How $\lim_{n\to\infty}n\left(\sin x\right)^{2n+1}\cos x$ is equal to 0 In my book, this limit $$\lim_{n\to\infty}n\left(\sin x\right)^{2n+1}\cos x=0$$
is used  to solve an integration problem. But I want to figure out how this limit works out. What special limit is used to derive it or is it itself a special limit. 
By special limits, I mean limits like 
$$\lim_{x\to0}\sin(x)/x=1$$
 A: If $\sin x\ne\pm 1$, then $|\sin x|<1$, and $\lim_{n\to\infty}\sin^nx=0$. Can you get it from there?
A: You simply want to break it into cases:  If $x$ is a multiple of $\pi/2$ the quantity is just 0.  Otherwise, $|\sin x|<1$, and $\cos x$ is just some constant.  The sequence is then something like $na^{2n+1}$ where $-1<a<1$. Now the "special limit" is the fact that $a^{2n+1}$ will be very small compared to $n$.
A: I'm supposing $\sin x\ne\pm 1$ and thus $|\sin x | < 1$
\begin{align}
\lim_{n\to\infty}n\left(\sin x\right)^{2n+1}\cos x&=\cos x \cdot \lim_{n\to\infty}n\left(\sin x\right)^{2n+1}\\ &< \cos x \cdot \lim_{n\to\infty}n(\left(\epsilon \right)^n)^2\\ &= 0
\end{align} since $|\epsilon|<1$. In fact the third passage is justified by noticing that applying L' Hospitals rule you get
$$\lim_{n\to\infty}n \epsilon^n = \lim_{n\to\infty} \frac{n}{\frac{1}{\epsilon^n}} = \lim_{n\to\infty} \frac{1}{\ln \left( \frac{1}{\epsilon} \right)\frac{1}{\epsilon^n}} = \lim_{n\to\infty} \frac{\epsilon^n}{\ln \left( \frac{1}{\epsilon} \right)} = 0$$
