show that $\lim_{n\to\infty}(\cos\frac{1}{n})^{n^{2}} = 1$ I need to find limit of $\lim_{n\to\infty}(\cos\frac{1}{n})^{n^{2}}$.
I tried going like this:
Let $x = \frac{1}{n}$, 
$\lim_{n\to\infty}(\cos\frac{1}{n})^{n^{2}} = \lim_{x\to0^{+}}(\cos(x))^{1/x^2} = \lim_{x\to0^{+}}e^{\ln(\cos(x))^{1/x^2}}= \lim_{x\to0^{+}}e^{\frac{1}{x^2}\ln(\cos(x))}$
The answer is $1$, therefore I need to show that $\lim_{x\to0^{+}}e^{\frac{1}{x^2}\ln(\cos(x))}$ = $e^{0} = 1$
Meaning, I need to show that $\lim_{x\to0^{+}}\frac{1}{x^2}\ln(\cos(x)) = 0$.
I know that $\lim_{x\to0^{+}}\cos(x) = 1$, therefore, $\ln(1) =0$,
so I'm stuck at $\lim_{x\to0^{+}}e^{\frac{1}{0^+}0} = ? $
If I could say that $\frac{1}{0^+}0$ is $0$ that would solve it but I guess I can't. 
Whats a better way to solve it?
 A: What is needed is that,
as $x \to 0$,
$\cos(x)
\approx 1-x^2/2
$.
Then
$(\cos(1/n))^{n^2}
\approx (1-1/(2n^2))^{n^2}
\to e^{-1/2}
$
since
$(1-1/n)^n
\to 1/e$
as
$n \to \infty$.
A: L'Hopital's rule can help.
$$\lim_{x \to 0} \frac{\ln(\cos x)}{x^2} = -\lim_{x \to 0} \frac{\tan x}{2x} = -\lim_{x\to 0} \frac{1}{2}\sec^2 x = -\frac{1}{2}.$$
A: First of all, your limit is wrong; your calculator probably can't handle such large numbers.  It should be $\frac{1}{\sqrt{e}}$, as you can see empirically.
So with that in mind, by the same logic you used, we're looking to show
$$\lim_{x\to0^+}\frac{\ln(\cos x)}{x^2} = -\frac{1}{2}$$
This is easy to do with l'Hôpital's rule, but you don't want to use that.  You could also use Taylor series, but usually the people who want an answer without l'Hôpital's rule don't want to use that, either.  So we'll do it this way.
$$\lim_{x\to0^+}\frac{\ln(\cos x)}{x^2} =\lim_{x\to0^+}\left(\frac{\ln(1 + (\cos x - 1))}{\cos x-1}\cdot\frac{\cos x - 1}{x^2}\right)$$
This can be shown to be equal to $-\frac{1}{2}$ using the familiar limits $\lim_{x\to0}\frac{\ln(1+x)}{x} = 1$ and $\lim_{x\to0}\frac{1 - \cos x}{x^2} = \frac{1}{2}$.
