Limits of Functions

By L'Hospital's Rule , $\frac{d}{dx}(\frac{\ln(n+1)}{n})=(\frac{1}{n+1})/1$ . Now substituting with $n=0$ we get the required limit as $\boxed{1}$

1ª.-

ln(n +1) ~ n when n $\rightarrow$ 0 $\Rightarrow$ $\lim_{n\to 0} \dfrac{ln(n +1)}{n} = 1$

2ª.-

f(x) = ln (1 + x) is a derivative function at x = 0, f(0) = ln 1 = 0 $\Rightarrow \lim_{n\to 0} \dfrac{ln(n +1)}{n} = \lim_{n\to 0} \dfrac{ln(n +1) - ln(1)}{n - 0} = f ' (0) = \frac{1}{1 + 0} = 1$