# How do we find the limit of periodic function as x approaches to infinity $\left(\cos x^{\frac{1}{\sin x}}\right)$

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

• Is $\cos x>0$?... – Mark Jan 20 '14 at 22:51
• @Mark, It's a periodic function so we can't say whether $cos(x)>0$ when x approaches to infinity. – k1ber Jan 20 '14 at 23:12
• His point is that ln(x) is not defined for negative x. – Vincent Pfenninger Jan 20 '14 at 23:21

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 $n\to\infty$.