Your $n^{-1}$ isn't correct when it appears.

The limit $\displaystyle \lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0$ where $a > 0$ and $b > 0$ is a standard result that is often simply quoted. Whether you are allowed to do that or not depends on what you have been given (or proved) as results.

\begin{alignat}{2}

0 &< \frac1t &&< t^{c-1} &&& (c > 0,\, t> 1) \\

\int_1^x 0 \,\mathrm dt &< \int_1^x \frac1t \,\mathrm dt &&< \int_1^xt^{c-1}\,\mathrm dt &&&(x > 1) \\

0 &< \ln x &&< x^c - \frac1c &&< x^c \\

0 &< \frac{\ln^a x}{x^b} &&< x^{ac-b} &&& (a > 0, \, b > 0) \quad \text{raising to the power $a$ and dividing by $x^b$}\\

\end{alignat}

With $c=\frac{b}{2a} > 0$ we have $ac-b=-\frac{b}2 < 0$ and so, the limit of the left- and right-hand expressions as $x \to \infty$ is zero and thus the limit $$\lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0 \qquad (a > 0, \, b > 0)$$