be 0 over 0 if you're careless? 395 00:22:09,060 --> 00:22:11,950 And the answer is that the probability is quite small, 396 00:22:11,950 --> 00:22:16,540 except in Calculus, in which case, every time, you take a 397 00:22:16,540 --> 00:22:17,460 derivative. 398 00:22:17,460 --> 00:22:21,470 Every time you are going to wind up with 0 over 0. 399 00:22:21,470 --> 00:22:24,700 This is why we said earlier that differential Calculus has 400 00:22:24,700 --> 00:22:28,360 been referred to as 'the study of 0 over 0'. 401 00:22:28,360 --> 00:22:31,950 You see, the whole point is, let's go back now to our basic 402 00:22:31,950 --> 00:22:36,190 definition of average rate of change of 'f', 'f of x1' plus 403 00:22:36,190 --> 00:22:40,970 'delta x' minus 'f of x1' over 'delta x'. 404 00:22:40,970 --> 00:22:43,430 I guess when you're writing in black, you have to use white 405 00:22:43,430 --> 00:22:46,150 for emphasis, so we'll use this. 406 00:22:46,150 --> 00:22:49,800 Notice that every time you let 'delta x' equal zero, you are 407 00:22:49,800 --> 00:22:51,230 going to wind up with what? 408 00:22:51,230 --> 00:22:55,790 'f of x1' minus 'f of x1' , which is 0, over 'delta x', 409 00:22:55,790 --> 00:22:56,720 which is 0. 410 00:22:56,720 --> 00:22:59,330 In other words, you are going to get 0 over 0. 411 00:23:02,850 --> 00:23:07,260 If you believe that all this means is to replace 'delta x' 412 00:23:07,260 --> 00:23:11,070 by 0, you see, and the point is, this is not the definition 413 00:23:11,070 --> 00:23:14,130 of this, and this is why you don't wind up with this 414 00:23:14,130 --> 00:23:15,436 particular thing. 415 00:23:15,436 --> 00:23:19,100 Well I've spent enough time, I think, trying to give a 416 00:23:19,100 --> 00:23:22,950 picture as to what limits mean intuitively, now what I'd like 417 00:23:22,950 --> 00:23:26,340 to do is to devote the remainder of this lecture, 418 00:23:26,340 --> 00:23:30,320 plus our next one, to investigating, more 419 00:23:30,320 --> 00:23:35,470 stringently, what limit means in a way that is unambiguous. 420 00:23:35,470 --> 00:23:37,650 And you know, it's like everything else in this world. 421 00:23:37,650 --> 00:23:41,400 When you want to get something that is ironclad, you don't 422 00:23:41,400 --> 00:23:43,150 get something for nothing. 423 00:23:43,150 --> 00:23:48,560 To get an ironclad expression that doesn't lend itself to 424 00:23:48,560 --> 00:23:52,020 misinterpretations usually involves a rigorous enough 425 00:23:52,020 --> 00:23:55,050 language that scares off the uninitiated. 426 00:23:55,050 --> 00:23:57,020 Sort of like a legal document. 427 00:23:57,020 --> 00:23:59,600 The chances are if you're not a trained lawyer, and the 428 00:23:59,600 --> 00:24:02,450 document is simple enough for you to understand, it must 429 00:24:02,450 --> 00:24:04,540 have thousands of loopholes in it. 430 00:24:04,540 --> 00:24:07,740 But that same rigorous language that makes it almost 431 00:24:07,740 --> 00:24:11,320 impossible for the layman to understand what's happening, 432 00:24:11,320 --> 00:24:14,660 is precisely the language that the lawyer needs to make the 433 00:24:14,660 --> 00:24:17,950 document well defined to him, hopefully. 434 00:24:17,950 --> 00:24:21,410 Well with that as a background, let's now try to 435 00:24:21,410 --> 00:24:24,560 give a more rigorous definition of limit. 436 00:24:24,560 --> 00:24:27,380 And let's try to do it in such a way that not only will the 437 00:24:27,380 --> 00:24:31,540 definition be rigorous, but it will agree, to the best of our 438 00:24:31,540 --> 00:24:34,470 ability, with what we intuitively believe a limit 439 00:24:34,470 --> 00:24:35,660 should mean. 440 00:24:35,660 --> 00:24:38,710 Now what I want to do here is scare you a little bit, and 441 00:24:38,710 --> 00:24:40,980 the reason I want to scare you a little bit is, I want to 442 00:24:40,980 --> 00:24:44,650 show you, once and for all, how in many cases, what 443 00:24:44,650 --> 00:24:48,120 appears to be ominous mathematical notation, is 444 00:24:48,120 --> 00:24:51,270 simply a rigorous way of saying something which was 445 00:24:51,270 --> 00:24:53,840 quite intuitive pictorially to do. 446 00:24:53,840 --> 00:24:56,630 The mathematical definition of limit is the following. 447 00:24:56,630 --> 00:25:00,330 The limit of 'f of x' as 'x' approaches 'a' equals 'l' 448 00:25:00,330 --> 00:25:04,660 means, and get this, given epsilon greater than 0, we can 449 00:25:04,660 --> 00:25:08,070 find delta greater than 0, such that when the absolute 450 00:25:08,070 --> 00:25:11,370 value of 'x minus a' is less than delta, but greater than 451 00:25:11,370 --> 00:25:15,750 0, then the absolute value of 'f of x' minus 'l' is less 452 00:25:15,750 --> 00:25:17,210 than epsilon. 453 00:25:17,210 --> 00:25:19,850 Now doesn't that kind of turn you on? 454 00:25:19,850 --> 00:25:21,460 See, this is the whole problem. 455 00:25:21,460 --> 00:25:25,650 I will now cut you in sort of on the trade secret that makes 456 00:25:25,650 --> 00:25:28,510 this thing very, very readable. 457 00:25:28,510 --> 00:25:32,170 First of all, to say that you want to get arbitrarily close 458 00:25:32,170 --> 00:25:36,210 in value, you want 'f of x' to be arbitrarily close to 'l', 459 00:25:36,210 --> 00:25:38,020 another way of saying that is what? 460 00:25:38,020 --> 00:25:41,460 You specify any tolerance limit whatsoever, which we 461 00:25:41,460 --> 00:25:45,775 call epsilon, and I can make 'f of x' get to within epsilon 462 00:25:45,775 --> 00:25:49,870 of 'l' just by choosing 'x' sufficiently close to 'a'. 463 00:25:49,870 --> 00:25:53,640 And by the way again, in terms of geometry, that's all the 464 00:25:53,640 --> 00:25:55,780 absolute value of a difference means. 465 00:25:55,780 --> 00:25:59,330 The absolute value of 'x minus a' simply means what? 466 00:25:59,330 --> 00:26:02,030 The distance between 'x' and 'a'. 467 00:26:02,030 --> 00:26:03,750 All this says is what? 468 00:26:03,750 --> 00:26:09,980 I can make 'f of x' fall within epsilon of 'l' by 469 00:26:09,980 --> 00:26:15,050 finding a delta such that 'x' is within delta of 'a', and by 470 00:26:15,050 --> 00:26:20,090 the way, this little tidbit over here is simply a fancy 471 00:26:20,090 --> 00:26:23,600 way of saying that you're not letting 'x' equal 'a'. 472 00:26:23,600 --> 00:26:25,860 You see, the only way the absolute value of a number can 473 00:26:25,860 --> 00:26:28,770 be 0 is if the number itself is 0. 474 00:26:28,770 --> 00:26:32,270 The only way this could be 0 is if 'x' equals 'a', so by 475 00:26:32,270 --> 00:26:35,360 saying this little tidbit over here, that this is greater 476 00:26:35,360 --> 00:26:38,800 than 0, that's our fancy way of saying that we are not 477 00:26:38,800 --> 00:26:42,050 going to allow 'x' to equal to 'a'. 478 00:26:42,050 --> 00:26:45,430 Now again, whereas this may sound a little bit simpler 479 00:26:45,430 --> 00:26:48,740 than the original definition, let's see how much easier the 480 00:26:48,740 --> 00:26:50,060 picture makes it. 481 00:26:50,060 --> 00:26:52,970 Let's go to a graph of 'y' equals 'f of x'. 482 00:26:56,860 --> 00:27:00,390 And in my diagram here, here is 'a', and here is 'l', and 483 00:27:00,390 --> 00:27:04,070 what I'm saying is, I intuitively suspect. 484 00:27:04,070 --> 00:27:05,530 Let me write down what I suspect. 485 00:27:05,530 --> 00:27:09,620 I suspect that the limit of 'f of x', as 'x' approaches 'a', 486 00:27:09,620 --> 00:27:11,580 is 'l' in this case. 487 00:27:11,580 --> 00:27:14,900 Namely I suspect that as 'x' gets closer and closer to 'a', 488 00:27:14,900 --> 00:27:18,950 no matter which side you come in on, that's why we use 489 00:27:18,950 --> 00:27:19,790 absolute value. 490 00:27:19,790 --> 00:27:22,550 It makes no difference whether it's positive or negative. 491 00:27:22,550 --> 00:27:25,950 As 'x' gets in tighter and tighter on 'a', I suspect that 492 00:27:25,950 --> 00:27:30,290 I can make 'f of x' come arbitrarily equal to 'l'. 493 00:27:30,290 --> 00:27:33,260 In other words, if 'x' is close enough to 'a', 'f of x', 494 00:27:33,260 --> 00:27:35,310 it seems, will be close to 'l'. 495 00:27:35,310 --> 00:27:38,310 And by the way, to complete this problem geometrically, 496 00:27:38,310 --> 00:27:39,990 pick your epsilon. 497 00:27:39,990 --> 00:27:43,220 Let's say that epsilon is this 'y', and since epsilon is 498 00:27:43,220 --> 00:27:45,960 positive, this will be called 'l plus epsilon'. 499 00:27:45,960 --> 00:27:49,920 This will be called 'l minus epsilon'. 500 00:27:49,920 --> 00:27:53,010 Now what we do is we come over to the curve here. 501 00:27:56,460 --> 00:27:59,190 All right, hopefully these will look like horizontal 502 00:27:59,190 --> 00:28:00,295 lines to you. 503 00:28:00,295 --> 00:28:03,330 When I get down to here, I then project down vertical 504 00:28:03,330 --> 00:28:04,580 lines to meet the x-axis. 505 00:28:07,230 --> 00:28:10,140 And now all I'm saying, and I think this is fairly intuitive 506 00:28:10,140 --> 00:28:12,790 clear, is what? 507 00:28:12,790 --> 00:28:18,810 That if 'x' is this close to 'a', 'f of x' will be within 508 00:28:18,810 --> 00:28:20,140 these tolerance limits of 'l'. 509 00:28:20,140 --> 00:28:23,750 In other words, for any value of 'x' in here, see, for this 510 00:28:23,750 --> 00:28:27,960 neighborhood of 'a', every input has its output come 511 00:28:27,960 --> 00:28:30,610 within epsilon of 'l'. 512 00:28:30,610 --> 00:28:33,500 By the way, just one brief note here, something that 513 00:28:33,500 --> 00:28:36,410 we'll have to come back to in more detail-- in fact, I'll do 514 00:28:36,410 --> 00:28:38,290 that in our next example-- 515 00:28:38,290 --> 00:28:47,560 notice here how important it was that our curve was not 516 00:28:47,560 --> 00:28:49,160 single value, but one to one. 517 00:28:49,160 --> 00:28:52,180 In other words, suppose our curve had doubled back, say 518 00:28:52,180 --> 00:28:53,780 something like this. 519 00:28:53,780 --> 00:28:57,710 Notice then, when you try to project over to the curve from 520 00:28:57,710 --> 00:29:01,090 'l plus epsilon', you don't know whether to stop at this 521 00:29:01,090 --> 00:29:02,810 point or at this point. 522 00:29:02,810 --> 00:29:05,260 You see, in other words, notice that there is another 523 00:29:05,260 --> 00:29:09,690 point down here which has the same 'l plus epsilon' value. 524 00:29:09,690 --> 00:29:12,560 In other words, whenever our function is not one to one, 525 00:29:12,560 --> 00:29:14,490 there is going to be a problem, when we work 526 00:29:14,490 --> 00:29:18,470 analytically, of being able to solve algebraic equations, and 527 00:29:18,470 --> 00:29:21,620 trying to distinguish this point of intersection from 528 00:29:21,620 --> 00:29:22,440 this point. 529 00:29:22,440 --> 00:29:26,650 You see, if we made a mistake and bypass this point and came 530 00:29:26,650 --> 00:29:29,920 all the way over to here, and thought that this was our 531 00:29:29,920 --> 00:29:32,840 tolerance limit, we would be in a great deal of difficulty. 532 00:29:32,840 --> 00:29:37,680 Because you see, for example, for this value of 'x', its 'f 533 00:29:37,680 --> 00:29:42,720 of x' value projects up here, which is outside of the 534 00:29:42,720 --> 00:29:45,120 tolerance limits that are given. 535 00:29:45,120 --> 00:29:47,950 You see, many things which are self evident from the picture 536 00:29:47,950 --> 00:29:52,180 are not nearly so self evident analytically, to which we then 537 00:29:52,180 --> 00:29:55,850 raise the question, why use analysis at all? 538 00:29:55,850 --> 00:29:57,560 Let's just use pictures. 539 00:29:57,560 --> 00:30:00,030 And again the answer is, in this case, we can get away 540 00:30:00,030 --> 00:30:02,640 with it, but in more complicated situations where 541 00:30:02,640 --> 00:30:04,760 either we can't draw the diagram because it's too 542 00:30:04,760 --> 00:30:08,300 complicated, or because we have several variables, and we 543 00:30:08,300 --> 00:30:11,400 get into a dimension problem, the point is that sometimes we 544 00:30:11,400 --> 00:30:14,490 have no recourse other than the analysis. 545 00:30:14,490 --> 00:30:18,820 Let me give you an exercise that we'll do in terms of 546 00:30:18,820 --> 00:30:22,990 putting the geometry and the analysis side by side. 547 00:30:22,990 --> 00:30:27,750 Let's look at the expression the limit of 'x squared minus 548 00:30:27,750 --> 00:30:30,590 2x' as 'x' approaches three. 549 00:30:30,590 --> 00:30:33,900 Now of course, we don't want 'x' to equal three here, but 550 00:30:33,900 --> 00:30:37,840 let's cheat again, and see what this thing means. 551 00:30:37,840 --> 00:30:41,950 Our intuition tells us that when 'x' is three, 'x squared' 552 00:30:41,950 --> 00:30:44,940 is nine, '2x' is six. 553 00:30:44,940 --> 00:30:48,810 So nine minus six is three, and we would suspect that the 554 00:30:48,810 --> 00:30:51,600 limit here should be three. 555 00:30:51,600 --> 00:30:56,990 Now to see what this means, let's draw a little diagram. 556 00:30:56,990 --> 00:31:00,610 The graph 'y' equals 'x squared minus 2x' crosses the 557 00:31:00,610 --> 00:31:03,040 x-axis at 0 and two. 558 00:31:03,040 --> 00:31:07,140 It has its low point at the 0.1 comma minus one. 559 00:31:07,140 --> 00:31:10,620 We can sketch these curves, it's a parabola. 560 00:31:10,620 --> 00:31:21,520 And now what we're saying is, is it true that when 'x' gets 561 00:31:21,520 --> 00:31:24,890 very close to three, that 'f of x' gets 562 00:31:24,890 --> 00:31:26,790 very close to three? 563 00:31:26,790 --> 00:31:29,640 And the answer is, from the diagram, it appears very 564 00:31:29,640 --> 00:31:31,550 obvious that this is the case. 565 00:31:31,550 --> 00:31:34,840 Fact, here's that example of non-single value again. 566 00:31:34,840 --> 00:31:37,860 If you call this thing 'l', and you now pick tolerance 567 00:31:37,860 --> 00:31:41,310 limits which we'll call 'l plus epsilon' and 'l minus 568 00:31:41,310 --> 00:31:44,740 epsilon', and now you want to see what interval you need on 569 00:31:44,740 --> 00:31:49,800 the x-axis, notice that in this particular diagram, you 570 00:31:49,800 --> 00:31:54,600 will get two intervals, one of which surrounds the value 'x' 571 00:31:54,600 --> 00:31:57,350 equals minus one, and the other of which surrounds the 572 00:31:57,350 --> 00:31:59,620 'x' value 'x' equals three. 573 00:31:59,620 --> 00:32:03,400 In other words, both when 'x' is minus one and 'x' is three, 574 00:32:03,400 --> 00:32:07,690 'x squared minus 2x' will equal three. 575 00:32:07,690 --> 00:32:11,350 At any rate, leaving the diagram as an aid, let's see 576 00:32:11,350 --> 00:32:14,650 what our epsilon delta definition says, and how we 577 00:32:14,650 --> 00:32:17,380 handle the stuff algebraically, and how we 578 00:32:17,380 --> 00:32:20,370 correlate the algebra with the geometry. 579 00:32:22,870 --> 00:32:25,970 Given epsilon greater than 0, what must I do? 580 00:32:25,970 --> 00:32:29,860 I must find a delta greater than 0 such that, and I'm 581 00:32:29,860 --> 00:32:31,770 going to read this colloquially, I'll write it 582 00:32:31,770 --> 00:32:34,600 formally, but read it colloquially, such that 583 00:32:34,600 --> 00:32:39,950 whenever 'x' is within delta of three, but not equal to 584 00:32:39,950 --> 00:32:44,340 three, then 'x squared minus 2x' will be 585 00:32:44,340 --> 00:32:47,080 within epsilon of three. 586 00:32:47,080 --> 00:32:48,350 Now here's the whole point. 587 00:32:48,350 --> 00:32:52,360 We know, algebraically, how to handle absolute values. 588 00:32:52,360 --> 00:32:56,700 Namely, the absolute value of 'x squared minus 2x minus 589 00:32:56,700 --> 00:32:59,730 three' being less than epsilon, is the same as saying 590 00:32:59,730 --> 00:33:04,180 that 'x squared minus 2x minus three' itself must be between 591 00:33:04,180 --> 00:33:07,100 epsilon and minus epsilon. 592 00:33:07,100 --> 00:33:10,300 Now again, we rewrite things in fancy ways if we wish. 593 00:33:10,300 --> 00:33:11,850 There other ways of doing this. 594 00:33:11,850 --> 00:33:13,760 I call this completing the square. 595 00:33:13,760 --> 00:33:16,215 Well, people for 2,000 years have called this completing 596 00:33:16,215 --> 00:33:17,280 the square. 597 00:33:17,280 --> 00:33:21,630 Namely, I rewrite minus three as one minus four, so that 'x 598 00:33:21,630 --> 00:33:24,580 squared minus 2x plus one' can be factored as 599 00:33:24,580 --> 00:33:26,500 'x minus one squared'. 600 00:33:26,500 --> 00:33:28,750 I'm now down to this form here. 601 00:33:28,750 --> 00:33:32,360 Then I add four to all sides of my inequality, and have 602 00:33:32,360 --> 00:33:36,210 that 'x minus one squared' is between four plus epsilon and 603 00:33:36,210 --> 00:33:37,460 four minus epsilon. 604 00:33:37,460 --> 00:33:41,330 Some fairly elementary Algebra of inequalities here. 605 00:33:41,330 --> 00:33:42,660 Now here's the key point. 606 00:33:42,660 --> 00:33:46,060 Remembering my diagram, I said to you, how do we know whether 607 00:33:46,060 --> 00:33:48,300 we're near three or near minus one? 608 00:33:48,300 --> 00:33:51,300 How do we distinguish, when we draw the lines to the curve, 609 00:33:51,300 --> 00:33:52,890 what neighborhood we want? 610 00:33:52,890 --> 00:33:55,510 And notice that all we're saying is that if 'x' is near 611 00:33:55,510 --> 00:33:58,560 three, that means what? 'x' is close to three. 612 00:33:58,560 --> 00:34:01,910 If 'x' is reasonably close to three, then 'x minus one' is 613 00:34:01,910 --> 00:34:06,190 going to be reasonably close to two, and hence be positive. 614 00:34:06,190 --> 00:34:08,920 Point is, that as long as you're dealing with positive 615 00:34:08,920 --> 00:34:11,940 numbers over here, see if epsilon is sufficiently small, 616 00:34:11,940 --> 00:34:14,070 these will then all be positive numbers. 617 00:34:14,070 --> 00:34:18,100 For positive numbers, if the squares obey a certain 618 00:34:18,100 --> 00:34:22,730 inequality, the square roots will obey the same inequality, 619 00:34:22,730 --> 00:34:26,090 as we have emphasized in one of our exercises. 620 00:34:26,090 --> 00:34:31,139 In other words, from here, we can now say this, and from 621 00:34:31,139 --> 00:34:34,409 this, we can now conclude, that if you want 'x squared 622 00:34:34,409 --> 00:34:39,710 minus 2x' to be within epsilon of three, 'x' itself is going 623 00:34:39,710 --> 00:34:42,850 to have to be between 'one plus the square root of 'four 624 00:34:42,850 --> 00:34:45,699 plus epsilon'', and 'one plus the square root 625 00:34:45,699 --> 00:34:47,190 of 'four minus epsilon''. 626 00:34:47,190 --> 00:34:50,030 And by the way, this is quite rigorous, but I don't think it 627 00:34:50,030 --> 00:34:50,639 turns you on. 628 00:34:50,639 --> 00:34:52,989 I don't think there's any picture that you associate 629 00:34:52,989 --> 00:34:56,159 with this, and so what I thought I'd like to do for our 630 00:34:56,159 --> 00:35:01,960 next little thing, is to come back to our diagram here, and 631 00:35:01,960 --> 00:35:04,800 show you what this really means. 632 00:35:04,800 --> 00:35:09,850 Namely, notice that if epsilon is a small positive number, 633 00:35:09,850 --> 00:35:11,060 let's take a look at this again. 634 00:35:11,060 --> 00:35:14,260 If epsilon is a small positive number, this is 635 00:35:14,260 --> 00:35:16,420 slightly less than four. 636 00:35:16,420 --> 00:35:19,280 Therefore, the square root is slightly less than two, 637 00:35:19,280 --> 00:35:22,560 therefore, this number will be slightly less than three. 638 00:35:22,560 --> 00:35:25,300 On the other hand, 'four plus epsilon' is slightly more than 639 00:35:25,300 --> 00:35:29,080 four, so its square root will be slightly more than two, and 640 00:35:29,080 --> 00:35:32,320 therefore, one plus that square root will be slightly 641 00:35:32,320 --> 00:35:33,840 more than three. 642 00:35:33,840 --> 00:35:36,890 And what these two numbers, as abstract as they look like, 643 00:35:36,890 --> 00:35:39,660 correspond to, is nothing more than this. 644 00:35:39,660 --> 00:35:43,870 Coming back to our diagram and choosing an epsilon, and 645 00:35:43,870 --> 00:35:47,980 coming over to the curve like this, and projecting down like 646 00:35:47,980 --> 00:35:51,990 this, all we're saying is, see, what are we saying 647 00:35:51,990 --> 00:35:52,970 pictorially? 648 00:35:52,970 --> 00:35:58,130 That for 'f of x' to be within epsilon of 'l', 'x' itself has 649 00:35:58,130 --> 00:36:00,260 to be in this range here. 650 00:36:00,260 --> 00:36:05,950 And all this says is, this very simple point to compute, 651 00:36:05,950 --> 00:36:09,740 just by projecting down like this, its rigorous name would 652 00:36:09,740 --> 00:36:13,560 be 'one plus the square root of 'four minus epsilon''. 653 00:36:13,560 --> 00:36:16,420 That's this number over here. 654 00:36:16,420 --> 00:36:20,460 And the furthest point, namely this point here, which again, 655 00:36:20,460 --> 00:36:24,490 in terms of the picture is very easy to find, that point 656 00:36:24,490 --> 00:36:28,220 corresponds to one plus the square root 657 00:36:28,220 --> 00:36:31,660 of 'four plus epsilon'. 658 00:36:31,660 --> 00:36:35,390 And again notice, in terms of the Algebra, I need no 659 00:36:35,390 --> 00:36:36,870 recourse to a picture. 660 00:36:36,870 --> 00:36:40,460 But in terms of the picture, I get a heck of a good feeling 661 00:36:40,460 --> 00:36:43,960 as to what these abstract symbols are telling me. 662 00:36:43,960 --> 00:36:48,830 Well let's continue on with the solution of this exercise. 663 00:36:48,830 --> 00:36:51,760 You see, we didn't want to find where 'x' was, we wanted 664 00:36:51,760 --> 00:36:54,360 to see where 'x minus three' had to be. 665 00:36:54,360 --> 00:36:56,400 In other words, we're trying to find the delta. 666 00:36:56,400 --> 00:37:00,290 We know that 'x minus three' is between these 667 00:37:00,290 --> 00:37:01,830 two extremes here. 668 00:37:01,830 --> 00:37:04,460 Consequently, and here's a place we have to be a little 669 00:37:04,460 --> 00:37:07,840 bit careful here, this of course, is a positive number, 670 00:37:07,840 --> 00:37:10,450 because this is more than two. 671 00:37:10,450 --> 00:37:13,200 This, on the other hand, is a negative number, because you 672 00:37:13,200 --> 00:37:16,640 see 'four minus epsilon' is less than four, so its square 673 00:37:16,640 --> 00:37:18,940 root is less than two. 674 00:37:18,940 --> 00:37:21,490 So if I subtract two from-- it's a negative number-- 675 00:37:21,490 --> 00:37:24,580 and since delta has to be positive, if this thing is 676 00:37:24,580 --> 00:37:28,880 negative, two minus the square root of 'four minus epsilon' 677 00:37:28,880 --> 00:37:30,730 will be positive. 678 00:37:30,730 --> 00:37:35,040 So what I do is, to solve this problem, is I simply choose 679 00:37:35,040 --> 00:37:39,530 delta to be the minimum of these two numbers. 680 00:37:39,530 --> 00:37:42,330 And then by definition, what do I have? 681 00:37:42,330 --> 00:37:45,280 That when the absolute value of 'x minus three' is less 682 00:37:45,280 --> 00:37:48,920 than delta, and at the same time, this is crucial, 683 00:37:48,920 --> 00:37:49,770 greater than 0. 684 00:37:49,770 --> 00:37:52,600 In other words, we never let 'x' equal three. 685 00:37:52,600 --> 00:37:56,570 For this choice of delta, by how we worked backwards, this 686 00:37:56,570 --> 00:37:59,920 will come out to be less than epsilon. 687 00:37:59,920 --> 00:38:03,200 And that is exactly what you mean by our formal definition 688 00:38:03,200 --> 00:38:06,140 of saying that the limit of 'x squared minus 2x' as 'x' 689 00:38:06,140 --> 00:38:08,980 approaches three, in this case, equals three. 690 00:38:08,980 --> 00:38:12,000 For someone who wants more concrete evidence, I think all 691 00:38:12,000 --> 00:38:17,860 one has to do is, for example, take a number like 'one plus 692 00:38:17,860 --> 00:38:20,480 the square root of 'four plus epsilon'', that is the widest 693 00:38:20,480 --> 00:38:24,240 point on our interval, and actually plug that in for 'x', 694 00:38:24,240 --> 00:38:27,830 and go through this computation of squaring 'one 695 00:38:27,830 --> 00:38:30,190 plus the square root of 'four plus epsilon''. 696 00:38:30,190 --> 00:38:32,890 Subtract twice, 'one plus the square root 697 00:38:32,890 --> 00:38:34,390 of 'four plus epsilon''. 698 00:38:34,390 --> 00:38:37,720 Carry out that computation and you get what? 699 00:38:37,720 --> 00:38:39,750 'Three plus epsilon'. 700 00:38:39,750 --> 00:38:43,150 That using the upper extreme, you wind up with what? 701 00:38:43,150 --> 00:38:48,570 In excess of three by epsilon, which by the way is exactly 702 00:38:48,570 --> 00:38:49,740 what's supposed to happen. 703 00:38:49,740 --> 00:38:52,560 In fact, I hope this doesn't give you an eye sore, I will 704 00:38:52,560 --> 00:38:56,180 pull down the top board again to show you also what these 705 00:38:56,180 --> 00:38:59,180 two distances mean, and what this means. 706 00:38:59,180 --> 00:39:02,390 You see, all these two numbers mean is the following, and 707 00:39:02,390 --> 00:39:04,910 I'll pull this down just far enough to see it. 708 00:39:04,910 --> 00:39:10,140 That these two numbers were the widths from this end point 709 00:39:10,140 --> 00:39:14,550 to three on the one extreme, and from this end point to 710 00:39:14,550 --> 00:39:17,270 three on the other extreme. 711 00:39:17,270 --> 00:39:21,170 In other words, those two numbers that we took the 712 00:39:21,170 --> 00:39:25,250 minimum of were the half-width intervals 713 00:39:25,250 --> 00:39:26,880 that surrounded three. 714 00:39:26,880 --> 00:39:29,550 You see the problem, as we mentioned before is, is that 715 00:39:29,550 --> 00:39:33,330 even though epsilon and minus epsilon are symmetric with 716 00:39:33,330 --> 00:39:37,340 respect to 'l', when you translate over to the curve, 717 00:39:37,340 --> 00:39:40,260 because this curve is not a straight line when you project 718 00:39:40,260 --> 00:39:43,610 down, these two widths will not be equal. 719 00:39:46,750 --> 00:39:49,030 By picking the minimum of these two widths, you 720 00:39:49,030 --> 00:39:52,740 guarantee that you have locked yourself inside 721 00:39:52,740 --> 00:39:54,000 the required area. 722 00:39:54,000 --> 00:39:57,710 You see, that's all this particular thing means. 723 00:39:57,710 --> 00:40:01,550 Now the trouble is that one might now get the idea that 724 00:40:01,550 --> 00:40:05,120 there are no shorter ways of doing the same problem. 725 00:40:05,120 --> 00:40:07,580 You see, I showed you a rigorous way. 726 00:40:07,580 --> 00:40:10,770 There was no law that said we had to find the biggest delta. 727 00:40:10,770 --> 00:40:14,485 In other words, if delta equals, a half will work, in 728 00:40:14,485 --> 00:40:16,670 other words, if you're within a half of three, something's 729 00:40:16,670 --> 00:40:19,060 going to happen, then certainly any smaller number 730 00:40:19,060 --> 00:40:21,410 will work just as well. 731 00:40:21,410 --> 00:40:23,020 In other words, I can get estimates. 732 00:40:23,020 --> 00:40:24,790 Let me show you what I mean by that. 733 00:40:24,790 --> 00:40:28,660 Another way of tackling how to make 'x squared minus 2x minus 734 00:40:28,660 --> 00:40:32,580 three' smaller than epsilon is to use our properties of 735 00:40:32,580 --> 00:40:35,180 absolute values, and to factor this. 736 00:40:35,180 --> 00:40:37,880 In other words, we know that 'x squared minus 2x minus 737 00:40:37,880 --> 00:40:41,040 three' is 'x minus three' times 'x plus one'. 738 00:40:41,040 --> 00:40:44,560 We know that the absolute value of a product is a 739 00:40:44,560 --> 00:40:47,360 product of the absolute values, so these two things 740 00:40:47,360 --> 00:40:54,050 here are synonyms. Then we know that near 'x' equals 741 00:40:54,050 --> 00:40:56,410 three, which we're interested in here, 'x plus 742 00:40:56,410 --> 00:40:59,550 one' is near four. 743 00:40:59,550 --> 00:41:02,700 Well what we mean more rigorously is this, choose an 744 00:41:02,700 --> 00:41:06,430 epsilon, and if the absolute value of 'x minus three' is 745 00:41:06,430 --> 00:41:09,560 less than epsilon, in other words, if 'x minus three' is 746 00:41:09,560 --> 00:41:13,040 less than epsilon but greater than minus epsilon, then by 747 00:41:13,040 --> 00:41:16,400 adding on four to all three sides of the inequality here, 748 00:41:16,400 --> 00:41:20,390 we see that 'x plus one' is less than 'four plus epsilon', 749 00:41:20,390 --> 00:41:22,140 but greater than 'four minus epsilon'. 750 00:41:31,650 --> 00:41:34,550 If the size of epsilon is less than one, then certainly this 751 00:41:34,550 --> 00:41:38,490 is less than five, and this is greater than three. 752 00:41:38,490 --> 00:41:41,200 Now of course, somebody may say to us, but what if epsilon 753 00:41:41,200 --> 00:41:42,070 is more than one? 754 00:41:42,070 --> 00:41:47,770 Well obviously, if a guy says make this thing within 10, and 755 00:41:47,770 --> 00:41:50,190 I can make it within one, certainly within 756 00:41:50,190 --> 00:41:51,600 10 is within one. 757 00:41:51,600 --> 00:41:53,870 I can always pick the smaller number. 758 00:41:53,870 --> 00:41:57,140 It's like that nonsense of the fellow asking for an ice cream 759 00:41:57,140 --> 00:41:59,240 cone, and the waitress said what flavor, and he said 760 00:41:59,240 --> 00:42:00,460 anything except chocolate. 761 00:42:00,460 --> 00:42:01,880 And she said, I'm all out of chocolate, will you take 762 00:42:01,880 --> 00:42:03,540 anything except vanilla? 763 00:42:03,540 --> 00:42:05,600 The idea here is that if somebody says make this less 764 00:42:05,600 --> 00:42:09,240 than 15, if you've made it less than 10, in particular, 765 00:42:09,240 --> 00:42:10,840 you've made it less than 15. 766 00:42:10,840 --> 00:42:13,140 And so all we do over here, you see, is 767 00:42:13,140 --> 00:42:13,890 something like this. 768 00:42:13,890 --> 00:42:17,780 We say look it, if we want to make this number here very 769 00:42:17,780 --> 00:42:22,170 small, we know that this number here is less than five, 770 00:42:22,170 --> 00:42:25,710 we know that 'epsilon over five' is a positive number if 771 00:42:25,710 --> 00:42:29,640 epsilon is small, so why don't we just pick the absolute 772 00:42:29,640 --> 00:42:34,100 value of 'x minus three' to be less than 'epsilon over five?' 773 00:42:34,100 --> 00:42:36,260 In fact, that's what this delta will be in that case. 774 00:42:36,260 --> 00:42:38,590 In other words, if the absolute value of x minus 775 00:42:38,590 --> 00:42:48,820 three is within 'epsilon over five', if 'x' is within 776 00:42:48,820 --> 00:42:51,620 'epsilon over five' of three, notice what 777 00:42:51,620 --> 00:42:53,040 this product becomes. 778 00:42:53,040 --> 00:42:56,160 This is less than 'epsilon over five', this we know from 779 00:42:56,160 --> 00:43:00,640 before is less than five, and therefore, this product is 780 00:43:00,640 --> 00:43:02,080 less than epsilon. 781 00:43:02,080 --> 00:43:05,640 In other words, we have exhibited the delta such that 782 00:43:05,640 --> 00:43:09,040 when this happens, the absolute value of what we want 783 00:43:09,040 --> 00:43:10,860 to make less than epsilon indeed 784 00:43:10,860 --> 00:43:13,090 becomes less than epsilon. 785 00:43:13,090 --> 00:43:18,210 Now because I recognize that this is a hard topic, you'll 786 00:43:18,210 --> 00:43:21,290 notice in the reading assignments that these 787 00:43:21,290 --> 00:43:24,120 problems are covered in great detail. 788 00:43:24,120 --> 00:43:28,220 Everything that I've said in the lesson so far is repeated 789 00:43:28,220 --> 00:43:31,920 in great computational depth in our learning exercises. 790 00:43:31,920 --> 00:43:35,350 The point is that even if I can make the epsilons and 791 00:43:35,350 --> 00:43:39,230 deltas seem a little bit more meaningful for you than by the 792 00:43:39,230 --> 00:43:43,400 formal definition, notice that it's simple only in comparison 793 00:43:43,400 --> 00:43:44,870 to what we had before. 794 00:43:44,870 --> 00:43:46,940 But it's still very, very difficult. 795 00:43:46,940 --> 00:43:51,370 The beauty of Calculus is that in many cases, we do not have 796 00:43:51,370 --> 00:43:54,500 to know what delta looks like for a given epsilon. 797 00:43:54,500 --> 00:43:57,400 What we shall do there therefore, in our next 798 00:43:57,400 --> 00:44:01,430 lecture, is to develop recipes that will allow us to get the 799 00:44:01,430 --> 00:44:04,970 answers to these limit problems without having to go 800 00:44:04,970 --> 00:44:09,580 through this genuinely difficult problem of finding a 801 00:44:09,580 --> 00:44:12,550 given delta to match a given epsilon. 802 00:44:12,550 --> 00:44:15,940 Though we must admit, in real life, in many cases where 803 00:44:15,940 --> 00:44:19,460 you're making approximations, we will want to know what the 804 00:44:19,460 --> 00:44:20,810 tolerance limits are. 805 00:44:20,810 --> 00:44:24,140 I am not belittling the epsilon delta approach. 806 00:44:24,140 --> 00:44:27,790 All I'm saying is that in certain problems, you do not 807 00:44:27,790 --> 00:44:31,360 need the epsilon delta approach to get nice results, 808 00:44:31,360 --> 00:44:34,560 and that will be the topic of our next lecture. 809 00:44:34,560 --> 00:44:36,090 So until next time, goodbye. 810 00:44:39,210 --> 00:44:41,740 GUEST SPEAKER: Funding for the publication of this video was 811 00:44:41,740 --> 00:44:46,460 provided by the Gabriella and Paul Rosenbaum Foundation. 812 00:44:46,460 --> 00:44:50,630 Help OCW continue to provide free and open access to MIT 813 00:44:50,630 --> 00:44:54,840 courses by making a donation at ocw.mit.edu/donate.