limit of $\lim_{n\to \infty}-\frac{1}{n}(\log(\frac{1}{n}))$
Solution 1:
With $x:=\ln n$ this becomes $\lim_{x\to\infty}xe^{-x}=0$, because $\int_0^\infty xe^{-x}dx=1$ is finite, with a continuous non-negative integrand.
Solution 2:
$$-\frac{1}{n}\ln\left(\frac{1}{n}\right) = \frac{\ln(n)}{n} \to 0 ~~~~~~~ \text{as}~~~ n\to +\infty$$
For example you can use De L'Hôpital rule, or simply the hierarchy of infinities.
(The property I have used is simply $\ln\left(\frac{1}{n}\right) = -\ln(n)$).