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

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  • $\begingroup$ Can you use that $\sum_{i=1}^\infty iy^i<\infty$ for $|y|<1$? $\endgroup$ – sinbadh Jan 13 '16 at 20:21
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If $\sin x\ne\pm 1$, then $|\sin x|<1$, and $\lim_{n\to\infty}\sin^nx=0$. Can you get it from there?

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  • $\begingroup$ The limits for x in the integration problem I suggested in the question were 0to π/2 $\endgroup$ – Onix Jan 14 '16 at 6:12
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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 $n$.

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  • $\begingroup$ Thanks @mathematician btw very nicely picked username. Suits you $\endgroup$ – Onix Jan 14 '16 at 6:05
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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 $$\lim_{n\to\infty}n \epsilon^n = \lim_{n\to\infty} \frac{n}{\frac{1}{\epsilon^n}} = \lim_{n\to\infty} \frac{1}{\ln \left( \frac{1}{\epsilon} \right)\frac{1}{\epsilon^n}} = \lim_{n\to\infty} \frac{\epsilon^n}{\ln \left( \frac{1}{\epsilon} \right)} = 0$$

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