lim n-> infinity (cos(4*x/(n^.5)))^n

I can plug this into mathematica and get e^-(8*x^2) but I don't know how that answer was obtained...

Printable View

- Sep 22nd 2009, 06:36 AMsomalim n->infinity for cos function
lim n-> infinity (cos(4*x/(n^.5)))^n

I can plug this into mathematica and get e^-(8*x^2) but I don't know how that answer was obtained... - Sep 22nd 2009, 06:45 AMChris L T521
Note that $\displaystyle \lim_{n\to\infty}\left(\cos\!\left(\frac{4x}{n^{.5 }}\right)\right)^{n}\to 1^{\infty}$

Let $\displaystyle L=\lim_{n\to\infty}\left(\cos\!\left(\frac{4x}{n^{ .5}}\right)\right)^{n}$. Applying natural log to both sides gives us

$\displaystyle \ln L=\lim_{n\to\infty}\frac{\ln\left(\cos\!\left(\fra c{4x}{n^{.5}}\right)\right)}{\frac{1}{n}}=\frac{0} {0}$

Now apply L'Hôpital's rule.

Can you try to take it from here? - Sep 22nd 2009, 07:08 AMsoma
For L'Hôpital's rule, I derived the top and bottom and got 0/0 again, and the second derivative looks daunting. Can you help me further with this?

- Sep 22nd 2009, 07:20 AMChris L T521
After applying L'Hôpital's rule once, we have

$\displaystyle \lim_{n\to\infty}\frac{2x\tan\left(\frac{4x}{n^{.5 }}\right)}{-\frac{1}{n^{.5}}}=\frac{0}{0}$

Applying it another time, we see that

$\displaystyle \lim_{n\to\infty}-\frac{\frac{4x^2}{n^{1.5}}\sec^2\left(\frac{4x}{n^ {.5}}\right)}{\frac{1}{2n^{1.5}}}=\lim_{n\to\infty }-8x^2\sec^2\left(\frac{4x}{n^{.5}}\right)=-8x^2$

Therefore, $\displaystyle \ln L=-8x^2\implies L=\lim_{n\to\infty}\left(\cos\left(\frac{4x}{n^{.5 }}\right)\right)^n=e^{-8x^2}$

Does this make sense?

(Note that when we differentiated, it was with respect to $\displaystyle n$, not $\displaystyle x$). - Sep 22nd 2009, 08:11 AMsoma
Perfect! :)

It took me a while to compute those derivatives myself, but I see how it all fits together now. Thanks so much! - Sep 22nd 2009, 08:12 AMsoma
+ rep

+ thanks

+ donation