Hi,
How can i solve this next limit?
lim (cos(x/√n))^n as n->infinity
Thanks for any help...
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Hi,
How can i solve this next limit?
lim (cos(x/√n))^n as n->infinity
Thanks for any help...
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:
$\displaystyle \lim_{\n\to\infty}cos^n(\frac{x}{\sqrt{n}})=e^{\li m_{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})}=e^{\fr ac{-x^2}{2}}$