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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.

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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....

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$$0 \leq \lim_{n \rightarrow \infty}\frac{\sqrt[n]{(3n+10)^{10}}}{n}\leq \lim_{n \rightarrow \infty}\frac{\sqrt[n]{(4n)^{10}}}{n} = \lim_{n \rightarrow\infty} \frac{((4n)^{\frac1n })^{10}}{n} =0$$

Note that $\lim_{n \rightarrow \infty} 4^{\frac1n}=1$ and $\lim_{n \rightarrow \infty} n^{\frac1n}=1$

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