Find, with proof, the following limit $$\lim_{x \to \infty} \, \cos \left(\dfrac{1}{x}\right)^{x} $$
So with this type of limit, does the value cos(1/x) take priority of the power of x as $x \rightarrow \infty$ ?
I checked it on wolfram and found the limit to be 1;
So would you realise that cos(1/x) as $x \rightarrow \infty$ goes to 1, and $1^{x}$ as $x \rightarrow \infty$ is just 1?
Any help on the correct approach would be appreciated.
 A: Let $y=\cos(1/x)^x\implies \ln y=x\ln(\cos(1/x))\to \infty\cdot 0$ as $x\to\infty$, but this is an indeterminate form.
Rewrite as $x\ln(\cos(1/x))={\ln(\cos(x^{-1}))\over x^{-1}}$ and apply L'Hopital's Rule to obtain 
$$
\lim_{x\to\infty}{\ln(\cos(x^{-1}))\over x^{-1}}=\lim_{x\to\infty} {1/\cos(x^{-1})\cdot (-\sin(x^{-1}))\cdot (-x^{-2})\over -x^{-2}}=\lim_{x\to\infty} -\tan(x^{-1})=0.
$$
That is, $\ln y\to 0$ as $x\to\infty$ which implies $y\to 1$ as $x\to\infty$. Thus, the limit you seek is $1$.
A: For $-\frac \pi 2 < \frac 1 x < \frac \pi 2 $
$$\lim_{x\to\infty}\left( 1 - \frac{1}{x^2}\right)^x \le \lim_{x\to\infty}\cos\left(\frac 1 x\right)^{x} \le 1^x = 1$$
The left side 
$$\lim_{x\to\infty}\left( 1 - \frac{1}{x^2}\right)^x = \lim_{x\to \infty} \left(\left( 1 - \frac 1 {x^2} \right)^{x^2 } \right)^{\frac 1 x}= e^{-1 \cdot 0} = 1$$
By Squeeze theorem, the limit is $1$.
A: Setting $\dfrac1x=2h$ 
$$\lim_{x \to \infty} \, \cos \left(\dfrac{1}{x}\right)^x=\lim_{h\to0}(\cos2h)^{\dfrac1{2h}}$$ 
$$=\left[\lim_{h\to0}\left(1+(-2\sin^2h)\right)^{\frac1{-2\sin^2h}}\right]^{\left(-\lim_{h\to0}\dfrac{2\sin^2h}{2h}\right)}$$
Now the inner limit converges to $e$ as $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n=e=\lim_{u\to0}(1+u)^{\frac1u}$
The limit in the exponent $\displaystyle\lim_{h\to0}\dfrac{2\sin^2h}{2h}=\left(\lim_{h\to0}\frac{\sin h}h\right)^2\cdot \lim_{h\to0}h=1^2\cdot0$
A: Write
$$\cos\left(\frac{1}{x}\right)^x=\exp \left[x \log \cos \left(\frac{1}{x}\right)\right]$$
Then,
$$\cos\left(\frac{1}{x}\right)=1-\frac{1}{2x^2}+O\left(\frac{1}{x^4}\right)$$
$$\cos\left(\frac{1}{x}\right)^x=\exp \left[x \log \left( 1-\frac{1}{2x^2}+O\left(\frac{1}{x^4}\right) \right) \right]$$
And since $\log (1-t) = t+O(t^2)$ for $t \rightarrow 0$,
$$\cos\left(\frac{1}{x}\right)^x=\exp \left[\frac{x}{2x^2}+O\left(\frac{1}{x^3}\right) \right] \stackrel{x\rightarrow\infty}\longrightarrow 1$$
A: You can get a lower approximation by starting with a series expansion for cos(1/x). This gives (1-1/(2x^2))^x. Write this as exp(x log(1-1/(2x^2))). Now approximate the log to first order and you get exp (x (-1/(2x^2))) or exp(-1/(2x)). This of course goes to 1 as x->infinity.
