calculus - Compute $limlimits_{ntoinfty}sqrt[n]{logleft|1+left(frac{1}{ncdotlog n}right)^kright|}$.

Compute $$\lim\limits_{n\to\infty}\left(\sqrt[n]{\log\left|1+\left(\dfrac{1}{n\cdot\log\left(n\right)}\right)^k\right|}\right).$$
What I have: $$\forall\ x\geq 0\ :\ x- \frac{x^2}{2}\leq \log(1+x)\leq x.$$

Apply to get that the limit equals $1$ for any real number $k$.

Is this correct? Are there any other proofs?

Answer

Yes it works, here's another proof using a little more sofisticate tool (in this case unnecessary, but sometimes more useful).

By Stolz-Cesaro if $(x_n)$ is a positive sequence and
$$\lim_n \dfrac{x_{n+1}}{x_n} = l$$
then
$$\lim_n \sqrt[n]{x_n} = l.$$

Taking as $(x_n)$ the sequence you defined, an easy calculation shows that $$\dfrac{x_{n+1}}{x_n} \rightarrow 1,$$
therefore the thesis.