Show work. This is a BC calculus problem.

Answer:
D.
Step-by-step explanation:
Recall that the limit definition of a derivative at a point is:
[tex]\displaystyle{\frac{d}{dx}[f(a)]= f'(a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}[/tex]
Hence, if we let f be ln(x + 1) and a be 1, this yields:
[tex]\displaystyle f'(1)= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}[/tex]
Hence, the limit is equivalent to the derivative of f at x = 1 or f’(1).
In conclusion, our answer is D.