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PROFESSOR: Hi.

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Welcome back to recitation.

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You've been talking about
computing limits of some

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indeterminate forms. In
particular, you used

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l'Hopital's rule to help you
out with some limits in the

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form 0 over 0 or infinity
over infinity.

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So these are the two
indeterminate ratios.

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And you also, so OK, so when
you have a limit that is in

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the form of an indeterminate
ratio, you've seen that one

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tool that you can use to help
compute the limit is

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l'Hopital's rule.

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There are other indeterminate
forms for limits as well.

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So you actually saw, in lecture,
another one of these,

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which was you saw a limit
of the form 0 to the 0.

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So that was the limit you saw,
was the limit as x goes to 0

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from the right to x to the x.

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So that was going to 0 over 0.

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And the two competing forces
are that as the base of the

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limit goes to 0.

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That wants to make the whole
thing get closer to 0.

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And when the exponent is going
to 0, that makes the whole

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thing want to get closer to 1.

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So you have those two
competing forces.

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That's why it is an
indeterminate form.

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When you were solving this
limit, you, first thing you

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did was you wrote it as an
exponential in base e.

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So you wrote x to the
x as e to the xlnx.

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So xlnx is also an indeterminate
form

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as x goes to 0.

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It's an indeterminate form of
the form 0 times infinity.

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Here I'm writing infinity to
mean either positive infinity

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or negative infinity.

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So in this case, this is an
indeterminate form because

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when you have one factor going
to 0, that makes the whole

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product want to get
closer to 0.

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Whereas when you have one factor
going to infinity,

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either positive or negative,
that makes the whole product

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want to get big.

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So that's why it becomes
indeterminate.

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And you were able to evaluate
the limit xlnx, in that case,

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by rewriting it as a quotient.

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There are a few other
indeterminate forms that I'd

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like to mention.

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So in particular, there are
three other indeterminate

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forms not, different
from these.

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And then I'll give you an
example of one of them to

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solve a problem.

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so the three other indeterminate
forms, there are

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two more that are an exponential
indeterminate

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form, and there's one
sort of outlier.

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So one of the other
indeterminate forms is

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infinity to the 0.

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So when you have an exponential
expression where

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the base is getting very very
large, and the exponent is

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going to 0, well the base
getting large makes the whole

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fraction want to be-- sorry--
the whole expression want to

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be large, whereas the exponent
going to 0 makes the whole

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expression want to
get closer to 1.

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And so those two forces are in
tension, and you can end up

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with limits that equal any value
when you have a limit

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like this, infinity to the 0.

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Another similar one is
1 to the infinity.

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And this is the one
I'll give you an

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exercise on in a minute.

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So when you have a limit of the
form 1 to the infinity, so

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something to the something where
the base is going to 1

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and the exponent is going to
infinity, the base going to 1

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makes this whole thing
want to go to 1.

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The exponent to infinity makes
this whole thing want to

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either blow up if it's a little
bigger than 1, or get

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really small if it's a little
smaller than 1.

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So you have, again, a tension
there, and the result is

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indeterminate.

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The one sort of unusual one
that you have is also--

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which I'm not going to talk much
more about-- is infinity

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minus infinity.

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So when you have two very large
things, so here I mean

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either positive infinity minus
positive infinity, or negative

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infinity minus negative
infinity.

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When you have two things that
are both getting very large,

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their difference could also be
getting very large, or it

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could be getting very
small, or could be

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doing anything in between.

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So that's also an indeterminate
form.

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So these are the seven
indeterminate forms. When you

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have a quotient, you can always
apply l'Hopital's rule.

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When you have a product, you
can always rewrite it as a

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quotient, by writing for
example, x squared plus 1

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times-- well, let's see--

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e to the minus x.

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That's a product.

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And you could always rewrite it
as a quotient, for example,

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e to the minus x over 1
over x squared plus 1.

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There might even be a smarter
way to rewrite this product as

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a quotient.

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But, OK.

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And so when you have an
exponential, as in this case,

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you can use this rewriting in
base e trick to turn it into a

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product, which you can then
turn into a quotient.

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For the difference case, the
reason I say it's unusual, is

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just that there's not a good,
general method for working

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with these.

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There are a lot of
special cases.

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And you sort of have to-- or
what I mean is you have to

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analyze them on a case
by case basis.

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There are different sort of
techniques that will work.

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So let me give you an example
of one of these 1 to the

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infinity kinds.

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So why don't we come
over here.

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So compute the limit as x goes
to 0 from the right of 1 plus

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3x to the 10 divided by x.

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So we see as x is going to 0,
this base is going to 1.

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And so that makes this whole
expression want to be close to

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1, whereas this exponent is
going to infinity, since this

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is just a little bit bigger than
1, that makes the whole

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expression want to be big when
the exponent is big.

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So you have a tension here
between the base going to 1

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and the exponent going
to infinity.

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So the question is, try and
actually compute this limit.

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So why don't you pause the
video, take a couple of

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minutes to work on this
question, come back, and we

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can work on it together.

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Welcome back.

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Hopefully you had some luck
working on this problem.

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As we said, this is a limit of
an indeterminate form of the 1

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to infinity type.

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As one of the three exponential
types of

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indeterminate forms, a really
promising first step almost

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every time is to rewrite this
as a exponential expression

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with the base e.

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So we just do, so first we
just do an algebraic

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manipulation on the thing we're
taking the limit of, and

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then often, very often, that
simplifies it into something

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that we can actually compute
the limit of.

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So in particular, we have that 1
plus 3x, if we want to write

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this in exponential form.

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This is equal to e to
the ln of 1 plus 3x.

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This is true of any
positive number.

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Any positive number is e to the
ln of it, because e and

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log are inverse functions.

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So OK, so we right
it like this.

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And so that means that 1 plus
3x to the 10 over x is equal

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to e to the ln of 1 plus
3x to the 10 over x.

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And now you can use your
exponent rules.

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So this is equal to e to
the ln of 1 plus 3x

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times 10 over x.

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So our original expression is
equal to this down here.

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So the limit of our original
expression is equal to the

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limit of this one.

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The other thing to notice is
that because exponentiation is

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a nice, continuous function,
in order to compute this

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limit, or the limit of this
expression, it suffices--

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just with base a constant, e--
it suffices to compute the

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limit of the exponent.

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All right, so let's do that.

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Let's compute the limit
of the exponent.

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So we have the limit, so it has
to be x going to the same

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place, which in this case is to
0 from the right of ln of 1

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plus 3x times 10 over x.

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Well, this is a product.

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It came to us as a product.

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But there's an obvious way to
rewrite this as a quotient.

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I should say it's an
indeterminate product.

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As x goes to 0, this goes
to ln of 1, which is 0.

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Whereas this goes to infinity,
positive infinity since we're

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coming from the right.

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So this is a 0 times positive
infinity form.

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So it is an indeterminate
limit-- or sorry-- an

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indeterminate product.

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And, right, and there's an
obvious way to rewrite this as

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an indeterminate quotient, which
is to rewrite it as the

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limit as x goes 0 plus of ten
ln of 1 plus 3x divided by x

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as x goes 0.

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So now, this is a limit where
a it's an infinity-- sorry--

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a 0 over 0.

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So OK, so good.

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So now we can apply
l'Hopital's rule.

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So by l'Hopital's rule, this
is equal to the limit as x

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goes to 0 on the right.

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Of, well we apply l'Hopital's
rule on the top, we get 10

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over 1 plus 3x, by the
chain rule, times 3.

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And on the bottom,
we just get one.

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So now, OK, so this is
true provided this

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second limit exists.

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And the second limit is no
longer indeterminate.

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It's easy to see what it is.

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You just plug in x equals
0, and we get that

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this is equal to 30.

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OK so this limit is equal to 30,
but this isn't the limit

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that we started out wanting
to compute.

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The limit we started out wanting
to compute is e to the

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ln of 1 plus 3x times
10 over x.

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So it's e to the this.

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So our original limit as x goes
to 0 from the right of 1

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plus 3x to the 10 over x is
equal to e to the thirtieth

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power, which is pretty huge.

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OK, So that's how we-- right--
and OK, so that's the answer

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to our question.

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So we took our original limit,
it was this indeterminate

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exponential form.

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So what we do to it is, when
you have an indeterminate

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exponential form, you do this
rewriting as an exponential in

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the base of e trick, and
then you pass the

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limit into the exponent.

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Because e is now a nice
constant, life is simple--

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00:11:12,460 --> 00:11:14,980

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life is simpler I should say--

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you pass the limit into the
constant, then you have an

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indeterminate ratio,
indeterminate product, which

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you rewrite as an indeterminate
ratio on which

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you can then apply
l'Hopital's rule.

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Or possibly, you know, you
rewrite it as a product.

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Then it's easy to see
what the value is.

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So, all right, so that's how
we deal with limits of

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indeterminate exponential forms.
And I'll end there.

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