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JOEL LEWIS: Hi, welcome
back to recitation.

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In lecture, Professor Jerison
was teaching about

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L'Hospital's Rule, and one
thing that he mentioned

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several times was 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, in order for the first

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limit to equal the
second limit.

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So I have a--

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but he didn't give you 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, see

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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, spend a couple minutes

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working on that, come
back, and we

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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 when we try to

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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, and you know,

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between minus 1 and 1, x
is going to infinity.

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So the top is going to infinity
and the bottom is

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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 try to apply L'Hospital's

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Rule, so that's fine.

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So it's an indeterminate ratio,
we can look at applying

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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 as x goes to

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infinity of our expression x
plus cosine x divided by x is

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equal-- and I'm going to put
a little question mark here

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because what L'Hospital's Rule
says is that it's equal

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provided, 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 to

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infinity of the ratio of the
derivative, so the derivative

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of the top divided by the
derivative of the bottom.

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So in this case that's 1 minus
sine x is the derivative of

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the top and 1 is 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 because of

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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, and

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it oscillates back and
forth in between.

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So in particular, because it's
oscillating, it's not

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approaching any value.

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So this limit doesn't
exist. It's not

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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 says that

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this equality is really wrong.

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OK?

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What it means is that
L'Hospital's

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Rule 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, we can't use

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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 though we

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don't have this tool.

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So if we try and solve this
limit without L'Hospital's

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Rule, so we want to look at
the limit as x goes to

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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 is getting

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big and what's cosine x doing?

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Well cosine x is behaving like a
constant but wigglier right,

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we could say.

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So cosine x is always between
minus 1 and 1, but, you know,

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it's oscillating, but it's in
some bounded interval there,

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whereas x and x are both going
to infinity at exactly the

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same speed, so this suggests
a manipulation that sort of

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allows us to quantify
that explicitly.

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And so one thing that we can
do is we can divide top and

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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, split it into

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two fractions.

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So we have--

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I'm just going to re-write it
one more time --the limit of x

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goes to infinity of x plus
cosine x divided by x is equal

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to the limit as x goes
to infinity of 1 plus

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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

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

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constant that's going to 1 --and
as x goes to infinity,

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cosine x again we can think
of it like a constant but

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wigglier, x is going to
infinity, so cosine x over x

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is going to 0 as x
goes to infinity.

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So I guess I wouldn't call this
plugging in, but it's,

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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, and that

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I'll say every time I use it,
which is L'Hospital's Rule

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says that the limit of the ratio
is equal to the limit of

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the ratio of the derivative
provided the

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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 equal to each

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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.

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