How can I calculate this (rather tricky) limit? $$\lim\limits_{n \to \infty} \cos^{n^2} \left (\frac{2x}{n} \right)$$
Any hints and/or help is greatly appreciated.
 A: When $n$ is large enough, $\frac {2x}n$ will be small, so $\cos\left(\frac{2x}n\right)$ will be very close to $1-\frac{2x^2}{n^2}$
A: Consider $$a_n=\cos^{n^2} \left (\frac{2x}{n} \right)\implies \log(a_n)=n^2\log\left( \cos\left (\frac{2x}{n} \right)\right)$$ Now use the series expansion
$$\cos\left (\frac{2x}{n} \right)=1-\frac{2 x^2}{n^2}+\frac{2 x^4}{3 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left( \cos\left (\frac{2x}{n} \right)\right)=-\frac{2 x^2}{n^2}-\frac{4 x^4}{3 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log(a_n)=-2 x^2-\frac{4 x^4}{3 n^2}+O\left(\frac{1}{n^4}\right)$$
$$a_n=e^{\log(a_n)}=e^{-2 x^2}\left( 1-\frac{4 x^4}{3 n^2}\right)+O\left(\frac{1}{n^4}\right)$$
A: $$L=\lim_{n\rightarrow}~(\cos(2x/n))^{n^2}=\lim_{n\rightarrow \infty} \left( 1-\frac{4x^2}{2n^2} \right)^{n^2}=\exp[\lim_{n\rightarrow \infty} n^2(1-2x^2/n^2-1)]=e^{-2x^2}.$$
A: Hint:
$$\lim_{n\to\infty}\cos^n\dfrac{2x}n=\left(\lim_n\left(1-2\sin^2\dfrac xn\right)^{-1/2\sin^2\dfrac xn}\right)^{\lim_n-2n^2\sin^2\dfrac xn}$$
The inner limit converges to $e$
Set $\dfrac xn=h\implies h\to0$ in the exponent
A: The limit is of $1^{\infty}$ form.
$$\therefore \lim_{n \to \infty} Cos^{n^2}(\frac{2x}{n})=e^{lim_{n \to \infty}(n^2(Cos\frac{2x}{n}-1))}$$
$$=e^{lim_{n \to \infty} (-n^2.2Sin^{2}(\frac{x}{n}))}$$
$$=e^{lim_{n \to \infty} (-2x^2.(\frac{Sin(\frac{x}{n})}{\frac{x}{n}})^2)}$$
$$=e^{-2x^2}$$
