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Stuck with a periodic function limit problem: $$\lim_{ x\to \infty}\cos x^{\frac{1}{\sin x}}$$ That's how I tried to solve it: $$\lim_{ x\to \infty}\cos x^{\frac{1}{\sin x}}=\exp \left(\lim_{x \to \infty } \left(\frac{1}{\sin x}\cdot \ln\left(\cos x\right)\right)\right)$$ Now I have to get rid of the fact that x $\to \infty$. So let $u=\frac{1}{x}, u\to 0:$ $$\exp \left(\lim_{u \to 0 } \left(\frac{1}{\sin\frac{1}{u}}\cdot \ln\left(\cos\frac{1}{u}\right)\right)\right)$$ But $\frac{1}{u}=\infty, $ as $ u \to 0$ and we have $sin(\infty)$. And I don't see the way to solve this problem. Any help/hint will be appreciated

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  • $\begingroup$ Is $\cos x>0$?... $\endgroup$ – Mark Jan 20 '14 at 22:51
  • $\begingroup$ @Mark, It's a periodic function so we can't say whether $cos(x)>0$ when x approaches to infinity. $\endgroup$ – k1ber Jan 20 '14 at 23:12
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    $\begingroup$ His point is that ln(x) is not defined for negative x. $\endgroup$ – Vincent Pfenninger Jan 20 '14 at 23:21
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We shouldn't expect this to have a limit, as both $\sin x$ and $\cos x$ oscillate without end as $x\to\infty$.

Perhaps the easiest way to show that there is no limit is to find to sequences $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ such that $x_n\to\infty$ and $y_n\to\infty$ as $n\to\infty$, but $(\cos x_n)^{1/\sin x_n}$ and $(\cos y_n)^{1/\sin y_n}$ don't have the same limit.

For instance, if we let $x_n=\frac{\pi}{2}+2\pi n$, then $\cos x_n=0$ and $\sin x_n=1$, so that $(\cos x_n)^{1/\sin x_n}\to 0$ as $n\to\infty$.

Alternatively, if $y_n=\frac{\pi}{4}+2\pi n$, then $\cos y_n=\frac{1}{\sqrt{2}}$ and $\sin x_n=\frac{1}{\sqrt{2}}$, so that $(\cos y_n)^{1/\sin y_n}\to 2^{-1/\sqrt{2}}$ as $n\to\infty$.

Since these aren't equal, the limit cannot exist.

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    $\begingroup$ I love your answers, to be honest (no homo). $\endgroup$ – Git Gud Jan 20 '14 at 23:12
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    $\begingroup$ @GitGud :-) Thank you, sir! $\endgroup$ – Nick Peterson Jan 20 '14 at 23:15

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