Question

Show work. This is a BC calculus problem.

Answer 1

Answer:

D.

Step-by-step explanation:

Remember that the limit definition of a derivative at a point is:

$\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}$

Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:

$\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}$

Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).

The answer will thus be D.