Limit computation

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Jun 24th 2010, 08:57 AM
rebecca

Limit computation

Hi,
How can i solve this next limit?

lim (cos(x/√n))^n as n->infinity

Thanks for any help...
Jun 24th 2010, 09:00 AM
Chris L T521

Quote:

Originally Posted by rebecca

Hi,
How can i solve this next limit?

lim (cos(x/√n))^n as n->infinity

Thanks for any help...

$\lim\limits_{n\to\infty}\cos^n\left(\frac{x}{\sqrt {n}}\right)\to 1^{\infty}$ , 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?
Jun 24th 2010, 11:22 AM
Also sprach Zarathustra

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

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

Now, lets concentrate only in power of e.

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

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

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

Hence we can use l'Hôpital's rule.

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

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

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

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

So:

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