Limit of $(\cos1/n)^{n^2}$ when $n\to\infty$ I've been trying to solve this limit:
$$\lim_{n\to \infty} \left(\cos\left(\frac{1}{n}\right)\right)^{n^2}$$
I solved it using l'hopital but I have to try in another way.
I tried to make the expression: 
$$ \lim_{n \to \infty}\exp\left(n^2\ln\left(\cos\left(\frac{1}{n}\right)\right)\right)$$
but it didn't work out for me...
 A: $$
\lim_{n\to\infty}n^2\ln\left(\cos{1\over n}\right)=
\lim_{n\to\infty}{\ln\left(1-\left(1-\cos{1\over n}\right)\right)
\over1-\cos{1\over n}}\cdot{1-\cos{1\over n}\over(1/n)^2}=
-1\cdot{1\over2}=-{1\over2}.
$$
A: How about this
\[
\cos(1/n)=1-\frac{1}{2n^2}+O\left(\frac{1}{n^4}\right)
\]
Then use the relation \[\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n=e^x\]
Combining the two one gets
\[
\lim_{n\rightarrow\infty}[\cos(1/n)]^{n^2}=\lim_{n\rightarrow\infty}\left(1-\frac{1}{2n^2}\right)^{n^2}=e^{-1/2}=\frac{1}{\sqrt{e}}
\]
A: Using $\cos(1/n)=1-\frac{1}{2n^2}+O\left(\frac1{n^4}\right)$, we have
$$\begin{align}
\cos^{n^2}(1/n)&=\left(1-\frac{1}{2n^2}+O\left(\frac1{n^4}\right)\right)^{n^2}\\\\
&=\left(1-\frac{1}{2n^2}\right)^{n^2}\,\left(1+\frac{O\left(\frac{1}{n^4}\right)}{1-\frac1{2n^2}}\right)^{n^2}
\end{align}$$
Next, we note that $\lim_{n\to \infty}\left(1-\frac{1}{2n^2}\right)^{n^2}=e^{-1/2}$ from the definition of the exponential function.
Finally, inasmuch as
$$\lim_{n\to \infty}\left(1+\frac{O\left(\frac{1}{n^4}\right)}{1-\frac1{2n^2}}\right)^{n^2}=\lim_{n\to \infty}\left(1+O\left(\frac{1}{n^4}\right)\right)^{n^2}=1$$
we arrive at the coveted limit
$$\lim_{n\to \infty}\cos^{n^2}(1/n)=e^{-1/2}$$
And we are done!
A: $$\lim_{n\to\infty}\left(\cos\dfrac1n\right)^{n^2}=\lim_{n\to\infty}\left(1-\sin^2\dfrac1n\right)^{n^2/2}$$
$$=\left(\lim_{n\to\infty}\left(1-\sin^2\dfrac1n\right)^{-\frac1{\sin^2\frac1n}}\right)^{-\lim_{n\to\infty}\frac{n^2\sin^2\frac1n}2}$$
Set $-\sin^2\dfrac1n=\dfrac1m$
$$\lim_{n\to\infty}\left(1-\sin^2\dfrac1n\right)^{-\frac1{\sin^2\frac1n}}=\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=?$$
For exponent set $\dfrac1n=h$ to get $$-\lim_{n\to\infty}\frac{n^2\sin^2\frac1n}2=-\dfrac12\left(\lim_{h\to0}\dfrac{\sin h}h\right)^2=?$$
A: If $L=\lim_{x \to a} f (x)^{g (x)} $ assumes the form $1^{\infty} $ we can always write it as $L=e^{lim_{x\to a}g (x)(f (x)-1)} $. Now here let $\frac {1}{n}=u $ so as $n \infty ,u\to 0$. So we have $L=e^{\lim_{x \to 0}\frac{\cos(u)-1}{u^2}}=e^{\frac {-1}{2}} $ . The proof of the first statememt can be added if you want.
A: $$\lim_{n \to \infty}\exp\underbrace{(n^2\ln(\cos(\frac{1}{n}))}_{a})\\$$exp is continous functio ,so find limit $a$
$$\lim_{n \to \infty}n^2\ln(\cos(\frac{1}{n})=\\
\lim_{n \to \infty}\frac{\ln(\cos(\frac{1}{n}))}{\frac{1}{n^2}}=\\hop\to 
\lim_{n \to \infty}\frac{\frac{\frac{-1}{n^2}\sin (\frac{1}{n})}{(\cos(\frac{1}{n})}}{\frac{-2}{n^3}}=\\
\lim_{n \to \infty}-\frac{\frac{\sin (\frac{1}{n})}{2\cos(\frac{1}{n})}}{\frac{1}{n}}=\\\frac{-1}{2}$$now 
$$a=\frac{-1}{2} \implies exp(a)=e^{\frac{-1}{2}}$$
