What is $\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$?

Put $y=\dfrac{1}{x} \to \displaystyle \lim_{x\to \infty}\ln(f(x))=\displaystyle \lim_{y\to 0}\dfrac{\ln\left(\sin y+\cos y\right)}{y}=\displaystyle \lim_{y\to 0} \dfrac{\cos y - \sin y}{\sin y+ \cos y}=1\to \displaystyle \lim_{x \to \infty} f(x) = e$.

Set $1/x=h$ to get

$$\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=\lim_{h\to 0^+}\left(\sin h+\cos h\right)^{\frac1h}$$

Now $(\sin h+\cos h)^2=1+2\sin h\cdot\cos h=1+\sin2h$

$$\lim_{h\to 0^+}\left(\sin h+\cos h\right)^{\frac1h}=\lim_{h\to 0^+}\left[\left(\sin h+\cos h\right)^2\right]^{\frac1{2h}}$$

$$=\left[\lim_{h\to 0^+}\left(1+\sin2h\right)^{\dfrac1{\sin2h}}\right]^{\lim_{h\to 0^+}\dfrac{\sin2h}{2h}}$$

Now for the inner limit, use $\lim_{ y\to0}(1+y)^{\dfrac1y}=\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$

How about the limit in the exponent?

$$\sin \frac1x + \cos \frac1x = 1 + \frac1x - \frac1{2x^2} + \cdots$$ therefore $$ \left( \sin \frac1x + \cos \frac1x \right)^x = \left( 1 + \frac1x+\cdots\right)^x \to e \text{ as } x \to \infty.$$