WEBVTT

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PROFESSOR: Hi.

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Welcome back to recitation.

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In lecture, you learned
about some new tools

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for computing certain limits.

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In particular, you learned
about l'Hopital's rule.

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So now you have the
ability to compute easily

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some limits that before might
have been difficult to do.

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So I have here four examples
of limits, some of which--

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or maybe all of
which, or maybe none

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of which or, well definitely
some of which-- will be,

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you could use
l'Hopital's rule on.

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So here they are.

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Why do you pause the video and
take a few minutes to work them

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all out, come back, and we
can work them out together.

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Welcome back.

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So we have these
four limits here.

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And let's start
solving them, shall we?

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So I'll just go from the first
one, and then I'll, you know,

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go through them in order.

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So limit (a) is the limit as x
goes to 1 of x to the a minus 1

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over x to the b minus 1.

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If we want to use
l'Hopital's rule,

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we have to first
check that this is

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a limit of an appropriate
kind of expression.

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So that's, in particular,
a limit that is a 0 over 0

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or infinity over infinity form,
so an indeterminate quotient.

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And as x goes to 1, we see
that x to the a goes to 1.

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So x to the a minus 1 goes to 0.

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And similarly x to the b goes
to 1, so x to the b minus 1

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goes to 0.

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So indeed, this is a 0 over
0 indeterminate form, so we

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can apply l'Hopital's rule.

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So we apply l'Hopital's rule.

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And it says that the limit as x
goes to 1 of x to the a minus 1

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over x to the b minus 1.

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OK, so what do we do?

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What l'Hopital's
rule says is we can

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take the derivative of
the top and the derivative

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of the bottom, and then
the limit of the left side

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will be equal to the
limit of the result

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provided the result actually
has an existing limit, either

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a real number or infinity
or minus infinity.

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So in this case, this
says that this expression

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is equal to the limit-- so
it's the same limit-- as x goes

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1 of-- OK, so the
derivative of the top

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is a x to the a minus 1, and
the derivative of the bottom

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is b x to the b minus 1.

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So l'Hopital's rule says
that these two limits

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are equal provided that
the second limit exists.

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So what is the second limit?

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Well, we-- OK, we can do
some simplifications here.

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Here there wasn't,
you know, a lot

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of obvious simplifying
to be done.

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But here there's an obvious
simplification step.

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So this is equal to the limit
as x goes to 1-- so a over b

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is just a constant, and
now the powers of x, this

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is x to the a minus b.

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But as x goes to 1,
x to the a minus b

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goes to 1 no matter
what a and b are.

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So at this point, this isn't
an indeterminate form anymore.

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This is nice simple limit
we can just plug in.

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So this is just
equal to a over b.

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OK so that's part a,
straightforward application

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of l'Hopital's rule.

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Let's go look it part b.

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So part b asks us
for the limit as x

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goes to 0 of sine 5x over x.

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So I'll put that
right here, maybe

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I'll draw a little border.

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So the limit as x goes to 0
of sine of 5x divided by x.

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So again there's a question
of, you know, is this limit,

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do we know what it is already?

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And the answer is, maybe we do
and maybe we don't, but suppose

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we don't.

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So in that case, can we
apply l'Hopital's rule?

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Well we check, and as x
goes to 0, sine of 5x.

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So as x goes to 0, 5x goes to
0, so sine of 5x goes to 0,

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and x goes to 0.

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So this is another
0 over 0 form.

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So we can apply
l'Hopital's rule.

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So l'Hopital's rule says that
this is equal to the limit

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as x goes to 0 of-- we take
the derivative of the top

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and divide by the
derivative of the bottom.

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So the derivative of the
top is just 5 cosine 5x.

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And the derivative
of the bottom is 1.

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So l'Hopital's rule
says the limit here

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is equal to the limit here
provided this limit exists.

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And now as x goes to 0,
well, OK, 5, 1 is constant.

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So we're just looking at
the limit of cosine 5x here.

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As x goes to 0, that just
approaches 5 times cosine of 0,

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which is 1.

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So this is equal to-- sorry,
the cosine of the 0 part is 1.

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So it's 5 times 1, which is 5.

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Now I should note that we
didn't actually really need

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l'Hopital's rule for this one.

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One thing you can remember is
that this limit is actually

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the definition of the derivative
of sine of 5x at x equals 0.

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So what we just did
with l'Hopital's rule

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was use the derivative to
compute the derivative,

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if you like.

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So we used the fact that we
already knew the derivative

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in order to compute it.

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We could also just
say by definition,

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this is the derivative
of sine 5x at 0,

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which is 5 cosine of 0, which
is 1, which is 5, rather.

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OK so that's part (b).

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Now part (c) asks
for the limit as x

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goes to 0 of x squared
minus 6x plus 2-- did

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I do that right,
yes-- over x plus 1.

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So this limit, you can't
apply l'Hopital's rule?

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Why can't you apply
l'Hopital's rule?

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Well, because it's not
an indeterminate form.

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This limit is really
easy to compute.

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You can just plug
in x equals 0 here.

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The top is going to 2, and
the bottom is going to 1.

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So the whole limit is going
to 2 divided by 1, which is 2.

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So there's no-- not only, you
don't need l'Hopital's rule,

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and also you can't
apply l'Hopital's rule.

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This is not a form in which
l'Hopital's rule applies.

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It's not an indeterminate
form, OK, so,

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but OK, but the result
is very easy to compute.

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So that's 2.

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And we've got our
last one now which

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is (d), the limit as x goes
to infinity of ln of 1 plus e

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to the 3x quantity
divided by 2x plus 5.

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OK, well is it obvious
what this limit is?

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Not to me.

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So OK, so then we can check.

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It's a quotient.

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So can we apply
l'Hopital's rule?

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Well we can apply
l'Hopital's rule

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if it's an
indeterminate quotient.

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So what's happening
is x goes to infinity?

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Well, let's do the bottom
first, because that's easy.

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As x goes to infinity, 2x
plus 5 goes to infinity.

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And, OK, as x goes to infinity,
e to the 3x goes to infinity.

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So 1 plus e to the
3x goes to infinity.

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So log of 1 plus 3 to
the 3x goes to infinity.

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So this is an infinity
over infinity form.

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That's an indeterminate form,
an indeterminate quotient.

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So we can apply
l'Hopital's rule.

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So l'Hopital's
rule says that this

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is equal to the
limit of the quotient

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of the ratio of the
derivatives, provided

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that that limit exists.

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So this is equal to the limit
as x goes to infinity of--

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OK, so we need to look at
the ratio of the derivatives.

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So the bottom one is easy.

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That's 2.

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For the top one, to
compute this derivative,

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we need to use the
chain rule here.

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So you take the derivative of
this, and you get, well, a log.

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So on the bottom we
get 1 plus e to the 3x.

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And then by the chain rule,
up top you get 3 e to the 3x.

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Or, just rewriting this a
little bit in a nicer form,

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this is the limit as
x goes to infinity

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of 3 over 2 times e to the
3x over 1 plus e to the 3x.

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All right, so now the question
is what do we do from here?

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Well is this limit obvious?

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Well, OK, so there are
two situations here.

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One is, you might look at
this and you might already

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know what this limit is.

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The reason you might know that,
is that x goes to infinity,

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e to the 3x is
going to infinity,

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and 1 plus e to the 3x
is going to infinity.

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So this is an infinity over
infinity indeterminate form.

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And so you could apply
l'Hopital's rule to it again.

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OK?

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That's one thing you could do.

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You might also look
at and say well,

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these two things are
going to infinity

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at exactly the same rate.

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Right?

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This plus 1 is almost
totally irrelevant.

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Actually for purposes
of estimating

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the magnitude of this fraction,
it is totally irrelevant.

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When you take-- you
know, when 3x is big,

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then e to the 3x is like
a billion or something.

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So e to 3x is a billion.

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So you have a billion
over a billion and 1.

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So it's very, very close to 1.

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It's going to get
closer and closer to 1.

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So, OK, so if you didn't
trust that analysis

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based on magnitudes,
I mean the idea here

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is just that the most
significant term in the top

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is the e to the 3x.

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The most significant term on
the bottom is e to the 3x.

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They dominate everything
else that appears,

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which is just this little 1.

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So that's going to
give you a ratio of 1.

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If you don't trust
that analysis,

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you can apply
l'Hopital's rule again,

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and l'Hopital's rule will tell
you exactly the equivalent

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thing.

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When you take a
derivative, you'll

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be left with 3 e to the 3x over
3 e to the 3x, which is just 1.

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So in either case,
this part goes to 1.

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So 3/2 times it
goes right to 3/2.

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All right, so there you have it.

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Four limits computed
with l'Hopital's rule.

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Well one of them was computed
without l'Hopital's rule.

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Remember, you always have
to check that you actually

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have an indeterminate
form before you

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apply l'Hopital's rule.

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And that's all I have
to say about that.

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So I'll end there.