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JOEL LEWIS: Hi, welcome
back to recitation.

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In lecture,
Professor Jerison was

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teaching about L'Hospital's
Rule, and one thing

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that he mentioned
several times was

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that when you apply
L'Hospital's Rule,

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it's really important that the
second limit exists in order

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for L'Hospital's
Rule to be true,

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in order for the first limit
to equal the second limit.

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So I have a-- but
he didn't give you

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any examples of
what can go wrong.

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So I have here a limit
for you to work on.

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So what I'd like you to do
is try L'Hospital's Rule,

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see what happens, and then try
to solve it a different way not

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using L'Hospital's Rule.

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So why don't you
pause the video,

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spend a couple minutes
working on that, come back,

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and we can work on it together.

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Welcome back.

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Hopefully you've have some
fun working on this limit.

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Let's go through it
and see what happens

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when we try to apply
L'Hospital's Rule to it.

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So the first thing to notice
is that as x goes to infinity,

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x plus cosine x-- well,
cosine x is small,

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and you know, between minus 1
and 1-- x is going to infinity.

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So the top is going to
infinity and the bottom

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is going to infinity.

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So this is an infinity over
infinity indeterminate form.

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So it is a context
in which we can

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try to apply L'Hospital's
Rule, so that's fine.

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So it's an
indeterminate ratio, we

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can look at applying
L'Hospital's Rule.

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So let's try to do that.

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So L'Hospital's Rule
says that the limit

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as x goes to infinity
of our expression x plus

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cosine x divided by
x is equal-- and I'm

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going to put a little question
mark here because what

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L'Hospital's Rule says is
that it's equal provided--

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it's equal to what I'm
going to write over here,

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provided that the
second limit exists.

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So what goes over here?

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Well it's the limit as x goes
to the same place, as x goes

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to infinity, of the
ratio of the derivative,

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so the derivative of the top
divided by the derivative

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of the bottom.

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So in this case
that's 1 minus sine

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x is the derivative
of the top and 1 is

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the derivative of the bottom.

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OK, now let's look
at this limit.

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What happens?

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Well, OK the over 1
part is irrelevant.

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The 1 minus part, OK, who cares?

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The only thing that's changing
in this limit is the sine x.

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So as x goes to infinity what
this does is it oscillates,

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just like sine x does.

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I mean, it's offset
and flipped upside down

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because of this 1 minus.

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So, in particular, you
know, sometimes it's near 1,

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sometimes it's near negative
1, sometimes it's near 0,

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and it oscillates back
and forth in between.

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So in particular,
because it's oscillating,

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it's not approaching any value.

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So this limit doesn't exist.

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It's not equal to
any real number.

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It's not equal to infinity.

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It's not equal to
minus infinity.

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It does not exist.

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OK, so the statement
of L'Hospital's theorem

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says that this equality
is really wrong.

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OK?

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What it means is that
L'Hospital's Rule

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tells you nothing.

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You don't learn anything here.

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So if we want to compute
this first limit,

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we can't use L'Hospital's Rule.

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We have to come up with some
better way to do it, OK.

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So that's what's going wrong.

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Now let's see about
doing this even

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though we don't have this tool.

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So if we try and
solve this limit

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without L'Hospital's Rule, so
we want to look at the limit

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as x goes to infinity of x
plus cosine x divided by x.

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Well, think about what's
important in this limit.

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As x is getting big, well
x is getting big and x

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is getting big and
what's cosine x doing?

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Well cosine x is behaving
like a constant but wigglier,

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right, we could say.

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So cosine x is always
between minus 1 and 1,

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but, you know, it's
oscillating, but it's

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in some bounded interval
there, whereas x and x are both

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going to infinity at
exactly the same speed,

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so this suggests a manipulation
that sort of allows

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us to quantify that explicitly.

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And so one thing
that we can do is

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we can divide top and
bottom of this limit by x.

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So if we do that, or
equivalently we can just

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divide this x into
the fraction above,

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split it into two fractions.

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So we have-- I'm just going
to re-write it one more

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time-- the limit as x goes
to infinity of x plus cosine

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x divided by x is
equal to the limit

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as x goes to infinity of
1 plus cosine x over x.

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All right, so what's
happening here?

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Well, all I've done is I've
divided that x in-- and now you

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see that this 1, that was
from the x over x, so that's

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a constant that's going to 1.

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And as x goes to
infinity, cosine x-- again

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we can think of it like
a constant but wigglier--

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x is going to
infinity, so cosine

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x over x is going to 0
as x goes to infinity.

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So I guess I wouldn't
call this plugging in,

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but it's, you know, sort
of just like plugging in.

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So this part is going to 0,
so we're left with just 1.

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OK, so we can evaluate
this limit fairly easily.

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It just took a couple steps.

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But we can't evaluate it
with L'Hospital's Rule.

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So this is why you have that
extra thing in the statement

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of the theorem that Professor
Jerison kept saying,

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and that I'll say
every time I use it,

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which is L'Hospital's Rule says
that the limit of the ratio

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is equal to the limit of
the ratio of the derivative

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provided the second
limit exists.

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OK, so here's an example where
the second limit didn't exist

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and so we can't say that
the two are actually

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equal to each other, and we need
some other tools to evaluate

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the limit in question.

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So I'll end there.