<s>[INST] Prove $\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=e$ I need to prove that $$\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=e.$$ I tried to activate the identity of $\sin^2x+\cos^2x=1$ but I'm still stuck with $$\left(\cfrac{1}{\sin\frac1x-\cos\frac1x}\right)^x.$$ 
Can I get a hint?
 [/INST] My attempted solution without using asymptoptic arguments: 
$$
\sin{\frac{1}{x} } = \frac{1}{x} - \frac{1}{3!x^3}+\frac{1}{5!x^5}-\ldots \\
\cos{\frac{1}{x} } = 1 -\frac{1}{2!x^2} + \frac{1}{4!x^4} -\ldots\\
 e^{x^{-1}}- \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = 2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right) \Longrightarrow \\
 e^{x^{-1}}+ \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = -2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\Longrightarrow \\
 \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = e^{x^{-1}}-2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\Longrightarrow \\
 \lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x = \\ \lim_{x\to\infty}\left( e^{x^{-1}}-2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\right)^x = \\  \lim_{x\to\infty} \left(e^{x^{-1}} \right)^x\left(1-2\left( \frac{1}{2!x^2e^{x^{-1}}} + \frac{1}{3!x^3e^{x^{-1}}} + \frac{1}{6!x^6e^{x^{-1}}} +\frac{1}{7!x^7e^{x^{-1}}} + \ldots \right)\right)^x = \\ e\lim_{x\to\infty} \left(1-2\left( \frac{1}{2!x^2e^{x^{-1}}} + \frac{1}{3!x^3e^{x^{-1}}} + \frac{1}{6!x^6e^{x^{-1}}} +\frac{1}{7!x^7e^{x^{-1}}} + \ldots \right)\right)^x = \ldots = e
$$
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