Indeterminate forms | MIT 18.01SC Single Variable Calculus, Fall 2010


Uploaded by MIT on 07.01.2011

Transcript:

PROFESSOR: Hi.
Welcome back to recitation.
You've been talking about computing limits of some
indeterminate forms. In particular, you used
l'Hopital's rule to help you out with some limits in the
form 0 over 0 or infinity over infinity.
So these are the two indeterminate ratios.

And you also, so OK, so when you have a limit that is in
the form of an indeterminate ratio, you've seen that one
tool that you can use to help compute the limit is
l'Hopital's rule.
There are other indeterminate forms for limits as well.
So you actually saw, in lecture, another one of these,
which was you saw a limit of the form 0 to the 0.
So that was the limit you saw, was the limit as x goes to 0
from the right to x to the x.
So that was going to 0 over 0.
And the two competing forces are that as the base of the
limit goes to 0.
That wants to make the whole thing get closer to 0.
And when the exponent is going to 0, that makes the whole
thing want to get closer to 1.
So you have those two competing forces.
That's why it is an indeterminate form.
When you were solving this limit, you, first thing you
did was you wrote it as an exponential in base e.
So you wrote x to the x as e to the xlnx.
So xlnx is also an indeterminate form
as x goes to 0.
It's an indeterminate form of the form 0 times infinity.
Here I'm writing infinity to mean either positive infinity
or negative infinity.
So in this case, this is an indeterminate form because
when you have one factor going to 0, that makes the whole
product want to get closer to 0.
Whereas when you have one factor going to infinity,
either positive or negative, that makes the whole product
want to get big.
So that's why it becomes indeterminate.
And you were able to evaluate the limit xlnx, in that case,
by rewriting it as a quotient.
There are a few other indeterminate forms that I'd
like to mention.
So in particular, there are three other indeterminate
forms not, different from these.
And then I'll give you an example of one of them to
solve a problem.
so the three other indeterminate forms, there are
two more that are an exponential indeterminate
form, and there's one sort of outlier.
So one of the other indeterminate forms is
infinity to the 0.
So when you have an exponential expression where
the base is getting very very large, and the exponent is
going to 0, well the base getting large makes the whole
fraction want to be-- sorry-- the whole expression want to
be large, whereas the exponent going to 0 makes the whole
expression want to get closer to 1.
And so those two forces are in tension, and you can end up
with limits that equal any value when you have a limit
like this, infinity to the 0.
Another similar one is 1 to the infinity.
And this is the one I'll give you an
exercise on in a minute.
So when you have a limit of the form 1 to the infinity, so
something to the something where the base is going to 1
and the exponent is going to infinity, the base going to 1
makes this whole thing want to go to 1.
The exponent to infinity makes this whole thing want to
either blow up if it's a little bigger than 1, or get
really small if it's a little smaller than 1.
So you have, again, a tension there, and the result is
indeterminate.
The one sort of unusual one that you have is also--
which I'm not going to talk much more about-- is infinity
minus infinity.

So when you have two very large things, so here I mean
either positive infinity minus positive infinity, or negative
infinity minus negative infinity.
When you have two things that are both getting very large,
their difference could also be getting very large, or it
could be getting very small, or could be
doing anything in between.
So that's also an indeterminate form.
So these are the seven indeterminate forms. When you
have a quotient, you can always apply l'Hopital's rule.
When you have a product, you can always rewrite it as a
quotient, by writing for example, x squared plus 1
times-- well, let's see--

e to the minus x.
That's a product.
And you could always rewrite it as a quotient, for example,
e to the minus x over 1 over x squared plus 1.
There might even be a smarter way to rewrite this product as
a quotient.

But, OK.
And so when you have an exponential, as in this case,
you can use this rewriting in base e trick to turn it into a
product, which you can then turn into a quotient.
For the difference case, the reason I say it's unusual, is
just that there's not a good, general method for working
with these.
There are a lot of special cases.
And you sort of have to-- or what I mean is you have to
analyze them on a case by case basis.
There are different sort of techniques that will work.
So let me give you an example of one of these 1 to the
infinity kinds.
So why don't we come over here.
So compute the limit as x goes to 0 from the right of 1 plus
3x to the 10 divided by x.
So we see as x is going to 0, this base is going to 1.
And so that makes this whole expression want to be close to
1, whereas this exponent is going to infinity, since this
is just a little bit bigger than 1, that makes the whole
expression want to be big when the exponent is big.
So you have a tension here between the base going to 1
and the exponent going to infinity.
So the question is, try and actually compute this limit.
So why don't you pause the video, take a couple of
minutes to work on this question, come back, and we
can work on it together.

Welcome back.
Hopefully you had some luck working on this problem.
As we said, this is a limit of an indeterminate form of the 1
to infinity type.
As one of the three exponential types of
indeterminate forms, a really promising first step almost
every time is to rewrite this as a exponential expression
with the base e.
So we just do, so first we just do an algebraic
manipulation on the thing we're taking the limit of, and
then often, very often, that simplifies it into something
that we can actually compute the limit of.
So in particular, we have that 1 plus 3x, if we want to write
this in exponential form.
This is equal to e to the ln of 1 plus 3x.
This is true of any positive number.
Any positive number is e to the ln of it, because e and
log are inverse functions.
So OK, so we right it like this.
And so that means that 1 plus 3x to the 10 over x is equal
to e to the ln of 1 plus 3x to the 10 over x.
And now you can use your exponent rules.
So this is equal to e to the ln of 1 plus 3x
times 10 over x.
So our original expression is equal to this down here.
So the limit of our original expression is equal to the
limit of this one.
The other thing to notice is that because exponentiation is
a nice, continuous function, in order to compute this
limit, or the limit of this expression, it suffices--
just with base a constant, e-- it suffices to compute the
limit of the exponent.
All right, so let's do that.
Let's compute the limit of the exponent.
So we have the limit, so it has to be x going to the same
place, which in this case is to 0 from the right of ln of 1
plus 3x times 10 over x.
Well, this is a product.
It came to us as a product.
But there's an obvious way to rewrite this as a quotient.
I should say it's an indeterminate product.
As x goes to 0, this goes to ln of 1, which is 0.
Whereas this goes to infinity, positive infinity since we're
coming from the right.
So this is a 0 times positive infinity form.
So it is an indeterminate limit-- or sorry-- an
indeterminate product.

And, right, and there's an obvious way to rewrite this as
an indeterminate quotient, which is to rewrite it as the
limit as x goes 0 plus of ten ln of 1 plus 3x divided by x
as x goes 0.
So now, this is a limit where a it's an infinity-- sorry--
a 0 over 0.
So OK, so good.
So now we can apply l'Hopital's rule.
So by l'Hopital's rule, this is equal to the limit as x
goes to 0 on the right.
Of, well we apply l'Hopital's rule on the top, we get 10
over 1 plus 3x, by the chain rule, times 3.
And on the bottom, we just get one.
So now, OK, so this is true provided this
second limit exists.
And the second limit is no longer indeterminate.
It's easy to see what it is.
You just plug in x equals 0, and we get that
this is equal to 30.
OK so this limit is equal to 30, but this isn't the limit
that we started out wanting to compute.
The limit we started out wanting to compute is e to the
ln of 1 plus 3x times 10 over x.
So it's e to the this.
So our original limit as x goes to 0 from the right of 1
plus 3x to the 10 over x is equal to e to the thirtieth
power, which is pretty huge.
OK, So that's how we-- right-- and OK, so that's the answer
to our question.
So we took our original limit, it was this indeterminate
exponential form.
So what we do to it is, when you have an indeterminate
exponential form, you do this rewriting as an exponential in
the base of e trick, and then you pass the
limit into the exponent.
Because e is now a nice constant, life is simple--

life is simpler I should say--
you pass the limit into the constant, then you have an
indeterminate ratio, indeterminate product, which
you rewrite as an indeterminate ratio on which
you can then apply l'Hopital's rule.
Or possibly, you know, you rewrite it as a product.
Then it's easy to see what the value is.
So, all right, so that's how we deal with limits of
indeterminate exponential forms. And I'll end there.