# Video: AP Calculus AB Exam 1 • Section I • Part A • Question 20

Find the lim_(ℎ → 0) (ln(7 + ℎ) − ln(7))/ℎ.

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### Video Transcript

Find the limit as ℎ approaches zero of ln of seven plus ℎ minus ln of seven all over ℎ.

If we try and evaluate this limit by direct substitution of ℎ equals zero, we get ln of seven minus ln of seven over zero. And that’s going to give us zero over zero, which is an indeterminate form. This doesn’t really help us to evaluate our limit. So instead, we use L’Hôpital’s rule.

Remember this says that if the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 of 𝑔 of 𝑥 are both zero or the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 of 𝑔 of 𝑥 are positive or negative infinity, then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥. In other words, this limit can be evaluated by finding the limit of the quotient of derivatives. Seen as we got zero over zero, we are in this situation here. So let’s apply L’Hôpital’s rule to our question.

Our numerator is ln of seven plus ℎ minus ln of seven. Now, ln of seven is a constant. So that differentiates to zero. But differentiating ln of seven plus ℎ is a bit trickier. For this, we let 𝑦 equal ln of seven plus ℎ. And we see that this is a function of a function. So we use the chain rule. The chain rule says that if 𝑦 equals 𝑓 of 𝑔 of 𝑥, then d𝑦 by d𝑥 equals d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥. And because we’re differentiating our function 𝑦 with respect to ℎ, we’ll replace 𝑥 in this formula with ℎ. So it makes a bit more sense to us.

Now with 𝑢 being equal to seven plus ℎ, we have that 𝑦 equals ln of 𝑢. And at this point, we recall that the derivative of ln of 𝑥 is one over 𝑥. And so with 𝑦 being equal to ln 𝑢, we have that d𝑦 by d𝑢 equals one over 𝑢. And seen as we let 𝑢 equal seven plus ℎ, d𝑢 by dℎ is the derivative of seven plus ℎ with respect to ℎ. Seven is a constant that differentiates to zero and ℎ differentiates to one.

And now applying the formula for the chain rule, d𝑦 by dℎ equals one over 𝑢 multiplied by one and this is just one over 𝑢. Remember we said that 𝑢 was equal to seven plus ℎ. So d𝑦 by dℎ is equal to one over seven plus ℎ. So now, we have our numerator differentiates to one over seven plus ℎ. When we differentiate our denominator ℎ with respect to ℎ, we just get one.

And so by L’Hôpital’s rule, the limit as ℎ approaches zero of ln of seven plus ℎ minus ln of seven over ℎ is equal to the limit as ℎ approaches zero of one over seven plus ℎ all over one. And anything over one is itself. So this is the limit as ℎ approaches zero of one over seven plus ℎ. And now, by direct substitution of ℎ equals zero, this gives us one over seven.

And before you finish, remember our conditions for L’Hôpital’s rule. We must have that the function on the numerator and the function on the denominator are both differentiable, which they were as we differentiated them. 𝑔 prime of 𝑥 should not be equal to zero. And this was true as the derivative of the denominator was not near zero at our limit point zero. And finally, the limit of the quotient of derivatives must exist, which it did because we found it. And so, one over seven is our final answer.

One final note is that because we were asked to evaluate the limit as ℎ approaches zero, we had to differentiate our numerator function and denominator function with respect to ℎ.