What does lim x→1 ln(x+1)−ln2 x−1 represent? (A) 0 (B) d dx [ln(x + 1)] (C) f ′ (1), if f(x) = ln(x + 1) (D) 1 (E) The limit does not exist.

Answer 1

f'(c) = lim_x→c [(f(x) - f(c)) / (x - c)]

So, lim_x→1 [(ln(x+1) - ln2) / (x - 1)] = f'(1), where f(x) = ln(x+1). (ANSWER C)