How find this limit $I=\lim_{x\to\infty}\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)^x$ Find this limit :
$$I=\displaystyle\lim_{x\to\infty}\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)^x$$
note $x=e^{\ln{x}}$
$$I=\exp\left(\lim_{x\to\infty}x\ln{\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)}\right)$$
and let $\frac{1}{x}=t$,then 
$$\lim_{t\to 0}\frac{\ln{(\sin{2t}+\cos{t})}}{t}=\lim_{t\to 0}\dfrac{2\cos{2t}-\sin{t}}{\sin{2t}+\cos{t}}=2$$
so
$$I=e^2$$
My question: have other methods? Thank you 
 A: $$\sin2h+\cos h=2\sin h\cos h+\cos h=\cos h(1+2\sin h)$$
$$\implies(\sin2h+\cos h)^{\frac1h}=(\cos h)^{\frac1h}\cdot(1+2\sin h)^{\frac1h}$$
$$\text{Now, }\displaystyle\lim_{h\to0}(1+2\sin h)^{\frac1h}=\left(\lim_{h\to0}(1+2\sin h)^{\frac1{2\sin h}}\right)^{2\frac{\lim_{h\to0}\sin h}h}=e^2$$
using $\displaystyle\frac1u=n,\lim_{u\to0}\left(1+u\right)^{\frac1u}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$
Again, $\displaystyle h=2v\implies\lim_{h\to0}(\cos h)^{\frac1h}=\lim_{v\to0}(\cos2v)^{\frac1{2v}}$
$$=\displaystyle \left(\lim_{v\to0}(1-2\sin^2v)^{-\frac1{2\sin^2v}}\right)^{(\lim_{v\to0}-\frac{2\sin^2v}{2v})}$$
Observe that the inner limit $=e$
and $\displaystyle\lim_{v\to0}-\frac{2\sin^2v}{2v}=-\left(\lim_{v\to0}\frac{\sin v}v\right)^2\cdot \lim_{v\to0} v=1\cdot 0$
A: Using Taylor expansion for $x\to\infty$, $\sin\left(\frac{2}{x}\right)\sim \frac{2}{x}$ and $\cos\left(\frac{1}{x}\right)\sim 1$ so that 
$$
\left[\sin\left(\frac{2}{x}\right)+\cos\left(\frac{1}{x}\right)\right]^x\sim \left(\frac{2}{x}+1\right)^x\to\operatorname{e}^2
$$
