Limit:$ \lim\limits_{n\rightarrow\infty}\left ( n\bigl(1-\sqrt[n]{\ln(n)} \bigr) \right )$ I find to difficult to evaluate with $$\lim_{n\rightarrow\infty}\left ( n\left(1-\sqrt[n]{\ln(n)} \right) \right )$$ I tried to use the fact, that $$\frac{1}{1-n} \geqslant \ln(n)\geqslant 1+n$$
what gives $$\lim_{n\rightarrow\infty}\left ( n\left(1-\sqrt[n]{\ln(n)} \right) \right ) \geqslant \lim_{n\rightarrow\infty} n(1-\sqrt[n]{1+n}) =\lim_{n\rightarrow\infty}n *\lim_{n\rightarrow\infty}(1-\sqrt[n]{1+n})$$ $$(1-\sqrt[n]{1+n})\rightarrow -1\Rightarrow\lim_{n\rightarrow\infty}\left ( n\left(1-\sqrt[n]{\ln(n)} \right) \right )\rightarrow-\infty$$Is it correct? If not, what do I wrong?

Use Taylor!
$$n(1-\sqrt[n]{\log n}) = n (1-e^{\frac{\log\log n}{n}}) \approx n\left(1-\left(1+\frac{\log\log n}{n}\right)\right) = - \log\log n$$
which clearly tends to $-\infty$.
