$\lim\limits_{x\to\infty}\left(\cos\left(\frac{1}{x}\right)\right)^x$

Question

I want to find $\lim\limits_{x\to\infty}\left(\cos\left(\frac{1}{x}\right)\right)^{\!x}$. I'd like to use the fact that $$\lim_{x\to\infty} e^{\ln(\cos(1/x)^x)}$$ but I am not sure what to do after this.

Answer 1

$$\ln\left[\cos(1/x)^{x}\right] =x\ln(\cos(1/x))=x\ln\left(1-\frac1{2x^2}+O(x^{-4})\right) =-\frac1{2x}+O(x^{-3})\to0$$ as $x\to\infty$. So your limit is $1$.

Answer 2

\begin{align*} \lim_{x \rightarrow \infty} x\log\left(\cos \frac{1}{x}\right) &= \lim_{x \rightarrow \infty}x \frac{\log \cos \frac{1}{x} -1 + 1}{ \cos \frac{1}{x} -1}\left(\cos \frac{1}{x} - 1 \right) \\ &=\lim_{x \rightarrow \infty} \frac{-2\sin ^2 \frac{1}{2x}}{x\frac{1}{x^2}} \\ &= 0 \end{align*} Hence the required limit is $e^0 = 1$.

Answer 3

Taylor expanding $\cos 1/x$ since as $x$ get's large, the argument of $\cos$ is small, your limit is $$ \lim_{x\rightarrow \infty}(1-\frac{1}{2x^2})^x=L $$ Then $$ \log L=\lim_{x\rightarrow \infty}x\log(1-\frac{1}{2x^2})\stackrel{\text{L'Hôpital's}}{=}\lim_{x\rightarrow \infty}\frac{2}{1/x-2x}=0 $$ which then implies that $L=e^0=1$

Answer 4

You are off to a good start!
Note that limits can pass through continuous functions, so starting from what you had we get: ${\displaystyle \lim_{x\to\infty}e^{\ln(\cos(1/x))^{x}}=e^{{\displaystyle \lim_{x\to\infty}}\ln(\cos(1/x))^{x}}}=e^{{\displaystyle \lim_{x\to\infty}}x\cdot \ln(\cos(1/x))}= e^{{\displaystyle \lim_{x\to\infty}}x \cdot \ln(\cos(1/x))\cdot \frac{1/x}{1/x}}=e^{{\displaystyle \lim_{x\to\infty}} \frac{\ln(\cos(1/x))}{1/x}}$

Now we can apply L'Hopital's Rule to this limit to get:

$= e^{{\displaystyle \lim_{x\to\infty}}\left( \frac{\frac{1}{\cos(1/x)}\cdot \sin(1/x)\cdot x^{-2}}{-x^{-2}}\right)} =e^{{\displaystyle \lim_{x\to\infty}} -\tan(1/x)}= e^{-\tan\left({\displaystyle \lim_{x\to\infty}}\frac{1}{x}\right)}=e^{-\tan(0)}=e^{-0}=1$