Proof of a generic limit with log and exponential How do I prove that

$\lim_{n \to \infty} \frac{\log_a n}{n^b}=0$
If $a&gt;1$, you could substitute $n=a^t$ and the limit becomes

$$\lim_{t\to\infty} \frac{t}{a^{bt}}.$$  

If $0&lt;a&lt;1$, use $n=a^{-t}.$  Now use L'hospital (or series) if that's allowed, or perhaps a concavity argument.
Just change basis to basis $\;e\;$ , say (and assume $\;b&gt;0\;$, as commented):

$$\frac{\log_an}{n^b}=\frac{\frac{\log n}{\log b}}{e^{b\log n}}=\frac{\log n}{\log b\cdot\left(e^b\right)^{\log n}}\xrightarrow[n\to\infty]{}0$$