Limit of $\lim\limits_{n\to+\infty}\left(\cos\frac{x}{n} + \sin\frac{2x}{n}\right)^n $ 
Given the limit:
  $$
\lim_{n \to \infty}\left[\cos\left(\frac{x}{n}\right) +
\sin\left(\frac{2x}{n}\right)\right]^{n} =
\alpha
$$
Find the value of $\alpha$.


I suppose there is a special technique of how to solve such expressions. Sadly, I am not familiar with it. 
Could anyone put me on point ?.
 A: By using the following Taylor expansions at $0$, $\cos(t)=1+o(t)$, $\sin(t)=t+o(t)$, and $\ln(1+t)=t+o(t)$, we have that
$$\begin{align}\left(\cos\left(\frac{x}{n}\right) 
+ \sin\left(\frac{2x}{n}\right)\right)^n &=\exp\left(n \ln\left(\cos\left(\frac{x}{n}\right) + \sin\left(\frac{2x}{n}\right)\right)\right)
\\&=\exp\left(n \ln\left(1 + \frac{2x}{n}+o(1/n)\right)\right)
\\&=\exp\left(n \left(\frac{2x}{n}+o(1/n)\right)\right).
\end{align}$$
Can you take it from here?
A: Short answer:
$$\cos\frac xn\approx1^*,\\\sin\frac{2x}n\approx\frac{2x}n^*$$ then
$$\left(1+\frac{2x}n\right)^n\to e^{2x}.$$
$^*$ the next terms are negligible.
A: Consider the sequence $$a_{n} =\frac{\cos(x/n) +\sin(2x/n)}{1+(2x/n)}$$ and then we have $$n(a_n-1)=n\cdot\frac{\cos(x/n) - 1+\sin(2x/n)-(2x/n)}{1+(2x/n)}$$ and it is easily seen that $n(a_n-1)\to 0$ and therefore $a_n^{n} \to 1$. Using the fact that $(1+(2x/n))^{n}\to e^{2x}$ the desired limit is $e^{2x}$.
A: Since $\sin(2a) =2\cos a\sin a $, $\sin x\sim x$  and $\cos x\sim 1-\frac{x^2}{2}$ $$\lim_{n\to+\infty}\left(\cos\frac{x}{n} + \sin\frac{2x}{n}\right)^n = \lim_{n\to+\infty}\left(1 + 2\sin\frac{x}{n}\right)^n\left(\cos\frac{x}{n} \right)^n \\ \sim \left(1 + \frac{2x}{n}\right)^n\left(1-\frac{x^2}{2n^2}\right)^n \\\overset{\color{red}{h=\frac{x}{n}}}{=}\underbrace{\exp\left(2x\frac{\ln(1+2h)}{2h}\right)}_\overset{h\to 0}{\to e^{2x}}  \underbrace{\exp\left(\frac{xh}{2}\frac{\ln(1-\frac{h^2}{2})}{\frac{h^2}{2}}\right)}_{\overset{h\to 0}{\to 1} }\overset{h\to 0}{\to} e^{2x}$$
Given that $$ \lim_{X\to 0}\frac{\ln(1+X)}{X} =0$$
