$$\lim_{n \to \infty} \left(\frac{\sin\left(\frac{1}{n^p}\right)}{\frac{1}{n^p}}\right)=1\ (p>0)$$ When it is converted to $$\lim_{n \to \infty} n^p\sin \left(\frac{1}{n^p}\right)$$ which doesn't make it simple to evaluate the value, how to find the limit?
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Where does p belongs ? – Theodoros Mpalis Mar 27 '16 at 08:24
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4Do you know that $\lim_{x\to0}\frac{\sin(x)}{x} = 1$? – kaedit Mar 27 '16 at 08:25
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To have the limit to be $1$ you must suppose $p>0$. – mrprottolo Mar 27 '16 at 08:25
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Also $|\sin(x)|\leq 1$ and $$\sin(x)\leq |x|$$ – Theodoros Mpalis Mar 27 '16 at 08:28
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See the article I mention in my answer. – Mar 27 '16 at 08:41
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If $p>0$ put $x=\frac{1}{n^p}$ then $$\lim_{n \to \infty} \left(\frac{sin\left(\frac{1}{n^p}\right)}{\frac{1}{n^p}}\right)=\lim_{x\to 0}\frac{sin(x)}{x}=1$$
For a nice proof see this article.
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$\dfrac1{n^p}$ tends to zero and you can use a change of variable,
$$\lim_{n \to \infty} \left(\frac{\sin\left(\frac{1}{n^p}\right)}{\frac{1}{n^p}}\right)=\lim_{x\to0}\frac{\sin(x)}x=1.$$
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@canoe: unfortunately, I can't answer that without knowing how your course is organized and how the trigonometric functions are introduced, because there is no single way to establish that limit. – Mar 27 '16 at 08:49
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