L'Hopital's Rule Example 3

We want to figure out the limit
as x approaches 1 of the

expression x over x minus
1 minus 1 over the

natural log of x.

So let's just see what
happens when we just

try to plug in the 1.

What happens if we evaluate
this expression at 1?

Well then, we're going to get
a one here, over 1 minus 1.

So we're going to get something
like a 1 over a 0, minus 1

over, and what's the
natural log of 1?

e to the what power
is equal to one?

Well, anything to the zeroth
power is equal to 1, so e to

the zeroth power is going to
be equal to 1, so the natural

log of 1 is going to be 0.

So we get the strange,
undefined 1 over

0 minus 1 over 0.

It's this bizarre-looking
undefined form.

But it's not the indeterminate
type of form that we looked

for in l'Hopital's rule.

We're not getting a 0 over
a 0, we're not getting an

infinity over an infinity.

So you might just say, hey,
OK, this is a non-l'Hopital's

rule problem.

We're going to have to figure
out this limit some other way.

And I would say, well
don't give up just yet!

Maybe we can manipulate this
algebraically somehow so that

it will give us the l'Hopital
indeterminate form, and then

we can just apply the rule.

And to do that, let's just
see, what happens if we

add these two expressions?

So if we add them, so this
expression, if we add it, it

will be, well, the common
denominator is going to be x

minus 1 times the
natural log of x.

I just multiplied
the denominators.

And then the numerator is going
to be, well, if I multiply

essentially this whole term by
natural log of x, so it's going

to be x natural log of x, and
then this whole term I'm going

to multiply by x minus one.

So minus x minus 1.

And you could break it apart
and see that this expression

and this expression
are the same thing.

This right here, that right
there, is the same thing as x

over x minus 1, because the
natural log of x's cancel out.

Let me get rid of that.

And then this right here is the
same thing as 1 over natural

log of x, because the x
minus 1's cancel out.

So hopefully you realize,
all I did is I added

these two expressions.

So given that, let's see what
happens if I take the limit as

x approaches 1 of this thing.

Because these are
the same thing.

Do we get anything
more interesting?

So what do we have here?

We have one times the
natural log of 1.

The natural log of 1 is 0, so
we have 0 here, so that is a 0.

Minus 1 minus 0, so that's
going to be another 0, minus 0.

So we get a 0 in the numerator.

And in the denominator we get a
1 minus 1, which is 0, times

the natural log of 1, which is
0, so 0 times 0, that is 0.

And there you have it.

We have indeterminate form that
we need for l'Hopital's rule,

assuming that if we take the
derivative of that, and put it

over the derivative of that,
that that limit exists.

So let's try to do it.

So this is going to be equal
to, if the limit exists, this

is going to be equal to the
limit as x approaches 1.

And let's take the derivative
in magenta, I'll take

the derivative of this
numerator right over here.

And for this first term,
just do the product rule.

Derivative of x is one, and
then so 1 times the natural log

of x, the derivative of the
first term times

the second term.

And then we're going to have
plus the derivative of the

second term plus 1 over
x times the first term.

It's just the product rule.

So 1 over x times x, we're
going to see, that's just 1,

and then we have minus the
derivative of x minus 1.

Well, the derivative of x minus
1 is just 1, so it's just

going to be minus 1.

And then, all of that is over
the derivative of this thing.

So let's take the derivative
of that, over here.

So the derivative of the first
term, of x minus 1, is just 1.

Multiply that times the second
term, you get natural log of x.

And then plus the derivative of
the second term, derivative

of natural log of x is one
over x, times x minus 1.

I think we can simplify
this a little bit.

This 1 over x times
x, that's a 1.

We're going to
subtract one from it.

So these cancel
out, right there.

And so this whole expression
can be rewritten as the limit

as approaches 1, the numerator
is just natural log of x, do

that in magenta, and the
denominator is the natural log

of x plus x minus 1 over x

So let's try to evaluate
this limit here.

So if we take x approaches one
of natural log of x, that

will give us a, well,
natural log of 1 is 0.

And over here, we get natural
log of 1, which is 0.

And then plus 1 minus 1 over
plus 1 minus 1 over 1, well,

that's just going
to be another 0.

1 minus 1 is zero.

So you're going to
have 0 plus 0.

So you're going to get
a 0 over 0 again.

0 over 0.

So once again, let's apply
l'Hopital's rule again.

Let's take the derivative
of that, put it over

the derivative of that.

So this, if we're ever going to
get to a limit, is going to be

equal to the limit as x
approaches 1 of the derivative

of the numerator, 1 over x,
right, the derivative of ln of

x is 1/x, over the derivative
of the denominator.

And what's that?

Well, derivative of natural
log of x is 1 over x plus

derivative of x minus 1 over x.

You could view it this way,
as 1 over x times x minus 1.

Well, derivative of x to the
negative 1, we'll take the

derivative of the first one
times the second thing, and

then the derivative of the
second thing times

the first thing.

So the derivative of the first
term, x to the negative 1, is

negative x to the negative 2
times the second term, times x

minus 1, plus the derivative of
the second term, which is

just 1 times the first
term, plus 1 over x.

So this is going to be equal
to, I just had a random thing

pop up on my computer.

Sorry for that little
sound, if you heard it.

But where was I?

Oh, let's just simplify
this over here.

We were doing our
l'Hopital's rule.

So this is going to be equal
to, let me, this is going to be

equal to, if we evaluate x as
equal to 1, the numerator is

just 1/1, which is just 1.

So we're definitely not going
to have an indeterminate or

at least a 0/0 form anymore.

And the denominator is going to
be, if you evaluate it at 1,

this is 1/1, which is 1, plus
negative 1 to the negative 2.

So, or you say, 1 to the
negative 2 is just 1, it's

just a negative one.

But then you multiply that
times 1 minus 1, which is

0, so this whole term's
going to cancel out.

And you have a plus
another 1 over 1.

So plus 1 And so this is
going to be equal to 1/2.

And there you have it.

Using L'Hopital's rule and a
couple of steps, we solved

something that at least
initially didn't look

like it was 0/0.

We just added the 2 terms, got
0/0, took derivatives of the

numerators and the denominators
2 times in a row to

eventually get our limit.