1
$\begingroup$

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

$\endgroup$
6
$\begingroup$

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 $1$.

$\endgroup$
2
$\begingroup$

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 $1$.

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.