I tried using the logarithmic method. But I couldn't work with it as I am not sure we can apply L'Hospital to the indeterminate form -infinity/infinity. Please help me prove the convergence.
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do you mean $$a_n=\left(\sin(1/n)\right)^{1/n}$$ – Dr. Sonnhard Graubner Sep 18 '16 at 13:24
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it is the same for $$n>0$$ – Dr. Sonnhard Graubner Sep 18 '16 at 13:26
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No i meant (1/n)^(sin(1/n)) – Ayush Kumar Sep 18 '16 at 13:31
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For any $x\in(0,1)$ we have $x\sin(1) \leq \sin(x) \leq x$ by concavity, hence $$ \lim_{n\to +\infty}\frac{1}{n^{\sin(1/n)}} = \lim_{n\to +\infty}\sqrt[n]{1/n}=1 $$ holds by squeezing.
Jack D'Aurizio
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Hint. One may write, as $n \to \infty$, $$ \left(\sin \frac1n \right)^{1/n}=e^{\large \frac1n\ln \left(\sin \frac1n \right)}\sim e^{\Large-\frac{\ln n}n}. $$
Olivier Oloa
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The result would follow if you are able to evaluate $$\lim_{x \to 0^+} x^{\sin x}.$$
Taking a logarithm is a good start, since \begin{align} \log (x^{\sin x}) = \sin x \log x = \frac{\log x}{1/\sin x}, \end{align} which is of type $\infty/\infty$. Thus L'Hospital rule applies: \begin{align*} \lim_{x \to 0^+} \frac{\log x}{1/\sin x} = \lim_{x \to 0^+}\frac{1/x}{-\frac{1}{\sin^2 x}\cos x} = -\lim_{x \to 0^+} \frac{\sin^2 x}{x\cos x} = 0. \end{align*} In the last step we used the relation $\sin x \sim x$ as $x \to 0^+$. Thus the answer to the original limit is $e^0 = 1$.
Zhanxiong
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I guess the form we obtain from sinx* logx is -(infinity)/(infinity).Can we still apply L'Hospitals? – Ayush Kumar Sep 18 '16 at 14:51
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