find $\lim_{n\to \infty}(\log(1+1/n))^{1/n}$ 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?
 A: \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 $1$. 
A: $\ln a_n = \dfrac{\ln(n+1)-\ln n}{n}= \dfrac{\ln (n+1)}{n} - \dfrac{\ln n}{n}\to 0-0=0\to a_n \to 1$
A: Note that $n^{1/n} \to 1$ as $n \to \infty$ and $n\log\left(1 + \dfrac{1}{n}\right) \to 1$ as $n \to \infty$. And therefore $$\left(\log\left(1 + \frac{1}{n}\right)\right)^{1/n} = \dfrac{\left(n\log\left(1 + \dfrac{1}{n}\right)\right)^{1/n}}{n^{1/n}} \to \frac{1^{0}}{1} = 1$$
Update: A simple proof of $n^{1/n} \to 1$ is based on the following theorem:
If $a_{n} > 0$ for all $n$ and $\lim_{n \to \infty}\dfrac{a_{n + 1}}{a_{n}} = L$ then $\lim_{n \to \infty}(a_{n})^{1/n} \to L$.
Simply put $a_{n} = n$ in the above theorem and get $n^{1/n} \to 1$.
A: $$\lim_{n\to \infty}\left[\log\left(1+\frac{1}{n}\right)\right]^{\frac{1}{n}}$$
Let, $\frac{1}{n}=t\implies t\to 0\ as \ n\to \infty$
$$\lim_{t\to 0}\left[\log\left(1+t\right)\right]^{t}$$ $$=exp\left[\lim_{t\to 0}t\log(\log\left(1+t\right))\right]$$ $$=exp\left[\lim_{t\to 0}\frac{\log(\log\left(1+t\right))}{\frac{1}{t}}\right]$$
$$=exp\left[\lim_{t\to 0}\frac{\frac{1}{(1+t)\log(1+t)}}{\frac{-1}{t^2}}\right]$$
$$=exp\left[\lim_{t\to 0}\frac{-t^2}{(1+t)\log(1+t)}\right]$$ $$=exp\left[\lim_{t\to 0}\frac{-2t}{\frac{(1+t)}{1+t}+\log(1+t)}\right]$$
$$=exp\left[\lim_{t\to 0}\frac{-2t}{1+\log(1+t)}\right]$$
$$=exp\left[\frac{-2(0)}{1+\log(1)}\right]$$
$$=e^{0}=1$$
A: We have that
$$\left(\log\left(1+\frac{1}{n}\right)\right)^{\frac{1}{n}}=\left(\frac1n\right)^{\frac{1}{n}}\left(\frac{\log\left(1+\frac{1}{n}\right)}{\frac1n}\right)^{\frac{1}{n}} \to 0 \cdot 1^0=0$$
indeed
$$\left(\frac1n\right)^{\frac{1}{n}}=e^{-\frac{\log n}{n}} \to e^0=1$$
and by standard limits
$$\frac{\log\left(1+\frac{1}{n}\right)}{\frac1n} \to 1$$
