1. Limit computation

Hi,
How can i solve this next limit?

lim (cos(x/√n))^n as n->infinity

Thanks for any help...

2. Originally Posted by rebecca
Hi,
How can i solve this next limit?

lim (cos(x/√n))^n as n->infinity

Thanks for any help...
$\displaystyle \lim\limits_{n\to\infty}\cos^n\left(\frac{x}{\sqrt {n}}\right)\to 1^{\infty}$, an indeterminate form.

What you'll need to do at this stage is take natural logs of both sides, and then apply L'Hopitals rule.

Can you proceed?

3. I will continue from the point where mr. chris stopped...

$\displaystyle \lim_{\n\to\infty}cos^n(\frac{x}{\sqrt{n}})=e^\lim _{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})$
(the last pass is from continuous of function e)

Now, lets concentrate only in power of e.

$\displaystyle lim_{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})=lim_{ \n\to\infty}nlncos(\frac{x}{\sqrt{n}})$

And now we do a substitution: $\displaystyle n=\frac{1}{t}$.

$\displaystyle lim_{t\to 0}\frac{lncos({x}\sqrt{t})}{t}$
So we got a $\displaystyle \frac{0}{0}$ form.

Hence we can use l'Hôpital's rule.

$\displaystyle lim_{t\to 0}\frac{-xsin(x\sqrt(t))}{2cos(x\sqrt(t))\sqrt(t)}$

So we got again a $\displaystyle \frac{0}{0}$ form.

Hence we can use again l'Hôpital's rule.

$\displaystyle lim_{t\to 0}\frac{\frac{-x^2cos(x)\sqrt{t}}{2\sqrt{t}}}{\frac{-(xsin(x\sqrt(t))\sqrt(t)-cos(x\sqrt(t)))}{\sqrt{t}}}=\frac{-x^2}{2}$

So:

$\displaystyle \lim_{\n\to\infty}cos^n(\frac{x}{\sqrt{n}})=e^{\li m_{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})}=e^{\fr ac{-x^2}{2}}$