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PROFESSOR: So we're through
with techniques of integration,

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which is really the most
technical thing that

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we're going to be doing.
 

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And now we're just clearing
up a few loose ends

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about calculus.
 

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And the one we're going to talk
about today will allow us

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to deal with infinity.
 

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And it's what's known
as L'Hopital's Rule.

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Here's L'Hopital's Rule.
 

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And that's what we're
going to do today.

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L'Hopital's Rule it's also
known as L'Hospital's Rule.

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That's the same name, since the
circumflex is what you put

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in French to omit the s.
 

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So it's the same thing,
and it's still pronounced

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L'Hopital, even if
it's got an s in it.

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Alright, so that's the
first thing you need

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to know about it.
 

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And what this method does
is, it's a convenient

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way to calculate limits
including some new ones.

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So it'll be convenient
for the old ones.

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There are going to be some new
ones and, as an example, you

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can calculate x ln x as
x goes to infinity.

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You could, whoops, that's
not a very interesting

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one, let's try x goes to
0 from the positive side.

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And you can calculate, for
example, x e^ - x, as

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x goes to infinity.
 

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And, well, maybe I should
include a few others.

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Maybe something like ln x /
x as x goes to infinity.

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So these are some examples of
things which, in fact, if you

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plug into your calculator,
you can see what's

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happening with these.
 

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But if you want to understand
them systematically, it's much

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better to have this tool
of L'Hopital's Rule.

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And certainly there isn't
a proof just based on a

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calculation in a calculator.
 

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So now here's the idea.
 

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I'll illustrate the idea
first with an example.

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And then we'll make
it systematic.

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And then we're going
to generalize it.

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We'll make it much more, so
when it includes these new

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limits, there are some little
pieces of trickiness that

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you have to understand.
 

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So, let's just take an example
that you could have done in the

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very first unit of this class.
 

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The limit as x goes to 1
of x ^ 10 - 1 / x ^2 - 1.

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So that's a limit that
we could've handled.

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And the thing that's
interesting, I mean, if you

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like this is in this category
that we mentioned at the

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beginning of the course
of interesting limits.

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What's interesting about it is
that if you do this silly

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thing, which is just plug in x
= 1, at x = 1 you're

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going to get 0 / 0.
 

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And that's what we call
an indeterminate form.

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It's just unclear what it is.
 

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From that plugging, in
you just can't get it.

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Now, on the other hand, there's
a trick for doing this.

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And this is the trick that
we did at the beginning

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of the class.
 

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And the idea is I can divide
in the numerator and

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denominator by x - 1.
 

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So this limit is unchanged, if
I try to cancel the hidden

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factor x - 1 in the
numerator and denominator.

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Now, we can actually carry out
these ratios of polynomials and

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calculate them by long
division in algebra.

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That's very, very long.
 

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We want to do this
with calculus.

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And we already have.
 

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We already know that this
ratio is what's called

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a difference quotient.
 

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And then in the limit, it
tends to the derivative

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of this function.
 

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So the idea is that this is
actually equal to, in the

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limit, now let's just
study one piece of it.

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So if I have a function f ( x),
which is x ^ 10 - 1, and the

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value at 1 happens to be equal
to 0, then this expression that

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we have, which is in disguise,
this is in disguise the

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difference quotient, tends
to, as x goes to 1, the

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derivative, which is f' ( 1).
 

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That's what it is.
 

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So we know what the numerator
goes to, and similarly

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we'll know what the
denominator goes to.

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But what is that?
 

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Well, f ' (x) = 10x ^ 9.
 

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So we know what the answer is.
 

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In the numerator it's 10x ^ 9.
 

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In the denominator, it's
going to be 2x, that's the

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derivative of x ^2 - 1.
 

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And then were going to have
to evaluate that at x = 1.

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And so it's going to
be 10/2, which is 5.

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So the answer is 5.
 

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And it's pretty easy to get
from our techniques and

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knowledge of derivatives,
using this rather

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clever algebraic trick.
 

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This business of
dividing by x - 1.

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What I want to do now is
just carry this method

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out systematically.
 

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And that's going to give
us the approach to what's

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known as L'Hopital's Rule.
 

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What my main subject for today.
 

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So here's the idea.
 

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Suppose we're considering, in
general, a limit as x goes to

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some number a of
f ( x) / g ( x).

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And suppose it's the bad
case where we can't decide.

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So it's in determinate.
f ( a) = g ( a) = 0.

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So it would be 0 / 0.
 

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Now we're just going to do
exactly the same thing

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we did over here.
 

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Namely, we're going to divide a
numerator and denominator, and

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we're going to repeat
that argument.

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So we have here f ( x) / x - a.
 

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And g (x) / x - a also.
 

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I haven't changed anything yet.
 

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And now I'm going to write
it in this suggestive form.

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Namely, I'm going to take
separately the limit in the

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numerator and the denominator.
 

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And I'm going to make
one more shift.

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So I'm going to take the
limit, as x goes to a in the

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numerator, but I'm going to
write it as f ( x)

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- f ( a) / x - a.
 

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So that's the way I'm going to
write the numerator, and I've

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got to draw a much
longer line here.

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So why am I allowed to do that?
 

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That's because f (a) = 0.
 

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So I didn't change this
numerator of the numerator any

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by subtracting. f ( a) = 0.
 

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And I'll do the same thing
to the denominator.

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Again, g ( a) = 0,
so this is OK.

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And lo and behold, I know
what these limits are.

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This is f ' (a) / g '(a).
 

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So that's it.
 

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That's the technique and
this evaluates the limit.

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And it's not so difficult.
 

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The formula's pretty
straightforward here.

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And it works, provided
that g ' (a) is not 0.

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Yeah, question.
 

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: The question is, is
there a more intuitive way of

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understanding this procedure.
 

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And I think the short
answer is that there are

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other, similar, ways.
 

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I don't consider them
to be more intuitive.

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I will be mentioning one of
them, which is the idea of

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linearization, which goes back
to what we did in Unit 2.

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I think it's very important
to understand all of these,

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more or less, at once.
 

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But I wouldn't claim that any
of these methods is a more

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intuitive one than the other.
 

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But basically what's happening
is, we're looking at the linear

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approximation to f, at a.
 

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And the linear
approximation to g at a.

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That's what underlies this.
 

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So now I get to formulate for
you L'Hopital's Rule at least

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in what I would call the easy
version or, if you

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like, Version 1.
 

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So here's L'Hopital's Rule.
 

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Version 1.
 

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It's not going to be quite the
same as what we just did.

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It's going to be
much, much better.

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And more useful.
 

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And what is going to take
care of is this problem that

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the denominator is not 0.
 

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So now here's what
we're going to do.

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We're going to say that it
turns out that the limit a x

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00:10:35 --> 00:10:43
goes to a of f(x) / g ( x) =
the limit as x goes to a

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00:10:43 --> 00:10:48
to a f '(x) / g' (x).
 

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Now, that looks practically the
same as what we said before.

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And I have to make sure that
you understand when it works.

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So it works provided
this is one of these

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undefined expressions.
 

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In other words, f (
a) = g ( a) = 0.

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00:11:06 --> 00:11:11
So we have a 0 / 0
expression, indeterminant.

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And, also, we need
one more assumption.

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And the right-hand side, the
right-hand limit exists.

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Now, this is practically
the same thing as what

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I said over here.
 

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Namely, I took the ratio
of these functions, x

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^ 10 - 1 and x ^2 - 1.
 

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I took their derivatives,
which is what I did

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right here, right.
 

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I just differentiated them
and I took the ratio.

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This is way easier than the
quotient rule, and is nothing

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like the quotient rule.
 

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Don't think quotient rule.
 

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Don't think quotient rule.
 

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So we differentiate
the numerator and

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denominator separately.
 

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And then I take the limit as
x goes to 1 and I get 5.

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So that's what I'm
claiming over here.

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I take these functions, I
replace them with this ratio of

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derivatives, and then I take
the limit instead, over here.

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And it turned out that the
functions got much simpler

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when I differentiated them.
 

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I started with this messy
object and I got this much

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easier object that I
could easily evaluate.

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So that's the big game
that's happening here.

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It works, if this limit makes
sense and this limit exists.

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Now, notice I didn't claim
that g, that the denominator,

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had to be non-0.
 

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So that's what's going
to help us a little

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bit in a few examples.
 

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00:12:43 --> 00:12:45
So let me give you a couple
of examples and then

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00:12:45 --> 00:12:46
we'll go further.
 

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Now, this is only Version 1.
 

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But first we have to understand
how this one works.

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00:12:51 --> 00:12:56
So here's another example.
 

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00:12:56 --> 00:13:06
Take the limit as x goes
to 0, of sin 5x / sin 2x.

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00:13:06 --> 00:13:11
This is another kind of example
of a limit that we discussed in

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00:13:11 --> 00:13:12
the first part of the course.
 

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00:13:12 --> 00:13:13
Unfortunately, now
we're reviewing stuff.

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00:13:13 --> 00:13:15
So this should reinforce
what you did there.

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00:13:15 --> 00:13:20
This will be an easier way
of thinking about it.

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00:13:20 --> 00:13:25
So by L'Hopital's Rule,
so here's the step.

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00:13:25 --> 00:13:27
We're going to take
one of these steps.

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00:13:27 --> 00:13:31
This is the limit, as x goes to
1, of the derivatives here.

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So that's 5 cos 5x / 2 cos 2x.
 

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The limit was 1 over
there, but now it's 0.

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00:13:46 --> 00:13:48
a is 0 in this case.
 

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00:13:48 --> 00:13:51
This is the number a.
 

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00:13:51 --> 00:13:54
Thank you.
 

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So the limit as x goes to 0
is the same as the limit

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00:13:58 --> 00:14:01
of the derivatives.
 

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00:14:01 --> 00:14:02
And that's easy to evaluate.
 

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00:14:02 --> 00:14:04
Cosine of 0 = 1, right.
 

232
00:14:04 --> 00:14:10
This is equal to 5 (cos (5* 0).
 

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And that's a
multiplication sign.

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00:14:12 --> 00:14:14
Maybe I should just
write this as 0.

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Divided by 2 cos 0.
 

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00:14:17 --> 00:14:24
But you know that that's 5/2.
 

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So this is how L'Hopital's
method works.

238
00:14:27 --> 00:14:33
It's pretty painless.
 

239
00:14:33 --> 00:14:36
I'm going to give you another
example, which shows that it

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00:14:36 --> 00:14:45
works a little better than the
method that I started out with.

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00:14:45 --> 00:14:48
Here's what happens if we
consider the function

242
00:14:48 --> 00:14:55
cos x - 1 / x ^2.
 

243
00:14:55 --> 00:14:57
That was a little
harder to deal with.

244
00:14:57 --> 00:15:03
And again, this is one of these
0 / 0 things near x = 0.

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00:15:03 --> 00:15:11
As x tends to 0, this goes to
an indeterminate form here.

246
00:15:11 --> 00:15:14
Now, according to our method,
this is equivalent to, now I'm

247
00:15:14 --> 00:15:16
going to use this little wiggle
because I don't want to

248
00:15:16 --> 00:15:20
write limit, limit, limit,
limit a million times.

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00:15:20 --> 00:15:22
So I'm going to use a
little wiggle here.

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00:15:22 --> 00:15:26
So as x goes to 0, this is
going to behave the same

251
00:15:26 --> 00:15:30
way as differentiating
numerator and denominator.

252
00:15:30 --> 00:15:33
So again this is going to be
- sin x in the numerator.

253
00:15:33 --> 00:15:42
In the denominator,
it's going to be 2x.

254
00:15:42 --> 00:15:47
Now, notice that we
still haven't won yet.

255
00:15:47 --> 00:15:51
Because this is still
of 0 / 0 type.

256
00:15:51 --> 00:15:54
When you plug in x =
0 you still get 0.

257
00:15:54 --> 00:15:57
But that doesn't
damage the method.

258
00:15:57 --> 00:16:00
That doesn't make
the method fail.

259
00:16:00 --> 00:16:10
This 0 / 0, we can apply
L'Hopital's Rule a second time.

260
00:16:10 --> 00:16:13
And as x goes to 0 this is
the same thing as, again,

261
00:16:13 --> 00:16:14
differentiating the
numerator and denominator.

262
00:16:14 --> 00:16:19
So here I get - cos x in
the numerator, and I get

263
00:16:19 --> 00:16:22
2 in the denominator.
 

264
00:16:22 --> 00:16:25
Again this is way easier
than differentiating

265
00:16:25 --> 00:16:26
ratios of functions.
 

266
00:16:26 --> 00:16:28
We're only differentiating
the numerator and the

267
00:16:28 --> 00:16:33
denominator separately.
 

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00:16:33 --> 00:16:35
And now this is the end.
 

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00:16:35 --> 00:16:48
As x goes to 0, this is -
cos 0 / 2, which is - 1/2.

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00:16:48 --> 00:16:53
Now, the justification for this
comes only when you win in

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00:16:53 --> 00:16:56
the end and get the limit.
 

272
00:16:56 --> 00:16:58
Because what the theorem says
is that if one of these

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00:16:58 --> 00:17:01
limits exists, then the
preceding one exists.

274
00:17:01 --> 00:17:03
And once the preceding
one exists, then the

275
00:17:03 --> 00:17:03
one before it exists.
 

276
00:17:03 --> 00:17:09
So once we know that this one
exists, that works backwards.

277
00:17:09 --> 00:17:11
It applies to the preceding
limit, which then applies

278
00:17:11 --> 00:17:15
to the very first one.
 

279
00:17:15 --> 00:17:18
And the logical structure here
is a little subtle, which is

280
00:17:18 --> 00:17:21
that if the right side exists,
then the left side

281
00:17:21 --> 00:17:25
will also exist.
 

282
00:17:25 --> 00:17:26
Yeah, question.
 

283
00:17:26 --> 00:17:32
STUDENT: [INAUDIBLE]
 

284
00:17:32 --> 00:17:34
PROFESSOR: Why does the
right-hand limit have to exist,

285
00:17:34 --> 00:17:37
isn't it just the derivative
that has to exist?

286
00:17:37 --> 00:17:38
No.
 

287
00:17:38 --> 00:17:40
The derivative of the
numerator has to exist.

288
00:17:40 --> 00:17:42
The derivative of the
denominator has to exist.

289
00:17:42 --> 00:17:45
And this limit has to exist.
 

290
00:17:45 --> 00:17:48
What doesn't have to exist, by
the way, I never said that

291
00:17:48 --> 00:17:50
f prime of a has to exist.
 

292
00:17:50 --> 00:17:53
In fact, it's much,
much more subtle.

293
00:17:53 --> 00:17:56
I'm not claiming that f ' (a)
exists, because in order to

294
00:17:56 --> 00:18:01
evaluate this limit, f
' (a) need not exist.

295
00:18:01 --> 00:18:04
What has to happen is that
nearby, for x not equal to

296
00:18:04 --> 00:18:06
a, these things exist.
 

297
00:18:06 --> 00:18:09
And then the limit
has to exist.

298
00:18:09 --> 00:18:12
So there's no requirements
that the limits exist.

299
00:18:12 --> 00:18:14
In fact, that's exactly going
to be the point when we

300
00:18:14 --> 00:18:16
evaluate these limits here.
 

301
00:18:16 --> 00:18:22
Is we don't have to evaluate
it right at the end.

302
00:18:22 --> 00:18:26
STUDENT: [INAUDIBLE]
 

303
00:18:26 --> 00:18:30
PROFESSOR: So the question that
you're asking is, why is this

304
00:18:30 --> 00:18:31
the hypothesis of the theorem?
 

305
00:18:31 --> 00:18:34
In other words, why
does this work?

306
00:18:34 --> 00:18:37
Well, the answer is that this
is a theorem that's true.

307
00:18:37 --> 00:18:40
If you drop this hypothesis,
it's totally false.

308
00:18:40 --> 00:18:42
And if you don't have this
hypothesis, you can't use

309
00:18:42 --> 00:18:44
the theorem and you will
get the wrong answer.

310
00:18:44 --> 00:18:48
I mean, it's hard to express
it any further than that.

311
00:18:48 --> 00:18:52
So look, in many cases
we tell you formulas.

312
00:18:52 --> 00:18:55
And in many cases it's so
obvious when they're true

313
00:18:55 --> 00:18:59
that we don't have to
worry about what we say.

314
00:18:59 --> 00:19:01
And indeed, there's
something implicit here.

315
00:19:01 --> 00:19:04
I'm saying well, you know, if I
wrote this symbol down, it must

316
00:19:04 --> 00:19:06
mean that the thing exists.
 

317
00:19:06 --> 00:19:08
So that's a subtle point.
 

318
00:19:08 --> 00:19:11
But what I'm emphasizing is
that you don't need to know in

319
00:19:11 --> 00:19:13
advance that this one exists.
 

320
00:19:13 --> 00:19:18
You do need to know in advance
that that one exists.

321
00:19:18 --> 00:19:19
Essentially, yeah.
 

322
00:19:19 --> 00:19:24
So that's the direction
that it goes.

323
00:19:24 --> 00:19:28
You can't get away with not
having this exist and still

324
00:19:28 --> 00:19:37
have the statement be true.
 

325
00:19:37 --> 00:19:38
Alright, another question.
 

326
00:19:38 --> 00:19:39
Thank you.
 

327
00:19:39 --> 00:19:47
STUDENT: [INAUDIBLE]
 

328
00:19:47 --> 00:19:52
PROFESSOR: So I'm getting
a little ahead of myself,

329
00:19:52 --> 00:19:54
but let me just say.
 

330
00:19:54 --> 00:19:59
In these situations here,
when x is going to 0 and

331
00:19:59 --> 00:20:00
x is going to infinity.
 

332
00:20:00 --> 00:20:02
For instance, here when x
goes to 0, the logarithm

333
00:20:02 --> 00:20:06
is undefined at x = 0.
 

334
00:20:06 --> 00:20:08
Nevertheless, this
theorem applies.

335
00:20:08 --> 00:20:10
And we'll be able to use it.
 

336
00:20:10 --> 00:20:12
Over here, as x goes to
infinity, neither of these --

337
00:20:12 --> 00:20:15
well, actually, come to think
of it, e^ -x, if you like,

338
00:20:15 --> 00:20:17
it's equal to 0 at infinity.
 

339
00:20:17 --> 00:20:21
If you want to say
that it has a value.

340
00:20:21 --> 00:20:25
But in fact, these expressions
don't necessarily have values.

341
00:20:25 --> 00:20:27
At the ends.
 

342
00:20:27 --> 00:20:33
And nevertheless, the
theorem applies.

343
00:20:33 --> 00:20:34
I mean, it can exist.
 

344
00:20:34 --> 00:20:36
It's perfectly OK
for it to exist.

345
00:20:36 --> 00:20:37
It's no problem.
 

346
00:20:37 --> 00:20:39
It just doesn't need to exists.
 

347
00:20:39 --> 00:20:45
It isn't forced to exist.
 

348
00:20:45 --> 00:20:50
So here's a calculation
which we just did.

349
00:20:50 --> 00:20:51
And we evaluated this.
 

350
00:20:51 --> 00:20:57
Now, I want to make a
comparison with the

351
00:20:57 --> 00:21:06
method of approximation.
 

352
00:21:06 --> 00:21:11
In the method of
approximations, this Example 2,

353
00:21:11 --> 00:21:15
which was the example with the
sine function, we would use the

354
00:21:15 --> 00:21:16
following property.
 

355
00:21:16 --> 00:21:19
We would use sin u
is approximately u.

356
00:21:19 --> 00:21:22
We would use that
linear approximation.

357
00:21:22 --> 00:21:29
And then what we would have
here is that sin 5x / sin 2x

358
00:21:29 --> 00:21:35
is approximately 5x / 2x,
which is of course 5/2.

359
00:21:35 --> 00:21:38
And this is true when u is
approximately 0, and this is

360
00:21:38 --> 00:21:45
true certainly as x goes to 0,
it's going to be a valid limit.

361
00:21:45 --> 00:21:50
So that's very similar
to Example 2.

362
00:21:50 --> 00:21:54
In Example 3, we managed to
look at this expression

363
00:21:54 --> 00:21:59
cos x - 1 / x ^2.
 

364
00:21:59 --> 00:22:03
And for this one, you have to
remember the approximation near

365
00:22:03 --> 00:22:07
x = 0 to the cosine function.
 

366
00:22:07 --> 00:22:15
And that's 1 - x ^2 / 2.
 

367
00:22:15 --> 00:22:18
So that was the approximation,
the quadratic approximation

368
00:22:18 --> 00:22:20
to the cosine function.
 

369
00:22:20 --> 00:22:22
And now, sure enough,
this simplifies.

370
00:22:22 --> 00:22:32
This becomes - x ^2 / 2
/ x ^2, which is - 1/2.

371
00:22:32 --> 00:22:34
So we get the same answer,
which is a good thing.

372
00:22:34 --> 00:22:36
Because both of these
methods are valid.

373
00:22:36 --> 00:22:39
They're consistent.
 

374
00:22:39 --> 00:22:42
You can see that neither
of them is particularly

375
00:22:42 --> 00:22:42
a lot longer.
 

376
00:22:42 --> 00:22:45
You may have trouble
remembering this property.

377
00:22:45 --> 00:22:51
But in fact it's something
that you can easily derive.

378
00:22:51 --> 00:22:54
And, indeed, it's related to
the second derivative of the

379
00:22:54 --> 00:22:56
cosine, as is this
calculation here.

380
00:22:56 --> 00:23:04
They're almost the same amount
of numerical content to them.

381
00:23:04 --> 00:23:12
So now what I'd like to do is
explain to you why L'Hopital's

382
00:23:12 --> 00:23:14
Rule works better
in some cases.

383
00:23:14 --> 00:23:20
And the real value that it
has is in handling these

384
00:23:20 --> 00:23:25
other more exotic limits.
 

385
00:23:25 --> 00:23:33
So now we're going to do
L'Hopital's Rule over again.

386
00:23:33 --> 00:23:35
And I'll handle
these functions.

387
00:23:35 --> 00:23:40
But I'll have to rewrite them,
but we'll just do that.

388
00:23:40 --> 00:23:42
So here's the property.
 

389
00:23:42 --> 00:23:48
That the limit as x goes to a
of f ( x) / g (x) is equal to

390
00:23:48 --> 00:23:54
the limit as x goes to a
of f ' ( x) / g '(x).

391
00:23:54 --> 00:23:55
That's the property.
 

392
00:23:55 --> 00:23:57
And this is what we'll
always be using.

393
00:23:57 --> 00:23:59
Very convenient thing.
 

394
00:23:59 --> 00:24:04
And remember it was
true provided that

395
00:24:04 --> 00:24:11
f ( a) = g (a) = 0.
 

396
00:24:11 --> 00:24:23
And that the right-hand
side exists.

397
00:24:23 --> 00:24:25
But I claim that it
works better, and I'll

398
00:24:25 --> 00:24:26
get rid of these.
 

399
00:24:26 --> 00:24:30
But I'll write them again
to show you that it

400
00:24:30 --> 00:24:30
works for these.
 

401
00:24:30 --> 00:24:43
So there are other cases.
 

402
00:24:43 --> 00:24:47
And the other cases that
are allowed are this.

403
00:24:47 --> 00:24:51
First of all, as indicated
by what I just erased, you

404
00:24:51 --> 00:24:53
can allow a to be equal to
plus or minus infinity.

405
00:24:53 --> 00:24:57
It's also OK.
 

406
00:24:57 --> 00:25:04
So you can take the limit
going to the far ends

407
00:25:04 --> 00:25:05
of the universe.
 

408
00:25:05 --> 00:25:06
Both left and right.
 

409
00:25:06 --> 00:25:11
And then the other thing that
you can do is, you can allow

410
00:25:11 --> 00:25:19
f ( a) and g (a) to be
plus or minus infinity.

411
00:25:19 --> 00:25:22
Is OK.
 

412
00:25:22 --> 00:25:25
So now, the point is that
we can handle not just the

413
00:25:25 --> 00:25:33
0 / 0 case, but also the
infinity / infinity case.

414
00:25:33 --> 00:25:36
That's a very powerful
tool, and quite different

415
00:25:36 --> 00:25:42
from the other cases.
 

416
00:25:42 --> 00:25:49
And the third thing is that the
right-hand side doesn't really

417
00:25:49 --> 00:25:56
quite have to exist, in
the ordinary sense.

418
00:25:56 --> 00:26:00
Or, it could be plus
or minus infinity.

419
00:26:00 --> 00:26:01
That's also OK.
 

420
00:26:01 --> 00:26:04
That's still information.
 

421
00:26:04 --> 00:26:10
So if we can see where it
goes, then we're still good.

422
00:26:10 --> 00:26:13
If it goes to plus infinity, if
it goes to 0, if it goes to a

423
00:26:13 --> 00:26:15
finite number, if it goes to
minus infinity, all

424
00:26:15 --> 00:26:16
of that will be OK.
 

425
00:26:16 --> 00:26:19
It just if it oscillates
wildly that we'll be lost.

426
00:26:19 --> 00:26:27
And those calculations
we'll never encounter.

427
00:26:27 --> 00:26:29
So this basically handles
everything that you could

428
00:26:29 --> 00:26:32
possibly hope for.
 

429
00:26:32 --> 00:26:37
And it's a very
convenient process.

430
00:26:37 --> 00:26:40
So let me carry out
a few examples.

431
00:26:40 --> 00:26:43
And, let's see, I guess the
first one that I wanted

432
00:26:43 --> 00:26:47
to do was x ln x.
 

433
00:26:47 --> 00:26:49
So what example are we up to.
 

434
00:26:49 --> 00:26:57
Example 3, so Example
4 is coming up.

435
00:26:57 --> 00:26:59
Example 4, this is one of the
ones that I wrote at the

436
00:26:59 --> 00:27:06
beginning of the
lecture, x ln x.

437
00:27:06 --> 00:27:12
This one was on our
homework problem.

438
00:27:12 --> 00:27:17
In the limits of
some calculation.

439
00:27:17 --> 00:27:26
But so this one, you have to
look at it first to think

440
00:27:26 --> 00:27:27
about what it's doing.
 

441
00:27:27 --> 00:27:29
It's an indeterminate form,
but it sort of looks like

442
00:27:29 --> 00:27:30
it's the wrong type.
 

443
00:27:30 --> 00:27:33
So why is it in an
indeterminate form.

444
00:27:33 --> 00:27:38
This one goes to 0, and this
one goes to minus infinity.

445
00:27:38 --> 00:27:40
So, excuse me, this
is a product.

446
00:27:40 --> 00:27:46
It's 0 times minus infinity.
 

447
00:27:46 --> 00:27:48
So that's an indeterminate
form, because we don't know

448
00:27:48 --> 00:27:50
whether the 0 wins or the
infinity this could keep

449
00:27:50 --> 00:27:51
getting smaller and smaller and
smaller, and this could be

450
00:27:51 --> 00:27:52
getting bigger and
bigger bigger.

451
00:27:52 --> 00:27:55
The product could be
anything in between.

452
00:27:55 --> 00:27:57
We just don't know.
 

453
00:27:57 --> 00:28:01
So the first step is to
write this as a ratio

454
00:28:01 --> 00:28:06
of things, rather than
a product of things.

455
00:28:06 --> 00:28:09
And it turns out that the way
to do that is to use the

456
00:28:09 --> 00:28:14
logarithm in the numerator, and
the 1 / x in the denominator.

457
00:28:14 --> 00:28:18
So this is a choice
that I'm making here.

458
00:28:18 --> 00:28:23
Now, I've just converted it to
a limit of the type minus

459
00:28:23 --> 00:28:28
infinity divided by infinity.
 

460
00:28:28 --> 00:28:30
Because the numerator is going
to minus infinity as x goes to

461
00:28:30 --> 00:28:37
0 plus and the denominator 1 /
x is going to plus infinity.

462
00:28:37 --> 00:28:40
Again, there's a competitions,
but now it's one of the forms

463
00:28:40 --> 00:28:44
to which L'Hopital's
Rule applies.

464
00:28:44 --> 00:28:49
Now I'm just going to
apply L'Hopital's Rule.

465
00:28:49 --> 00:28:54
And what it says is that
I differentiate here.

466
00:28:54 --> 00:28:56
So I just differentiate a
numerator and denominator.

467
00:28:56 --> 00:28:58
Applying L'Hopital's
Rule is a breeze.

468
00:28:58 --> 00:29:03
You just differentiate,
differentiate.

469
00:29:03 --> 00:29:06
And now it just simplifies
and we're done.

470
00:29:06 --> 00:29:12
This is the limit as
x goes to 0 plus of,

471
00:29:12 --> 00:29:14
well, the x^2's cancel.
 

472
00:29:14 --> 00:29:20
This is the same as just
- x. x factors cancel.

473
00:29:20 --> 00:29:21
And so that's 0.
 

474
00:29:21 --> 00:29:24
The answer is that it's 0.
 

475
00:29:24 --> 00:29:30
So x goes to 0 faster then ln
n goes to minus infinity.

476
00:29:30 --> 00:29:36
This 0 was the winner.
 

477
00:29:36 --> 00:29:44
Something you can't necessarily
predict in advance.

478
00:29:44 --> 00:29:49
So let's do the other two
examples that I wrote down.

479
00:29:49 --> 00:29:53
I'm going to do them in
slightly more generality,

480
00:29:53 --> 00:29:58
because they're the most
fundamental rate properties

481
00:29:58 --> 00:30:01
that you're going to need to
know for the next section.

482
00:30:01 --> 00:30:03
Which is improper integrals.
 

483
00:30:03 --> 00:30:06
And also they're just very
important for physical

484
00:30:06 --> 00:30:10
math, and any other kind
of thing, basically.

485
00:30:10 --> 00:30:12
So here, let's just do these.
 

486
00:30:12 --> 00:30:16
So let's see, which one
do I want to do first.

487
00:30:16 --> 00:30:21
So I wrote down the limit of
x e^2 - x, but I'm going to

488
00:30:21 --> 00:30:22
make it even more general.
 

489
00:30:22 --> 00:30:26
I'm going to make it any
negative power here, where p

490
00:30:26 --> 00:30:30
is some positive constant.
 

491
00:30:30 --> 00:30:35
Now again, this is a product
of functions, not a quotient,

492
00:30:35 --> 00:30:37
a ratio, of functions.
 

493
00:30:37 --> 00:30:41
And so I need to rewrite it.
 

494
00:30:41 --> 00:30:50
I'm going to write
it as x / e ^ p x.

495
00:30:50 --> 00:30:52
And now I'm going to apply,
well, so it's of this

496
00:30:52 --> 00:30:58
form infinity / infinity.
 

497
00:30:58 --> 00:31:00
And now that's the same as
the limit as x goes to

498
00:31:00 --> 00:31:07
infinity of 1 / p e^ px.
 

499
00:31:07 --> 00:31:08
So where does that go?
 

500
00:31:08 --> 00:31:10
As x goes to infinity.
 

501
00:31:10 --> 00:31:12
Now we can decide.
 

502
00:31:12 --> 00:31:14
The 1 stays where it is.
 

503
00:31:14 --> 00:31:23
And this, as x goes to
infinity, goes to infinity.

504
00:31:23 --> 00:31:27
So the answer is 0.
 

505
00:31:27 --> 00:31:48
And the conclusion is that x
grows more slowly then e ^ px.

506
00:31:48 --> 00:31:49
As x goes to infinity.
 

507
00:31:49 --> 00:31:50
Remember, p is positive
here, of course.

508
00:31:50 --> 00:31:53
It's the increasing
exponentials.

509
00:31:53 --> 00:32:03
Not the decreasing ones.
 

510
00:32:03 --> 00:32:08
Let's do a variant of this.
 

511
00:32:08 --> 00:32:10
I'll do it the opposite way.
 

512
00:32:10 --> 00:32:13
So I'm going to call
this example 5 '.

513
00:32:13 --> 00:32:15
It really doesn't give us any
more information, but it

514
00:32:15 --> 00:32:18
gives you just a little
bit more practice.

515
00:32:18 --> 00:32:27
So suppose I look at things
the other way. e^ px

516
00:32:27 --> 00:32:29
divided by, say, x^ 100.
 

517
00:32:35 --> 00:32:42
Now, this is an infinity /
infinity example, again.

518
00:32:42 --> 00:32:44
And you can work out
what it's doing.

519
00:32:44 --> 00:32:48
But there are two ways
of thinking about this.

520
00:32:48 --> 00:32:49
There's the slow way
and the fast way.

521
00:32:49 --> 00:32:54
The slow way is to
differentiate this 100 times.

522
00:32:54 --> 00:32:55
That is, right?
 

523
00:32:55 --> 00:32:58
Apply L'Hopital's Rule over and
over and over and over again.

524
00:32:58 --> 00:33:00
All the way.
 

525
00:33:00 --> 00:33:02
It's clear that you
could do it, but it's

526
00:33:02 --> 00:33:03
kind of a nuisance.
 

527
00:33:03 --> 00:33:06
So there's a much
cleverer trick here.

528
00:33:06 --> 00:33:12
Which is to change this to (the
limit as x goes to infinity of

529
00:33:12 --> 00:33:19
the e ^ px / 100 / x) ^ 100.
 

530
00:33:25 --> 00:33:31
So if you do that, then we
just have one L'Hopital's

531
00:33:31 --> 00:33:34
Rule step here.
 

532
00:33:34 --> 00:33:44
And that one is that this is
the same as (x goes to infinity

533
00:33:44 --> 00:33:53
of, well it's p / 100 e^
p x / 100 / 1) ^ 100.

534
00:33:55 --> 00:34:02
That's our L'Hopital's step.
 

535
00:34:02 --> 00:34:07
And of course, that's
(infinity / 1 ) ^ 100.

536
00:34:09 --> 00:34:10
Which is infinity.
 

537
00:34:10 --> 00:34:14
Now, again I did this in a
slightly different way to

538
00:34:14 --> 00:34:16
show you that it works
with infinity as well.

539
00:34:16 --> 00:34:18
So that was this other case.
 

540
00:34:18 --> 00:34:21
The right-hand side can
exist, or it can be

541
00:34:21 --> 00:34:22
plus or minus infinity.
 

542
00:34:22 --> 00:34:25
And that applies to this limit.
 

543
00:34:25 --> 00:34:27
And therefore, to
the original limit.

544
00:34:27 --> 00:34:39
And the conclusion here is that
e ^ px, p > 0, grows faster

545
00:34:39 --> 00:34:46
than any power of x.
 

546
00:34:46 --> 00:34:49
I picked x ^ 100, but
obviously it didn't matter

547
00:34:49 --> 00:34:52
what power I picked.
 

548
00:34:52 --> 00:35:02
The exponents beat
all the powers.

549
00:35:02 --> 00:35:04
So we have one more of the
ones that I gave at the

550
00:35:04 --> 00:35:07
beginning to take care of.
 

551
00:35:07 --> 00:35:11
And that one is the logarithm.
 

552
00:35:11 --> 00:35:15
And its behavior at infinity.
 

553
00:35:15 --> 00:35:18
So I'll do a slightly
variant on that one, too.

554
00:35:18 --> 00:35:25
So we have Example 6, which is
ln x, and instead of dividing

555
00:35:25 --> 00:35:27
by x, I'm going to
divide by x^ 1/3.

556
00:35:27 --> 00:35:30
I could divide by any positive
power of x, we'll just

557
00:35:30 --> 00:35:32
do this example here.
 

558
00:35:32 --> 00:35:38
So now this, as x goes to
infinity, is of the form

559
00:35:38 --> 00:35:43
infinity / infinity.
 

560
00:35:43 --> 00:35:46
And so it's equivalent
to what happens when I

561
00:35:46 --> 00:35:49
differentiate numerator and
denominator separately.

562
00:35:49 --> 00:36:00
And that's 1 / x, and here
I have 1/3 x ^ - 2/3.

563
00:36:00 --> 00:36:03
1 / x, and then 1/3 x^ - 2/3.
 

564
00:36:03 --> 00:36:06
Now, when the dust settles here
and you get your exponents

565
00:36:06 --> 00:36:10
right, we have an x^ - 1, and
this is an x ^ + 2/3, and

566
00:36:10 --> 00:36:12
that's a 1/3 becomes a 3.
 

567
00:36:12 --> 00:36:19
So this is what it is.
 

568
00:36:19 --> 00:36:26
And that's equal to 3x ^ - 1/3.
 

569
00:36:26 --> 00:36:27
Which we can decide.
 

570
00:36:27 --> 00:36:30
It goes to 0.
 

571
00:36:30 --> 00:36:37
As x goes to infinity.
 

572
00:36:37 --> 00:36:49
And so the conclusion is that
ln x grows more slowly as x

573
00:36:49 --> 00:37:08
goes to infinity, than x ^ 1/3
or any positive power of x.

574
00:37:08 --> 00:37:15
So any x ^ p, p
positive, will work.

575
00:37:15 --> 00:37:17
So ln is really slow,
going to infinity.

576
00:37:17 --> 00:37:20
It's very, very gradual.
 

577
00:37:20 --> 00:37:21
Yeah, question.
 

578
00:37:21 --> 00:37:45
STUDENT: [INAUDIBLE]
 

579
00:37:45 --> 00:37:48
PROFESSOR: The question
is, how many hypotheses

580
00:37:48 --> 00:37:50
do you need here?
 

581
00:37:50 --> 00:37:57
So I said that, and I think
what you were asking is, if I

582
00:37:57 --> 00:38:02
have this hypothesis, can I
also have this hypothesis.

583
00:38:02 --> 00:38:04
That's OK.
 

584
00:38:04 --> 00:38:08
I can have this hypothesis
combined with this one.

585
00:38:08 --> 00:38:11
I need something about
f (a) and g ( a).

586
00:38:11 --> 00:38:14
I can't assume nothing
about f(a) and g(a).

587
00:38:14 --> 00:38:18
So in other words, I have to be
faced with either an infinity /

588
00:38:18 --> 00:38:24
infinity, or a 0 / 0 situation.
 

589
00:38:24 --> 00:38:26
So let's see.
 

590
00:38:26 --> 00:38:35
A rule applies in the 0 / 0,
or infinity / infinity case.

591
00:38:35 --> 00:38:40
These are the only two
cases that it applies in.

592
00:38:40 --> 00:38:45
And a can be anything.
 

593
00:38:45 --> 00:38:48
Including infinity.
 

594
00:38:48 --> 00:38:51
Plus or minus infinity.
 

595
00:38:51 --> 00:38:53
The rule applies in
these two cases.

596
00:38:53 --> 00:38:58
So in other words, this is
what f ( a) / g ( a) is.

597
00:38:58 --> 00:39:00
Either one of these.
 

598
00:39:00 --> 00:39:02
And in fact, it can
be plus or minus.

599
00:39:02 --> 00:39:06
STUDENT: [INAUDIBLE]
 

600
00:39:06 --> 00:39:10
PROFESSOR: And the right-hand
side has to be something.

601
00:39:10 --> 00:39:21
It has to be either finite
or plus or minus infinity.

602
00:39:21 --> 00:39:23
So you need something.
 

603
00:39:23 --> 00:39:26
You need a specific value of
a, you need to decide whether

604
00:39:26 --> 00:39:28
it's an indeterminate form.
 

605
00:39:28 --> 00:39:30
And you need the right-hand
limit to exist.

606
00:39:30 --> 00:39:33
It's not hard to impose this.
 

607
00:39:33 --> 00:39:36
Because when you look at the
right-hand side, you'll

608
00:39:36 --> 00:39:37
want to be calculating it.
 

609
00:39:37 --> 00:39:38
So you want to know what it is.
 

610
00:39:38 --> 00:39:47
So you'll never have problems
confirming this hypothesis.

611
00:39:47 --> 00:39:51
Alright.
 

612
00:39:51 --> 00:39:54
Let me give you one
more example here.

613
00:39:54 --> 00:39:56
Which is just
slightly trickier.

614
00:39:56 --> 00:40:15
Which involves, so here's
another indeterminate form.

615
00:40:15 --> 00:40:16
That's going to be 0 ^ 0.
 

616
00:40:16 --> 00:40:20
 


617
00:40:20 --> 00:40:22
So there are lots of these
things where you just

618
00:40:22 --> 00:40:23
don't know what to do.
 

619
00:40:23 --> 00:40:27
And they come out in
various different ways.

620
00:40:27 --> 00:40:32
The simplest example of this
is the limit as x goes to

621
00:40:32 --> 00:40:41
0 from above of x ^ x.
 

622
00:40:41 --> 00:40:45
In order to work out what's
happening with this one,

623
00:40:45 --> 00:40:47
we have to use a trick.
 

624
00:40:47 --> 00:40:52
And the trick is this
is a moving exponent.

625
00:40:52 --> 00:40:56
And so it's appropriate
to use base e.

626
00:40:56 --> 00:40:59
This is something that we did
way back in the first unit.

627
00:40:59 --> 00:41:06
So, since we have a
moving exponent, we're

628
00:41:06 --> 00:41:11
going to use base e.
 

629
00:41:11 --> 00:41:13
That's the good base to
use whenever you have

630
00:41:13 --> 00:41:15
a moving exponent.
 

631
00:41:15 --> 00:41:21
And so rewrite this as
x^ x = e ^ x ln x.

632
00:41:21 --> 00:41:24
And now, in order to figure out
what's happening, we really

633
00:41:24 --> 00:41:32
only have to know what's
going on with the exponent.

634
00:41:32 --> 00:41:34
So remember, actually
we already did this.

635
00:41:34 --> 00:41:36
But I'm going to do it
once more for you.

636
00:41:36 --> 00:41:39
This is ln x / (1 / x).
 

637
00:41:39 --> 00:41:44
And that's equivalent, as x
goes to 0, to using L'Hopital's

638
00:41:44 --> 00:41:51
Rule to 1 / x, and this is -
1 / x ^2, which is -

639
00:41:51 --> 00:41:54
x, which goes to 0.
 

640
00:41:54 --> 00:41:58
As x goes to 0.
 

641
00:41:58 --> 00:42:01
And so what we have here is
that this one is going to be

642
00:42:01 --> 00:42:06
equivalent to, well, it's
going to tend to what

643
00:42:06 --> 00:42:07
we got over here.
 

644
00:42:07 --> 00:42:10
It's e ^ 0.
 

645
00:42:10 --> 00:42:13
That exponent is what we want.
 

646
00:42:13 --> 00:42:18
As x goes to 0.
 

647
00:42:18 --> 00:42:27
So that's the answer This
limit happens to be 1.

648
00:42:27 --> 00:42:30
That's actually relatively easy
to do, given all of the power

649
00:42:30 --> 00:42:42
that we have at our hands.
 

650
00:42:42 --> 00:42:49
Now, let me give you
one more example.

651
00:42:49 --> 00:42:52
Suppose you're trying to
understand the limit

652
00:42:52 --> 00:42:59
of sin x / x ^2.
 

653
00:42:59 --> 00:43:06
If you apply L'Hopital's
Rule, as x goes to 0, you're

654
00:43:06 --> 00:43:11
going to get cos x / 2x.
 

655
00:43:11 --> 00:43:19
And if you apply L'Hopital's
Rule again, as x goes to 0,

656
00:43:19 --> 00:43:24
you're going to get
the - sin x / 2.

657
00:43:24 --> 00:43:35
And this, as x goes
to 0, goes to 0.

658
00:43:35 --> 00:43:39
On the other hand, if you look
at the linear approximation

659
00:43:39 --> 00:43:49
method, linear approximation
says that sin x is

660
00:43:49 --> 00:43:55
approximately x near 0.
 

661
00:43:55 --> 00:43:59
So that should be x / x ^2.
 

662
00:43:59 --> 00:44:04
Which is 1 / x, which
goes to infinity.

663
00:44:04 --> 00:44:08
As x goes to 0, at least from
one side, minus infinity

664
00:44:08 --> 00:44:13
to the other side.
 

665
00:44:13 --> 00:44:17
So there's something fishy
going on here, right?

666
00:44:17 --> 00:44:19
So this is fishy.
 

667
00:44:19 --> 00:44:21
Or maybe this is
fishy, I don't know.

668
00:44:21 --> 00:44:26
So, tell me what's wrong here.
 

669
00:44:26 --> 00:44:26
Yeah.
 

670
00:44:26 --> 00:44:37
STUDENT: [INAUDIBLE]
 

671
00:44:37 --> 00:44:38
PROFESSOR: OK.
 

672
00:44:38 --> 00:44:43
So the claim is that the second
application of L'Hopital's

673
00:44:43 --> 00:44:51
Rule, this one, is wrong.
 

674
00:44:51 --> 00:44:54
And that's correct.
 

675
00:44:54 --> 00:44:56
And this is where you
have to watch out,

676
00:44:56 --> 00:44:58
with L'Hopital's Rule.
 

677
00:44:58 --> 00:44:59
This is exactly where
you have to watch out.

678
00:44:59 --> 00:45:02
You have to apply the test.
 

679
00:45:02 --> 00:45:03
Here it's an
indeterminate form.

680
00:45:03 --> 00:45:08
It's 0 / 0 before I
applied the rule.

681
00:45:08 --> 00:45:10
But in order to apply the
rule the second time, it

682
00:45:10 --> 00:45:12
still has to be 0 / 0.
 

683
00:45:12 --> 00:45:14
But this one isn't.
 

684
00:45:14 --> 00:45:19
This one is 1 / 0.
 

685
00:45:19 --> 00:45:20
It's no longer an
indeterminate form.

686
00:45:20 --> 00:45:22
It's actually infinite.
 

687
00:45:22 --> 00:45:25
Either plus or minus, depending
on the sign of the denominator.

688
00:45:25 --> 00:45:27
Which is just what
this answer is.

689
00:45:27 --> 00:45:30
So the linear
approximation is safe.

690
00:45:30 --> 00:45:35
And we just applied
L'Hopital's Rule wrong.

691
00:45:35 --> 00:45:55
So the moral of the story here
is look before you L'Hop.

692
00:45:55 --> 00:45:58
Alright.
 

693
00:45:58 --> 00:46:09
Now, let me say one more thing.
 

694
00:46:09 --> 00:46:22
I need to pile it on just
a little bit, sorry.

695
00:46:22 --> 00:46:36
So don't use it as a crutch.
 

696
00:46:36 --> 00:46:39
We don't want to just get
ourselves so weak, after being

697
00:46:39 --> 00:46:42
in the hospital for all this
time, that we can't

698
00:46:42 --> 00:46:55
use, I'm sorry.
 

699
00:46:55 --> 00:47:00
So remember that you shouldn't
have lost your senses.

700
00:47:00 --> 00:47:09
If you have something
like this, so we'll

701
00:47:09 --> 00:47:12
do this one here.
 

702
00:47:12 --> 00:47:15
Suppose you're trying to
understand what this does

703
00:47:15 --> 00:47:18
as x goes to infinity.
 

704
00:47:18 --> 00:47:25
Now, you could L'Hopital's Rule
five times, or four times.

705
00:47:25 --> 00:47:30
And get the answer here.
 

706
00:47:30 --> 00:47:33
But really, you should realize
that the main terms are sitting

707
00:47:33 --> 00:47:34
there right in front of you.
 

708
00:47:34 --> 00:47:36
And that there's some
algebra that you can

709
00:47:36 --> 00:47:38
do to simplify this.
 

710
00:47:38 --> 00:47:45
Namely, it's the same as
1 + 2 / x + 1 / x^ 5.

711
00:47:45 --> 00:47:48
 


712
00:47:48 --> 00:47:51
And then in the denominator,
well, let's see.

713
00:47:51 --> 00:47:53
It's x.
 

714
00:47:53 --> 00:47:57
So this would be dividing by
1 / x^ 5 in both numerator

715
00:47:57 --> 00:47:58
and denominator.
 

716
00:47:58 --> 00:48:04
And here you have 1 / x +
2 over, sorry I overshot.

717
00:48:04 --> 00:48:06
But that's OK.
 

718
00:48:06 --> 00:48:09
2 / x^ 5 here.
 

719
00:48:09 --> 00:48:12
So these are the main
term, if you like.

720
00:48:12 --> 00:48:18
And it's the same as 1 / 1 /
x, which is the same as x,

721
00:48:18 --> 00:48:21
and it goes to infinity.
 

722
00:48:21 --> 00:48:22
As x goes to infinity.
 

723
00:48:22 --> 00:48:26
Or, if you like, much more
simply, just x ^ 5 /

724
00:48:26 --> 00:48:29
x^ 4 is the main term.
 

725
00:48:29 --> 00:48:30
Which is x.
 

726
00:48:30 --> 00:48:31
Which goes to infinity.
 

727
00:48:31 --> 00:48:35
So don't forget your basic
algebra when you're doing

728
00:48:35 --> 00:48:37
this kind of stuff.
 

729
00:48:37 --> 00:48:40
Use these things and don't
use L'Hopital's Rule.

730
00:48:40 --> 00:48:42
OK, see you next time.
 

731
00:48:42 --> 00:48:42