# Show that $\lim\limits_{n \to \infty} \frac {\sqrt[n]{(3n+10)^{10}}} {2n} = 0$

By just looking at it I'd say that $\lim\limits_{n \to \infty} \frac {\sqrt[n]{(3n+10)^{10}}} {2n}$ is $0$. However I do not know a way to show that. I kind of feel like there is a rule I don't know. A hint would be nice.

Squeeze theorem is your friend. For sufficiently large $n$, we have:
$$0 < \frac{ \sqrt[n]{(3n+10)^{10}}}{2n} < \frac{ \sqrt[n]{(3n+10)^{n/2}}}{2n} = \frac{\sqrt{3n+10}}{2n}$$
There are a number of ways to approach calculating $\displaystyle \lim_{n \rightarrow \infty} \frac{ \sqrt{3n+10}}{2n}$. For instance, this is an indeterminant form....
$$0 \leq \lim_{n \rightarrow \infty}\frac{\sqrt[n]{(3n+10)^{10}}}{n}\leq \lim_{n \rightarrow \infty}\frac{\sqrt[n]{(4n)^{10}}}{n} = \lim_{n \rightarrow\infty} \frac{((4n)^{\frac1n })^{10}}{n} =0$$
Note that $\lim_{n \rightarrow \infty} 4^{\frac1n}=1$ and $\lim_{n \rightarrow \infty} n^{\frac1n}=1$