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PROFESSOR: So we're through
with techniques of
integration, which is really the
most technical thing that
we're going to be doing.
And now we're just
clearing up a few
loose ends about calculus.
And the one we're going to talk
about today will allow us
to deal with infinity.
And it's what's known
as L'Hopital's Rule.
Here's L'Hopital's Rule.
And that's what we're
going to do today.
L'Hopital's Rule it's also known
as L'Hospital's Rule.
That's the same name, since the
circumflex is what you put
in French to omit the s.
So it's the same thing, and
it's still pronounced
L'Hopital, even if it's
got an s in it.
Alright, so that's the
first thing you need
to know about it.
And what this method does is,
it's a convenient way to
calculate limits including
some new ones.
So it'll be convenient
for the old ones.
There are going to be some new
ones and, as an example, you
can calculate x ln x as
x goes to infinity.
You could, whoops, that's not a
very interesting one, let's
try x goes to 0 from
the positive side.
And you can calculate, for
example, x e^ - x, as x goes
to infinity.
And, well, maybe I should
include a few others.
Maybe something like ln x /
x as x goes to infinity.
So these are some examples of
things which, in fact, if you
plug into your calculator,
you can see what's
happening with these.
But if you want to understand
them systematically, it's much
better to have this tool
of L'Hopital's Rule.
And certainly there isn't
a proof just based on a
calculation in a calculator.
So now here's the idea.
I'll illustrate the idea
first with an example.
And then we'll make
it systematic.
And then we're going
to generalize it.
We'll make it much more, so
when it includes these new
limits, there are some little
pieces of trickiness that you
have to understand.
So, let's just take an example
that you could have done in
the very first unit
of this class.
The limit as x goes to 1 of
x ^ 10 - 1 / x ^2 - 1.
So that's a limit that
we could've handled.
And the thing that's
interesting, I mean, if you
like this is in this category
that we mentioned at the
beginning of the course
of interesting limits.
What's interesting about it is
that if you do this silly
thing, which is just plug in
x = 1, at x = 1 you're
going to get 0 / 0.
And that's what we call
an indeterminate form.
It's just unclear what it is.
From that plugging, in you
just can't get it.
Now, on the other
hand, there's a
trick for doing this.
And this is the trick
that we did at the
beginning of the class.
And the idea is I can divide
in the numerator and
denominator by x - 1.
So this limit is unchanged, if
I try to cancel the hidden
factor x - 1 in the numerator
and denominator.
Now, we can actually carry out
these ratios of polynomials
and calculate them by long
division in algebra.
That's very, very long.
We want to do this
with calculus.
And we already have. We already
know that this ratio
is what's called a difference
quotient.
And then in the limit, it tends
to the derivative of
this function.
So the idea is that this is
actually equal to, in the
limit, now let's just study
one piece of it.
So if I have a function f ( x),
which is x ^ 10 - 1, and
the value at 1 happens to
be equal to 0, then this
expression that we have, which
is in disguise, this is in
disguise the difference
quotient, tends to, as x goes
to 1, the derivative,
which is f' ( 1).
That's what it is.
So we know what the numerator
goes to, and similarly we'll
know what the denominator
goes to.
But what is that?
Well, f ' (x) = 10x ^ 9.
So we know what the answer is.
In the numerator it's 10x ^ 9.
In the denominator, it's going
to be 2x, that's the
derivative of x ^2 - 1.
And then were going to have
to evaluate that at x = 1.
And so it's going to be
10/2, which is 5.
So the answer is 5.
And it's pretty easy to get
from our techniques and
knowledge of derivatives,
using this rather clever
algebraic trick.
This business of dividing
by x - 1.
What I want to do now is just
carry this method out
systematically.
And that's going to give us the
approach to what's known
as L'Hopital's Rule.
What my main subject
for today.
So here's the idea.
Suppose we're considering, in
general, a limit as x goes to
some number a of f
( x) / g ( x).
And suppose it's the bad case
where we can't decide.
So it's in determinate.
f ( a) = g ( a) = 0.
So it would be 0 / 0.
Now we're just going to
do exactly the same
thing we did over here.
Namely, we're going to divide
a numerator and denominator,
and we're going to repeat
that argument.
So we have here f
( x) / x - a.
And g (x) / x - a also.
I haven't changed
anything yet.
And now I'm going to write it
in this suggestive form.
Namely, I'm going to take
separately the limit in the
numerator and the denominator.
And I'm going to make
one more shift.
So I'm going to take the limit,
as x goes to a in the
numerator, but I'm going
to write it as f ( x) -
f ( a) / x - a.
So that's the way I'm going to
write the numerator, and I've
got to draw a much
longer line here.
So why am I allowed
to do that?
That's because f (a) = 0.
So I didn't change this
numerator of the numerator any
by subtracting. f ( a) = 0.
And I'll do the same thing
to the denominator.
Again, g ( a) = 0,
so this is OK.
And lo and behold, I know
what these limits are.
This is f ' (a) / g '(a).
So that's it.
That's the technique and this
evaluates the limit.
And it's not so difficult.
The formula's pretty
straightforward here.
And it works, provided that
g ' (a) is not 0.
Yeah, question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is, is
there a more intuitive way of
understanding this procedure.
And I think the short answer
is that there are other,
similar, ways.
I don't consider them to
be more intuitive.
I will be mentioning one of
them, which is the idea of
linearization, which goes back
to what we did in Unit 2.
I think it's very important to
understand all of these, more
or less, at once.
But I wouldn't claim that any
of these methods is a more
intuitive one than the other.
But basically what's happening
is, we're looking at the
linear approximation
to f, at a.
And the linear approximation
to g at a.
That's what underlies this.
So now I get to formulate for
you L'Hopital's Rule at least
in what I would call the
easy version or, if you
like, Version 1.
So here's L'Hopital's Rule.
Version 1.
It's not going to be quite the
same as what we just did.
It's going to be much,
much better.
And more useful.
And what is going to take care
of is this problem that the
denominator is not 0.
So now here's what we're
going to do.
We're going to say that it turns
out that the limit a x
goes to a of f(x) / g ( x) = the
limit as x goes to a to a
f '(x) / g' (x).
Now, that looks practically
the same as
what we said before.
And I have to make sure that you
understand when it works.
So it works provided this is
one of these undefined
expressions.
In other words, f (
a) = g ( a) = 0.
So we have a 0 / 0 expression,
indeterminant.
And, also, we need one
more assumption.
And the right-hand side, the
right-hand limit exists.
Now, this is practically
the same thing as
what I said over here.
Namely, I took the ratio of
these functions, x ^ 10 - 1
and x ^2 - 1.
I took their derivatives, which
is what I did right
here, right.
I just differentiated them
and I took the ratio.
This is way easier than the
quotient rule, and is nothing
like the quotient rule.
Don't think quotient rule.
Don't think quotient rule.
So we differentiate the
numerator and denominator
separately.
And then I take the limit as
x goes to 1 and I get 5.
So that's what I'm claiming
over here.
I take these functions, I
replace them with this ratio
of derivatives, and then
I take the limit
instead, over here.
And it turned out that the
functions got much simpler
when I differentiated them.
I started with this messy object
and I got this much
easier object that I could
easily evaluate.
So that's the big game that's
happening here.
It works, if this limit makes
sense and this limit exists.
Now, notice I didn't claim that
g, that the denominator,
had to be non-0.
So that's what's going
to help us a little
bit in a few examples.
So let me give you a couple
of examples and
then we'll go further.
Now, this is only Version 1.
But first we have to understand
how this one works.
So here's another example.
Take the limit as x goes to
0, of sin 5x / sin 2x.
This is another kind of example
of a limit that we
discussed in the first
part of the course.
Unfortunately, now we're
reviewing stuff.
So this should reinforce
what you did there.
This will be an easier way
of thinking about it.
So by L'Hopital's Rule,
so here's the step.
We're going to take one
of these steps.
This is the limit, as x goes to
1, of the derivatives here.
So that's 5 cos 5x / 2 cos 2x.
The limit was 1 over there,
but now it's 0. a
is 0 in this case.
This is the number a.
Thank you.
So the limit as x goes to 0 is
the same as the limit of the
derivatives.
And that's easy to evaluate.
Cosine of 0 = 1, right.
This is equal to
5 (cos (5* 0).
And that's a multiplication
sign.
Maybe I should just
write this as 0.
Divided by 2 cos 0.
But you know that that's 5/2.
So this is how L'Hopital's
method works.
It's pretty painless.
I'm going to give you another
example, which shows that it
works a little better
than the method that
I started out with.
Here's what happens if we
consider the function
cos x - 1 / x ^2.
That was a little harder
to deal with.
And again, this is one of these
0 / 0 things near x = 0.
As x tends to 0, this goes to
an indeterminate form here.
Now, according to our method,
this is equivalent to, now I'm
going to use this little wiggle
because I don't want to
write limit, limit, limit,
limit a million times.
So I'm going to use a
little wiggle here.
So as x goes to 0, this is going
to behave the same way
as differentiating numerator
and denominator.
So again this is going to be
- sin x in the numerator.
In the denominator, it's
going to be 2x.
Now, notice that we still
haven't won yet.
Because this is still
of 0 / 0 type.
When you plug in x =
0 you still get 0.
But that doesn't damage
the method.
That doesn't make
the method fail.
This 0 / 0, we can
apply L'Hopital's
Rule a second time.
And as x goes to 0 this is
the same thing as, again,
differentiating the numerator
and denominator.
So here I get - cos x in the
numerator, and I get 2 in the
denominator.
Again this is way easier than
differentiating ratios of functions.
We're only differentiating the
numerator and the denominator
separately.
And now this is the end.
As x goes to 0, this is - cos
0 / 2, which is - 1/2.
Now, the justification for this
comes only when you win
in the end and get the limit.
Because what the theorem says is
that if one of these limits
exists, then the preceding
one exists.
And once the preceding
one exists, then the
one before it exists.
So once we know that this one
exists, that works backwards.
It applies to the preceding
limit, which then applies to
the very first one.
And the logical structure here
is a little subtle, which is
that if the right side exists,
then the left side will also
exist. Yeah, question.
STUDENT: [INAUDIBLE]
PROFESSOR: Why does the
right-hand limit have to
exist, isn't it just the
derivative that has to exist?
No.
The derivative of the numerator
has to exist. The
derivative of the denominator
has to exist. And this limit
has to exist. What doesn't have
to exist, by the way, I
never said that f prime of a
has to exist. In fact, it's
much, much more subtle.
I'm not claiming that f ' (a)
exists, because in order to
evaluate this limit, f ' (a)
need not exist. What has to
happen is that nearby, for x not
equal to a, these things
exist. And then the limit has
to exist. So there's no
requirements that the limits
exist. In fact, that's exactly
going to be the point when we
evaluate these limits here.
Is we don't have to evaluate
it right at the end.
STUDENT: [INAUDIBLE]
PROFESSOR: So the question that
you're asking is, why is
this the hypothesis
of the theorem?
In other words, why
does this work?
Well, the answer is that this
is a theorem that's true.
If you drop this hypothesis,
it's totally false.
And if you don't have this
hypothesis, you can't use the
theorem and you will get
the wrong answer.
I mean, it's hard to express
it any further than that.
So look, in many cases
we tell you formulas.
And in many cases it's so
obvious when they're true that
we don't have to worry
about what we say.
And indeed, there's something
implicit here.
I'm saying well, you know, if
I wrote this symbol down, it
must mean that the
thing exists.
So that's a subtle point.
But what I'm emphasizing is that
you don't need to know in
advance that this one exists.
You do need to know in advance
that that one exists.
Essentially, yeah.
So that's the direction
that it goes.
You can't get away with not
having this exist and still
have the statement be true.
Alright, another question.
Thank you.
STUDENT: [INAUDIBLE]
PROFESSOR: So I'm getting a
little ahead of myself, but
let me just say.
In these situations here, when x
is going to 0 and x is going
to infinity.
For instance, here when x goes
to 0, the logarithm is
undefined at x = 0.
Nevertheless, this
theorem applies.
And we'll be able to use it.
Over here, as x goes to
infinity, neither of these --
well, actually, come to think
of it, e^ -x, if you like,
it's equal to 0 at infinity.
If you want to say that
it has a value.
But in fact, these expressions
don't necessarily have values.
At the ends.
And nevertheless, the
theorem applies.
I mean, it can exist. It's
perfectly OK for it to exist.
It's no problem.
It just doesn't need
to exists.
It isn't forced to exist.
So here's a calculation
which we just did.
And we evaluated this.
Now, I want to make a comparison
with the method of
approximation.
In the method of approximations,
this Example
2, which was the example with
the sine function, we would
use the following property.
We would use sin u is
approximately u.
We would use that linear
approximation.
And then what we would have here
is that sin 5x / sin 2x
is approximately 5x / 2x,
which is of course 5/2.
And this is true when u is
approximately 0, and this is
true certainly as x goes
to 0, it's going
to be a valid limit.
So that's very similar
to Example 2.
In Example 3, we managed to look
at this expression cos x
- 1 / x ^2.
And for this one, you have to
remember the approximation
near x = 0 to the
cosine function.
And that's 1 - x ^2 / 2.
So that was the approximation,
the quadratic approximation to
the cosine function.
And now, sure enough,
this simplifies.
This becomes - x ^2 / 2 /
x ^2, which is - 1/2.
So we get the same answer,
which is a good thing.
Because both of these
methods are valid.
They're consistent.
You can see that neither
of them is
particularly a lot longer.
You may have trouble remembering
this property.
But in fact it's something that
you can easily derive.
And, indeed, it's related to the
second derivative of the
cosine, as is this
calculation here.
They're almost the same amount
of numerical content to them.
So now what I'd like to do is
explain to you why L'Hopital's
Rule works better
in some cases.
And the real value that it has
is in handling these other
more exotic limits.
So now we're going to do
L'Hopital's Rule over again.
And I'll handle these
functions.
But I'll have to rewrite them,
but we'll just do that.
So here's the property.
That the limit as x goes to a of
f ( x) / g (x) is equal to
the limit as x goes to a
of f ' ( x) / g '(x).
That's the property.
And this is what we'll
always be using.
Very convenient thing.
And remember it was true
provided that f (
a) = g (a) = 0.
And that the right-hand
side exists.
But I claim that it
works better, and
I'll get rid of these.
But I'll write them again
to show you that
it works for these.
So there are other cases.
And the other cases that
are allowed are this.
First of all, as indicated by
what I just erased, you can
allow a to be equal to plus
or minus infinity.
It's also OK.
So you can take the
limit going to the
far ends of the universe.
Both left and right.
And then the other thing that
you can do is, you can allow f
( a) and g (a) to be plus
or minus infinity.
Is OK.
So now, the point is that we can
handle not just the 0 / 0
case, but also the infinity
/ infinity case.
That's a very powerful tool, and
quite different from the
other cases.
And the third thing is that the
right-hand side doesn't
really quite have to exist,
in the ordinary sense.
Or, it could be plus
or minus infinity.
That's also OK.
That's still information.
So if we can see where it goes,
then we're still good.
If it goes to plus infinity, if
it goes to 0, if it goes to
a finite number, if it goes to
minus infinity, all of that
will be OK.
It just if it oscillates wildly
that we'll be lost. And
those calculations we'll
never encounter.
So this basically handles
everything that you could
possibly hope for.
And it's a very convenient
process.
So let me carry out
a few examples.
And, let's see, I guess the
first one that I wanted to do
was x ln x.
So what example are we up to.
Example 3, so Example
4 is coming up.
Example 4, this is one of the
ones that I wrote at the
beginning of the lecture,
x ln x.
This one was on our
homework problem.
In the limits of some
calculation.
But so this one, you have to
look at it first to think
about what it's doing.
It's an indeterminate form, but
it sort of looks like it's
the wrong type.
So why is it in an indeterminate
form.
This one goes to 0, and this
one goes to minus infinity.
So, excuse me, this
is a product.
It's 0 times minus infinity.
So that's an indeterminate form,
because we don't know
whether the 0 wins or the
infinity this could keep
getting smaller and smaller and
smaller, and this could be
getting bigger and
bigger bigger.
The product could be anything
in between.
We just don't know.
So the first step is to write
this as a ratio of things,
rather than a product
of things.
And it turns out that the way
to do that is to use the
logarithm in the numerator,
and the 1 / x in the
denominator.
So this is a choice that
I'm making here.
Now, I've just converted it to
a limit of the type minus
infinity divided by infinity.
Because the numerator is going
to minus infinity as x goes to
0 plus and the denominator 1 /
x is going to plus infinity.
Again, there's a competitions,
but now it's one of the forms
to which L'Hopital's
Rule applies.
Now I'm just going to apply
L'Hopital's Rule.
And what it says is that
I differentiate here.
So I just differentiate a
numerator and denominator.
Applying L'Hopital's
Rule is a breeze.
You just differentiate,
differentiate.
And now it just simplifies
and we're done.
This is the limit as x
goes to 0 plus of,
well, the x^2's cancel.
This is the same as just
- x. x factors cancel.
And so that's 0.
The answer is that it's 0.
So x goes to 0 faster then ln
n goes to minus infinity.
This 0 was the winner.
Something you can't
necessarily predict in advance.
So let's do the other two
examples that I wrote down.
I'm going to do them in slightly
more generality,
because they're the most
fundamental rate properties
that you're going to need to
know for the next section.
Which is improper integrals.
And also they're just very
important for physical math,
and any other kind of
thing, basically.
So here, let's just do these.
So let's see, which one do I
want to do first. So I wrote
down the limit of x e^2 - x, but
I'm going to make it even
more general.
I'm going to make it any
negative power here, where p
is some positive constant.
Now again, this is a product of
functions, not a quotient,
a ratio, of functions.
And so I need to rewrite it.
I'm going to write it
as x / e ^ p x.
And now I'm going to apply,
well, so it's of this form
infinity / infinity.
And now that's the same as the
limit as x goes to infinity of
1 / p e^ px.
So where does that go?
As x goes to infinity.
Now we can decide.
The 1 stays where it is.
And this, as x goes to infinity,
goes to infinity.
So the answer is 0.
And the conclusion is that x
grows more slowly then e ^ px.
As x goes to infinity.
Remember, p is positive
here, of course.
It's the increasing
exponentials.
Not the decreasing ones.
Let's do a variant of this.
I'll do it the opposite way.
So I'm going to call
this example 5 '.
It really doesn't give us any
more information, but it gives
you just a little bit
more practice.
So suppose I look at things the
other way. e^ px divided
by, say, x^ 100.
Now, this is an infinity /
infinity example, again.
And you can work out
what it's doing.
But there are two ways of
thinking about this.
There's the slow way
and the fast way.
The slow way is to differentiate
this 100 times.
That is, right?
Apply L'Hopital's Rule
over and over and
over and over again.
All the way.
It's clear that you
could do it, but
it's kind of a nuisance.
So there's a much cleverer
trick here.
Which is to change this to
(the limit as x goes to
infinity of the e ^ px
/ 100 / x) ^ 100.
So if you do that, then
we just have one
L'Hopital's Rule step here.
And that one is that this is
the same as (x goes to
infinity of, well it's p / 100
e^ p x / 100 / 1) ^ 100.
That's our L'Hopital's step.
And of course, that's (infinity
/ 1 ) ^ 100.
Which is infinity.
Now, again I did this in a
slightly different way to show
you that it works with
infinity as well.
So that was this other case.
The right-hand side can
exist, or it can be
plus or minus infinity.
And that applies
to this limit.
And therefore, to the
original limit.
And the conclusion here is
that e ^ px, p > 0, grows
faster than any power of x.
I picked x ^ 100, but obviously
it didn't matter
what power I picked.
The exponents beat
all the powers.
So we have one more of the
ones that I gave at the
beginning to take care of.
And that one is the logarithm.
And its behavior at infinity.
So I'll do a slightly variant
on that one, too.
So we have Example 6, which is
ln x, and instead of dividing
by x, I'm going to
divide by x^ 1/3.
I could divide by any positive
power of x, we'll just do this
example here.
So now this, as x goes to
infinity, is of the form
infinity / infinity.
And so it's equivalent to what
happens when I differentiate
numerator and denominator
separately.
And that's 1 / x, and here
I have 1/3 x ^ - 2/3.
1 / x, and then 1/3 x^ - 2/3.
Now, when the dust settles
here and you get your
exponents right, we have an x^ -
1, and this is an x ^ + 2/3,
and that's a 1/3 becomes a 3.
So this is what it is.
And that's equal
to 3x ^ - 1/3.
Which we can decide.
It goes to 0.
As x goes to infinity.
And so the conclusion is that
ln x grows more slowly as x
goes to infinity, than x ^ 1/3
or any positive power of x.
So any x ^ p, p positive,
will work.
So ln is really slow,
going to infinity.
It's very, very gradual.
Yeah, question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
is, how many
hypotheses do you need here?
So I said that, and I think what
you were asking is, if I
have this hypothesis, can I
also have this hypothesis.
That's OK.
I can have this hypothesis
combined with this one.
I need something about
f (a) and g ( a).
I can't assume nothing
about f(a) and g(a).
So in other words, I have to
be faced with either an
infinity / infinity, or
a 0 / 0 situation.
So let's see.
A rule applies in the 0 / 0, or
infinity / infinity case.
These are the only two cases
that it applies in.
And a can be anything.
Including infinity.
Plus or minus infinity.
The rule applies in
these two cases.
So in other words, this is
what f ( a) / g ( a) is.
Either one of these.
And in fact, it can
be plus or minus.
STUDENT: [INAUDIBLE]
PROFESSOR: And the right-hand
side has to be something.
It has to be either finite or
plus or minus infinity.
So you need something.
You need a specific value of a,
you need to decide whether
it's an indeterminate form.
And you need the right-hand
limit to exist. It's not hard
to impose this.
Because when you look at the
right-hand side, you'll want
to be calculating it.
So you want to know
what it is.
So you'll never have problems
confirming this hypothesis.
Alright.
Let me give you one
more example here.
Which is just slightly
trickier.
Which involves, so here's
another indeterminate form.
That's going to be 0 ^ 0.
So there are lots of these
things where you just don't
know what to do.
And they come out in various
different ways.
The simplest example of this
is the limit as x goes to 0
from above of x ^ x.
In order to work out what's
happening with this one, we
have to use a trick.
And the trick is this is
a moving exponent.
And so it's appropriate
to use base e.
This is something that we did
way back in the first unit.
So, since we have a moving
exponent, we're
going to use base e.
That's the good base
to use whenever you
have a moving exponent.
And so rewrite this as
x^ x = e ^ x ln x.
And now, in order to figure
out what's happening, we
really only have to know what's
going on with the exponent.
So remember, actually
we already did this.
But I'm going to do it
once more for you.
This is ln x / (1 / x).
And that's equivalent, as
x goes to 0, to using
L'Hopital's Rule to 1 / x, and
this is - 1 / x ^2, which is -
x, which goes to 0.
As x goes to 0.
And so what we have here is that
this one is going to be
equivalent to, well, it's
going to tend to
what we got over here.
It's e ^ 0.
That exponent is what we want.
As x goes to 0.
So that's the answer This
limit happens to be 1.
That's actually relatively easy
to do, given all of the
power that we have
at our hands.
Now, let me give you
one more example.
Suppose you're trying to
understand the limit
of sin x / x ^2.
If you apply L'Hopital's Rule,
as x goes to 0, you're going
to get cos x / 2x.
And if you apply L'Hopital's
Rule again, as x goes to 0,
you're going to get
the - sin x / 2.
And this, as x goes
to 0, goes to 0.
On the other hand, if you look
at the linear approximation
method, linear approximation
says that sin x is
approximately x near 0.
So that should be x / x ^2.
Which is 1 / x, which
goes to infinity.
As x goes to 0, at least from
one side, minus infinity to
the other side.
So there's something fishy
going on here, right?
So this is fishy.
Or maybe this is fishy,
I don't know.
So, tell me what's wrong here.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: OK.
So the claim is that the
second application of
L'Hopital's Rule, this
one, is wrong.
And that's correct.
And this is where you
have to watch out,
with L'Hopital's Rule.
This is exactly where you
have to watch out.
You have to apply the
test. Here it's an
indeterminate form.
It's 0 / 0 before I
applied the rule.
But in order to apply the rule
the second time, it still has
to be 0 / 0.
But this one isn't.
This one is 1 / 0.
It's no longer an indeterminate
form.
It's actually infinite.
Either plus or minus, depending
on the sign of the
denominator.
Which is just what
this answer is.
So the linear approximation
is safe.
And we just applied L'Hopital's
Rule wrong.
So the moral of the story here
is look before you L'Hop.
Alright.
Now, let me say one
more thing.
I need to pile it on just
a little bit, sorry.
So don't use it as a crutch.
We don't want to just get
ourselves so weak, after being
in the hospital for all
this time, that we
can't use, I'm sorry.
So remember that you shouldn't
have lost your senses.
If you have something
like this, so
we'll do this one here.
Suppose you're trying to
understand what this does as x
goes to infinity.
Now, you could L'Hopital's
Rule five
times, or four times.
And get the answer here.
But really, you should realize
that the main terms are
sitting there right
in front of you.
And that there's some
algebra that you can
do to simplify this.
Namely, it's the same as
1 + 2 / x + 1 / x^ 5.
And then in the denominator,
well, let's see.
It's x.
So this would be dividing by 1
/ x^ 5 in both numerator and
denominator.
And here you have 1 / x + 2
over, sorry I overshot.
But that's OK.
2 / x^ 5 here.
So these are the main
term, if you like.
And it's the same as 1 / 1 / x,
which is the same as x, and
it goes to infinity.
As x goes to infinity.
Or, if you like, much more
simply, just x ^ 5 / x^ 4 is
the main term.
Which is x.
Which goes to infinity.
So don't forget your basic
algebra when you're doing this
kind of stuff.
Use these things and don't
use L'Hopital's Rule.
OK, see you next time.
