# 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 $$\lim_{x\to0}\sin(x)/x=1$$

• Can you use that $\sum_{i=1}^\infty iy^i<\infty$ for $|y|<1$? – sinbadh Jan 13 '16 at 20:21

If $\sin x\ne\pm 1$, then $|\sin x|<1$, and $\lim_{n\to\infty}\sin^nx=0$. Can you get it from there?
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 $n$.
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 $$\lim_{n\to\infty}n \epsilon^n = \lim_{n\to\infty} \frac{n}{\frac{1}{\epsilon^n}} = \lim_{n\to\infty} \frac{1}{\ln \left( \frac{1}{\epsilon} \right)\frac{1}{\epsilon^n}} = \lim_{n\to\infty} \frac{\epsilon^n}{\ln \left( \frac{1}{\epsilon} \right)} = 0$$