TITLE: How find this limit $I=\lim_{x\to\infty}\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)^x$
QUESTION [5 upvotes]: 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

REPLY [4 votes]: 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 
$$
\left[\sin\left(\frac{2}{x}\right)+\cos\left(\frac{1}{x}\right)\right]^x\sim \left(\frac{2}{x}+1\right)^x\to\operatorname{e}^2
$$