WEBVTT
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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.
00:10:54.950 --> 00:10:57.160
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
00:11:23.990 --> 00:11:26.020
on which you can then
apply l'Hopital's rule.
00:11:26.020 --> 00:11:29.730
Or possibly, you know, you
rewrite it as a product.
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Then it's easy to see
what the value is.
00:11:34.270 --> 00:11:35.900
So, all right, so
that's how we deal
00:11:35.900 --> 00:11:38.860
with limits of indeterminate
exponential forms.
00:11:38.860 --> 00:11:40.601
And I'll end there.