Question is to find the limit of following as n tends to infinity :
$\lim_\limits{n\to \infty}(\log(1+1/n))^{1/n}$
my attempt:
took expansion of $\log(1+x)$
so
$\lim_{n\to \infty}(\frac{1}{n^{1/n}}(1-\frac{1}{2n}+\frac{1}{3n^2}….)^{1/n}) $ is of form $1^0$ which is not indeterminate and hence limit is 1.
is it right?
Best Answer
\begin{align} & (\log(1 + 1/n))^{1/n} \\ = & \exp\left[\frac{1}{n}\log\left(\log\left(1 + \frac{1}{n}\right)\right)\right] \\ \end{align} So let's use L'Hospital's rule to evaluate the exponent: \begin{align} & \lim_{x \to 0} x\log(\log(1 + x)) \\ = & \lim_{x \to 0} \frac{\frac{1}{(1 + x)\log(1 + x)}}{-\frac{1}{x^2}} \\ = & -\lim_{x \to 0} \frac{x^2}{(1 + x)\log(1 + x)} \\ = & - \lim_{x \to 0} \frac{2x}{1 + \log(1 + x)} \\ = & 0. \end{align} Therefore the original limit is $1$.