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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.