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?
 A: Equivalently, we want the limit as $t$ approaches $0$ from the right of $(\sin t+\cos t)^{1/t}$.
Take the logarithm. We get $\dfrac{\log(\sin t +\cos t)}{t}$.  Using L'Hospital's Rule, we find that we need the limit of $\dfrac{\cos t-\sin t}{\sin t+\cos t}$ as $t\to 0^+$. This is $1$, so by continuity the original limit is $e^1$.
A: let $\dfrac{1}{x}=t\longrightarrow 0$
$$I=\lim_{t\to 0}(\sin t+\cos t)^{\frac{1}{t}}
=\lim_{t\to 0}(1+\sin2t))^{\frac{1}{2}}=e$$
the last we use
$$\lim_{x\to 0}(1+f(x))^{g(x)}=e,\text{since }\lim_{x\to 0}f(x)g(x)=1$$
A: $$\sin {1 \over x} = {1 \over x} + \mathcal{O}({1 \over x^3})$$
$$\cos {1 \over x} = 1 + \mathcal{O}({1 \over x^2})$$
$$\left(\sin {1 \over x} + \cos {1 \over x}\right)^x = \exp \left( x \ln \left(1+{1 \over x} + \mathcal{O}({1 \over x^2})\right)\right)$$
$$=\exp \left( 1 + \mathcal{O}({1 \over x})\right) \xrightarrow[x \to \infty]{} e$$
A: My attempted solution without using asymptoptic arguments: 
$$
\sin{\frac{1}{x} } = \frac{1}{x} - \frac{1}{3!x^3}+\frac{1}{5!x^5}-\ldots \\
\cos{\frac{1}{x} } = 1 -\frac{1}{2!x^2} + \frac{1}{4!x^4} -\ldots\\
 e^{x^{-1}}- \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = 2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right) \Longrightarrow \\
 e^{x^{-1}}+ \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = -2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\Longrightarrow \\
 \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = e^{x^{-1}}-2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\Longrightarrow \\
 \lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x = \\ \lim_{x\to\infty}\left( e^{x^{-1}}-2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} + \frac{1}{6!x^6} +\frac{1}{7!x^7} + \ldots \right)\right)^x = \\  \lim_{x\to\infty} \left(e^{x^{-1}} \right)^x\left(1-2\left( \frac{1}{2!x^2e^{x^{-1}}} + \frac{1}{3!x^3e^{x^{-1}}} + \frac{1}{6!x^6e^{x^{-1}}} +\frac{1}{7!x^7e^{x^{-1}}} + \ldots \right)\right)^x = \\ e\lim_{x\to\infty} \left(1-2\left( \frac{1}{2!x^2e^{x^{-1}}} + \frac{1}{3!x^3e^{x^{-1}}} + \frac{1}{6!x^6e^{x^{-1}}} +\frac{1}{7!x^7e^{x^{-1}}} + \ldots \right)\right)^x = \ldots = e
$$
A: Using $(\sin y+\cos y)^2=1+\sin2y,$
$$\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x$$
$$=\left(\lim_{x\to\infty}\left(1+\sin\dfrac2x\right)^{1/\sin\frac2x}\right)^{\lim_{x\to\infty}\dfrac{\sin\frac2x}{\frac2x}}$$
The inner limit converges to $\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$
For the exponent, choose $1/x=h$
