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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

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of some indeterminate forms.

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In particular, you
used l'Hopital's rule

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to help you out with some
limits in the form 0 over 0

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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

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have a limit that is in the
form of an indeterminate ratio,

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you've seen that
one tool that you

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can use to help compute the
limit is 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

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of these, 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

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as x goes to 0 from the
right of 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 limit

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goes to 0, that wants to make
the whole thing get closer

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to 0.

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And when the exponent
is going to 0,

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that makes the whole 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,

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you-- first thing you did was
you wrote it as an exponential

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in base e.

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So you wrote x to the
x as e to the x*ln x.

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So x*ln x is also an
indeterminate form as x goes

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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

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either positive infinity
or negative infinity.

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So in this case, this
is an indeterminate form

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because when you have
one factor going to 0,

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that makes the whole 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 x*ln x, in that case,

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by rewriting it as a quotient.

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There are a few other
indeterminate forms

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that I'd like to mention.

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So in particular, there are
three other indeterminate forms

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different from these.

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

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to solve a problem.

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so the three other
indeterminate forms,

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there are two more that are sort
of 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

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is 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,

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and the exponent is going to
0, well the base getting large

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makes the whole
fraction want to be

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sorry the whole expression
want to be large,

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whereas the exponent going to 0
makes the whole expression want

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to get closer to 1.

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And so those two
forces are in tension,

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and you can end up
with limits that

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equal any value when you
have a limit like this,

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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 exercise on in a minute.

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So when you have a
limit of the form 1

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to the infinity, so something to
the something where the base is

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going to 1 and the exponent
is going to infinity,

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the base going to 1 makes this
whole thing want to go to 1.

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The exponent to infinity
makes this whole thing

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want to either blow up if
it's a little bigger than 1,

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

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So you have, again,
a tension there,

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and the result is 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--

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is infinity minus infinity.

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So when you have two
very large things,

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so here I mean either
positive infinity

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minus positive infinity,
or negative infinity

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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,

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or it could be
getting very small,

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or could be 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.

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When you have a
quotient, you can always

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apply l'Hopital's rule.

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When you have a
product, you can always

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rewrite it as a
quotient, by writing

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for example, x squared
plus 1 times-- well,

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let's see-- 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,

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for example, e to the minus x
over 1 over x squared plus 1.

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There might even
be a smarter way

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to rewrite this
product as 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

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to turn it into a product,
which you can then

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turn into a quotient.

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

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is just that there's not
a good, general method

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for working 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

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is you have to 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

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of these 1 to the
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

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from the right of 1 plus
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

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to be close to 1, whereas this
exponent is going to infinity,

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since this is just a
little bit bigger than 1,

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that makes the whole
expression want to be big

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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 minutes

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to work on this
question, come back,

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and we 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 indeterminate forms,

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a really promising first
step almost every time

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is to rewrite this as a
exponential expression

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with the base e.

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So we just do-- first we just
do an algebraic manipulation

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on the thing we're taking
the limit of, and then often,

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very often, that simplifies
it into something that we can

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actually compute the limit of.

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So in particular, we
have that 1 plus 3x,

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if we want to write 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

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and log are inverse functions.

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So OK, so we write it like this.

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

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is equal 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 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

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is equal to the
limit of this one.

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The other thing to notice is
that because exponentiation

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is a nice, continuous
function, in order

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to compute this limit, or
the limit of this expression,

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it suffices-- just
with base a constant,

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e-- it suffices to compute
the 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

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x going to the same
place, which in this case

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is to 0 from the right, of ln
of 1 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

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an indeterminate product.

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And, right, and
there's an obvious way

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to rewrite this as an
indeterminate quotient,

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which is to rewrite
it as the limit

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as x goes 0 plus of
10 ln of 1 plus 3x

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divided by x, as x goes 0.

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

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a 0 over 0 form.

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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

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as x goes to 0 on
the right of-- well,

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we apply l'Hopital's rule
on the top, we get 10 over 1

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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

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this 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

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get that 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

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is e to the 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

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to the thirtieth power,
which is pretty huge.

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OK, so that's how we--
right-- and OK, so that's

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the answer to our question.

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So we took our
original limit, it

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was this indeterminate
exponential form.

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So what we do to it
is, when you have

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an indeterminate
exponential form,

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you do this rewriting as an
exponential in the base of e

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trick, and then you pass
the limit into the exponent.

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Because e is now
a nice constant,

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life is simple-- life is
simpler I should say-- you pass

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the limit into
the constant, then

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you have an indeterminate ratio,
indeterminate product, which

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you rewrite as an
indeterminate ratio

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on which 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

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with limits of indeterminate
exponential forms.

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And I'll end there.