The value of the limit as x approaches 0 of 1
over x to the third times the definite integral from 0
to x of t natural log of 1 plus t, all
of that over t to the fourth plus 4dt is.
Now, when you see an integral like this,
you're first temptation is to try to solve the integral,
try to find the antiderivative of this expression here.
And you might try to experiment maybe with inverse trig
functions, maybe arctangent looks tempting over here.
But you have this natural log.
Maybe you can do it with integration by parts.
But very quickly, or maybe not so very quickly,
you'll realize that this is a nearly impossible integral
to simply solve using analytical methods.
But, lucky for us, this problem isn't just
about solving this integral.
It's actually about evaluating this a limit.
And if we rearrange this expression a little bit,
it might become a little obvious to you
how we can approach this problem.
So this is the limit as x approaches 0 of.
Let me write this in the numerator,
because this is could be multiplying by 1 over x
to the third is the same thing as dividing by x to the third.
So this is the integral from 0 to x of t natural log of 1
plus t, all of that over t to the fourth plus 4dt.
And all of that over x to the third.
And now, this is probably ringing bells
in some of your brains.
And what we see here is we don't know
how to evaluate this integral over here.
But what does this integral approach as x approaches 0?
What is the limit of this numerator expression
as it approaches 0?
So let me just write it.
The limit as it approaches 0, so the limit
as x approaches 0 of 0 to x, all of that stuff inside dt
is going to equal the integral from 0.
It's going to approach the integral
from 0 to 0 of that stuff dt.
And a definite integral from 0 to 0,
we're not moving at all along the horizontal or the t axis.
And so, this is just going to be 0.
And if we do the same thing in the denominator,
that's much simpler.
The limit as x approaches 0 of x to the third, that's
clearly also going to be 0.
So we took the limits of the numerator and denominator
and we got 0/0.
That's the indeterminate form.
And so now the bells should be ringing very loudly
in your brain and screaming L'Hopital's rule.
And I've done several videos on this in the calculus playlist.
But this just tells us if we evaluate
the limit of the numerator as it approaches
0 and the limit of the denominator approaches 0,
and we get an indeterminate form,
then this limit is going to be the exact same thing,
assuming the limit exist, of the limit
as x approaches 0 of the derivative of this expression
over the derivative of this expression.
Now let's do the denominator first, since that's easier.
What's the derivative of x to the third?
Well, that's just 3x squared.
And what's the derivative of the numerator?
And just as a reminder, and I'm just
going to break out the fundamental theorem
of calculus.
The derivative with respect to x from some constant to x
of f of t dt is just equal to f of x.
You're undoing the integration here.
You just get f of x, it's just the function with respect to x.
So the derivative of this with respect to x
is just going to be this function with respect
to x instead of t.
So it's going to be x natural log-- that
was an ugly natural log-- of 1 plus x,
all of that over x to the fourth plus 4.
Now, when we try to evaluate the limit here, what happens?
If we do the numerator, if we do the same thing we did up here,
this numerator still evaluates to 0.
You get 0 times natural log of 1.
That's 0 as well.
So that's all 0.
The denominator still evaluates to 0.
So we have indeterminate form again.
So we might be tempted to just go straight, do L'Hopital's
rule again, take the derivative of the numerator
and then take the derivative of the denominator.
And that would work.
Or maybe you'll see that this is a pretty hairy thing
to take a derivative of.
And maybe we might have to take a derivative again after that.
So this could get very, very, very hairy.
So, instead of just straight up just
taking a derivative of the numerator and derivative
of the denominator again, let's see
if we can simplify this a little bit.
And we can.
So this is the same thing.
This is equal to the limit as x approaches 0.
I'm just going to put the x to the fourth
plus 4 in the denominator.
So it's x natural log of 1 plus x, all of that over this 3x
squared.
And if I'm dividing by something and dividing by something else,
that's just like dividing by that first thing.
So this is x to the fourth plus 4.
I just put this in the denominator,
and we can simplify this.
We can divide the numerator and denominator by x.
And now, let's try to evaluate this limit.
As x approaches 0 in the numerator,
we get natural log of 1.
So that's 0.
So this is approaching 0 over.
And if we do it in the denominator,
we have 3 times 0 times 4.
So that's going to be 0.
Once again, indeterminate form.
So now let's apply L'Hopital's rule again.
But we can apply it on a simpler expression,
on natural log of 1 plus x over 3x times x
to the fourth plus 4.
So let's do that.
So this is going to be equal to, assuming the limit
exists, this is equal to the limit
as x approaches 0 of the derivative of the numerator.
The derivative of natural log of 1 plus x, that's just 1 over 1
plus x over the derivative of the denominator.
We just have to do a little product rule here.
Derivative of the first function is derivative of 3x
is just 3 times the second function, which
is x to the fourth plus 4 plus the first function, 3x, it's
just the product rule, times the derivative
of the second function.
The derivative of this is 4x to the third.
Now, when we approach 0, what happens to this value?
What is the numerator and the denominator approach?
The numerator approaches 1 over 1 plus 0, or just 1.
The denominator approaches 3 times 0
plus 4 plus this is going to be 0, plus 0.
So this is equal to 1/12.
And we're done.
This crazy looking limit with this integral
that we couldn't solve is just equal to 1/12.
So thank goodness we had L'Hopital's rule
so we didn't have to actually take
the antiderivative of this expression.