How to determine $\lim \limits_ {n \rightarrow \infty} \frac{31^ { \frac{n}{5} }}{17^{ \frac{n}{2} } }?$ How to compute

$\lim \limits_ {n \rightarrow \infty} \frac{31^ { \frac{n}{5} }}{17^{ \frac{n}{2} } }?$

Wolframalpha gave me the solution $0$ but I dont get how to determine that.

Hope somebody can help.
$$\lim \limits_ {n \rightarrow \infty} \frac{31^ { \frac{n}{5} }}{17^{ \frac{n}{2} } } = \lim \limits_ {n \rightarrow \infty} \left(\frac{31^ { \frac{1}{5} }}{17^{ \frac{1}{2} } }\right)^n = \lim \limits_ {n \rightarrow \infty} \left(\frac{{\left(31^2\right)}^ { \frac{1}{10} }}{{\left(17^5\right)}^{ \frac{1}{10} } }\right)^n  = \lim \limits_ {n \rightarrow \infty} \left(\frac{{31^2}}{17^5}\right)^{n\over10} = 0 $$

since $\left|\frac{{31^2}}{17^5} \right|&lt; 1$
HINT:

$$31^2=961&lt;(17^2)^2&lt;17^5$$

$$\implies31^{1/5}&lt;17^{1/2}$$

Now use $\lim_{n\to\infty}c^n\to0$ for $|c|&lt;1$