### Video Transcript

Determine the limit of negative 11 ln π₯ over negative nine π₯ plus nine as π₯
approaches one.

The first thing we can do is to take the minus sign in front and absorb it into the
denominator. Now, weβve written the function inside the limit as a single fraction, whose
denominator is the additive inverse of the previous denominator.

Now, the first thing we always try is direct substitution. π₯ is approaching one in our limit. And so direct substitution of one for π₯ gives 11 times ln one over nine times one
minus nine. Now, ln one, the natural logarithm of one, is zero. This is because by definition π to the power of ln one is equal to one. And we know that the thing that you raise π to the power of to get one is zero.

Okay, as ln one is zero, 11 times ln one, the numerator is zero. And itβs also straightforward to see that the denominator is zero too. Direct substitution, therefore, gives us zero over zero, which is an indeterminate
form.

This tells us that we need to use some other methods to evaluate this limit. The same problem occurs when trying to use the fact that the limit of a quotient of
functions is the quotient of their limits. The function 11 times ln π₯ is continuous at π₯ equals one. And so the limit of the function as π₯ approaches one can be evaluated by direct
substitution. We get zero in the numerator. And similarly, the limit in the denominator really is zero.

So what went wrong? Well, our fact about the limit of a quotient of functions only applies when the limit
of the denominator function is nonzero. So how do we evaluate this limit? Well, as the limits of the numerator and the denominator are both zero, we can apply
LβHospitalβs rule.

Essentially, LβHospitalβs rule states that if the limits of two functions π of π₯
and π of π₯ as π₯ approaches π are both zero, then the limit of the quotient as π₯
approaches π is equal to the limit of the quotient of the derivatives.

Some terms and conditions apply. π and π must both be differentiable. Otherwise, the right-hand side doesnβt make that much sense. The derivative of the function in the denominator canβt be zero near the limit value
π, but it can be zero at π; thatβs okay. And the right-hand side limit should exist.

Okay, so can we apply our rule with π of π₯ equal to 11 ln π₯ and π of π₯ equal to
nine π₯ minus nine? Well, the first thing to note is that the limit of π of π₯ is equal to zero as is
the limit of π of π₯. Weβve shown that. And so assuring the terms and conditions are satisfied, the limit that weβre looking
for should be the limit of the derivative of 11 ln π₯ over the derivative of nine π₯
minus nine.

Okay, so what is the derivative of 11 ln π₯? Well, the derivative of ln π₯ is one over π₯. And so the derivative of 11 ln π₯ is going to be 11 times this, 11 over π₯. How about the derivative of nine π₯ minus nine? Well, this is more straightforward. This is just nine. We can simplify this compound fraction. 11 over π₯ divided by nine is 11 over nine π₯. And now, this is the limit that we can evaluate using direct substitution. We get 11 over nine times one, which is 11 over nine.

We should just mention here that the terms and conditions of LβHospitalβs rule were
satisfied. Both π which was 11 ln π₯ and π which was nine π₯ minus nine are
differentiable. And π prime of π₯ which was nine is not zero near the limit point one. And finally, as we manage to evaluate the limit on the right-hand side, it
existed. And actually, weβd have settled for it being plus or minus infinity. This limit doesnβt only work when both the limit of the numerator and that of the
denominator are zero, it also works if both limits are infinite.