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$$\lim_{x \to \infty} \, \cos \left(\dfrac{1}{x}\right)^{x} $$
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.

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up vote 6 down vote accepted

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 $1$.

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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 $1$.

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Indeed, the Squeeze Theorem requires a squeeze, i.e. two sides to the inequality. – JohnD Apr 30 '14 at 21:25
@JohnD I was going to write what you wrote, but you posted before I did, anyway (+1) to you. – Santosh Linkha Apr 30 '14 at 21:26


$$\cos\left(\frac{1}{x}\right)^x=\exp \left[x \log \cos \left(\frac{1}{x}\right)\right]$$



$$\cos\left(\frac{1}{x}\right)^x=\exp \left[x \log \left( 1-\frac{1}{2x^2}+O\left(\frac{1}{x^4}\right) \right) \right]$$

And since $\log (1-t) = t+O(t^2)$ for $t \rightarrow 0$, $$\cos\left(\frac{1}{x}\right)^x=\exp \left[\frac{x}{2x^2}+O\left(\frac{1}{x^3}\right) \right] \stackrel{x\rightarrow\infty}\longrightarrow 1$$

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Setting $\dfrac1x=2h$ $$\lim_{x \to \infty} \, \cos \left(\dfrac{1}{x}\right)^x=\lim_{h\to0}(\cos2h)^{\dfrac1{2h}}$$


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 $\displaystyle\lim_{h\to0}\dfrac{2\sin^2h}{2h}=\left(\lim_{h\to0}\frac{\sin h}h\right)^2\cdot \lim_{h\to0}h=1^2\cdot0$

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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.

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