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...
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?
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 $\displaystyle x$).