Limit of trigonometric functions

Question

Use the following identities:

1. $\lim_{n \to \infty}(1-\frac{x}{n})^{-n} = \exp{x}$ for all $x \in \mathbb{R}$ with uniform convergence on any bounded interval
2. $\cos{x} = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$
3. $\sin{x} = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$

Prove that $\lim_{n \to \infty} \cos^n \frac{x}{\sqrt{n}} = \exp(-\frac{x^2}{2})$ and $\lim_{n \to \infty} (\frac{\sin \frac{x}{\sqrt{n}}}{\frac{x}{\sqrt{n}}})^n = \exp(-\frac{x^2}{6})$ uniformly on all bounded intervals in $\mathbb{R}$.

It seems that I should link the form $(1-x/n)^n$ with the trigonometric functions, but didn't get much progress after trying. Any hints will be appreciated.

Answer 1

HINT:

$\cos^m{\frac{x}{\sqrt {m}}} = (\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{m^n(2n)!})^m$

Separate the first to elements of the sum (i.e. write the sum as when $n=0$ + when $n=1$+ $\sum_{n=2}^{\infty}$)

Also keep in mind what is the definition of $\exp (\frac{-x^2}{2})$ by part $1$

Answer 2

The first limit is generally solved using $$ \cos^n\frac{x}{\sqrt{n}}\approx\left(1-\frac{x^2}{2n}\right)^n. $$ The limit, implicit here, that $n\rightarrow\infty$ makes the rest. The other one can be obtained in a similar manner.