JOEL LEWIS: Hi, welcome back to recitation.
In lecture, Professor Jerison was teaching about
L'Hospital's Rule, and one thing that he mentioned
several times was that when you apply L'Hospital's Rule,
it's really important that the second limit exists in order
for L'Hospital's Rule to be true, in order for the first
limit to equal the second limit.
So I have a--
but he didn't give you any examples of what can go wrong.
So I have here a limit for you to work on.
So what I'd like you to do is try L'Hospital's Rule, see
what happens, and then try to solve it a different way not
using L'Hospital's Rule.
So why don't you pause the video, spend a couple minutes
working on that, come back, and we
can work on it together.
Welcome back.
Hopefully you've have some fun working on this limit.
Let's go through it and see what happens when we try to
apply L'Hospital's Rule to it.
So the first thing to notice is that as x goes to infinity,
x plus cosine x-- well, cosine x is small, and you know,
between minus 1 and 1, x is going to infinity.
So the top is going to infinity and the bottom is
going to infinity.
So this is an infinity over infinity indeterminate form.
So it is a context in which we can try to apply L'Hospital's
Rule, so that's fine.
So it's an indeterminate ratio, we can look at applying
L'Hospital's Rule.
So let's try to do that.
So L'Hospital's Rule says that the limit as x goes to
infinity of our expression x plus cosine x divided by x is
equal-- and I'm going to put a little question mark here
because what L'Hospital's Rule says is that it's equal
provided, it's equal to what I'm going to write over here
provided that the second limit exists.
So what goes over here?
Well it's the limit as x goes to the same place as x goes to
infinity of the ratio of the derivative, so the derivative
of the top divided by the derivative of the bottom.
So in this case that's 1 minus sine x is the derivative of
the top and 1 is the derivative of the bottom.
OK, now let's look at this limit.
What happens?
Well, OK the over 1 part is irrelevant.
The 1 minus part, OK, who cares?
The only thing that's changing in this limit is the sine x.
So as x goes to infinity what this does is it oscillates,
just like sine x does.
I mean, it's offset and flipped upside down because of
this 1 minus.
So, in particular, you know, sometimes it's near 1,
sometimes it's near negative 1, sometimes it's near 0, and
it oscillates back and forth in between.
So in particular, because it's oscillating, it's not
approaching any value.
So this limit doesn't exist. It's not
equal to any real number.
It's not equal to infinity.
It's not equal to minus infinity.
It does not exist.
OK, so the statement of L'Hospital's theorem says that
this equality is really wrong.
OK?
What it means is that L'Hospital's
Rule tells you nothing.
You don't learn anything here.
So if we want to compute this first limit, we can't use
L'Hospital's Rule.
We have to come up with some better way to do it, OK.
So that's what's going wrong.
Now let's see about doing this even though we
don't have this tool.
So if we try and solve this limit without L'Hospital's
Rule, so we want to look at the limit as x goes to
infinity of x plus cosine x divided by x.
Well, think about what's important in this limit.
As x is getting big, well x is getting big and x is getting
big and what's cosine x doing?
Well cosine x is behaving like a constant but wigglier right,
we could say.
So cosine x is always between minus 1 and 1, but, you know,
it's oscillating, but it's in some bounded interval there,
whereas x and x are both going to infinity at exactly the
same speed, so this suggests a manipulation that sort of
allows us to quantify that explicitly.
And so one thing that we can do is we can divide top and
bottom of this limit by x.
So if we do that, or equivalently we can just
divide this x into the fraction above, split it into
two fractions.
So we have--
I'm just going to re-write it one more time --the limit of x
goes to infinity of x plus cosine x divided by x is equal
to the limit as x goes to infinity of 1 plus
cosine x over x.
All right, so what's happening here?
Well all I've done is I've divided that x in-- and now
you see that this 1, that was from the x over x, so that's a
constant that's going to 1 --and as x goes to infinity,
cosine x again we can think of it like a constant but
wigglier, x is going to infinity, so cosine x over x
is going to 0 as x goes to infinity.
So I guess I wouldn't call this plugging in, but it's,
you know, sort of just like plugging in.
So this part is going to 0, so we're left with just 1.
OK, so we can evaluate this limit fairly easily.
It just took a couple steps.
But we can't evaluate it with L'Hospital's Rule.
So this is why you have that extra thing in the statement
of the theorem that Professor Jerison kept saying, and that
I'll say every time I use it, which is L'Hospital's Rule
says that the limit of the ratio is equal to the limit of
the ratio of the derivative provided the
second limit exists.
OK, so here's an example where the second limit didn't exist
and so we can't say that the two are actually equal to each
other, and we need some other tools to evaluate
the limit in question.
So I'll end there.