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What is the limit of $n {(\sin(x))^{2n+1}} \cos(x)$ as $n$ goes to infinity?
The value of $\sin(x)$ and $\cos (x)$ are between $-1$ and $1$ so the limit is $\infty$?

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    You need to look at it much more closely than that. For nearly every value of $x$ you have $\sin^{2n+1}(x)$ is **incredibly** close to zero or equal to zero and for the other values of $x$ you have $\cos(x)=0$, so you are in an $\infty \cdot 0$ situation where simple plug-and-see does not work. – 2017-02-06

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For every $q\in(-1,1)$ and $a\in\mathbb{R}$, we have $\lim_{n\to\infty}n^aq^n=0$.

This can be proved, for example, by considering the series $\sum_{n\ge1}n^aq^n$ and applying the ratio test :

$$\left|\frac{(n+1)^aq^{n+1}}{n^aq^n}\right|=\left(1+\frac{1}{n}\right)^a\vert q\vert\to\vert q\vert<1$$ This shows that the series converges, hence the conclusion.

Now, if $\cos(x)\neq0$, we see that $\sin(x)\in(-1,1)$ and we can aply the above result. And if $\cos(x)=0$, the result is obvious. In all cases :

$$\lim_{n\to\infty}n\sin^{2n+1}(x)\cos(x)=0$$