What is $\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$? $$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$$
It is about the $(\to1)^{(\to\infty)}$ situation. Can we find its limit using the formula $\lim_{x\to\infty}(1+\frac 1x)^x=e$? If yes, then how?
 A: Put $y=\dfrac{1}{x} \to \displaystyle \lim_{x\to \infty}\ln(f(x))=\displaystyle \lim_{y\to 0}\dfrac{\ln\left(\sin y+\cos y\right)}{y}=\displaystyle \lim_{y\to 0} \dfrac{\cos y - \sin y}{\sin y+ \cos y}=1\to \displaystyle \lim_{x \to \infty} f(x) = e$.
A: 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?
A: $$\sin \frac1x + \cos \frac1x  = 1 + \frac1x - \frac1{2x^2} + \cdots$$ therefore $$ \left( \sin \frac1x + \cos \frac1x \right)^x =  \left( 1 + \frac1x+\cdots\right)^x \to e \text{ as } x \to \infty.$$
A: $$
\lim_{x\to\infty}(\sin{\frac 1x}+\cos{\frac 1x})^x
$$
Take $x=\dfrac{1}{t}$ ,
$$
\lim_{t\to 0}(\sin t+\cos t)^\frac{1}{t}
$$
$$
\large \lim_{t\to 0}e^\frac{\ln(\sin t+\cos t)}{t}
$$
$$
\large e^{\lim_{t\to 0}\frac{\ln(\sin t+\cos t)}{t}}
$$
Then use L'Hôpital's rule.
A: Hint:
$$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x=
\lim_{x\to\infty}\exp\left(x\log\left(\sin{\frac 1x}+\cos{\frac 1x}\right)\right)=
$$
$$=
\exp\lim_{x\to\infty}\left(x\log\left(\sin{\frac 1x}+\cos{\frac 1x}\right)\right)=
$$
$$
=\exp\lim_{x\to\infty} \dfrac{\log\left(\sin{\frac 1x}+\cos{\frac 1x}\right)}{\frac 1x}=
$$
Now use l'Hopital rule.
A: $$\begin{array}{lll}
\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x&=&\lim_{h\to 0}\left(\sin{h}+\cos{h}\right)^{\frac{1}{h}}\\
&=&\lim_{h\to 0}(\cos h)^{\frac{1}{h}}\left(\tan{h}+1\right)^{\frac{1}{h}}\\
&=&\lim_{h\to 0}((1+\cos h-1)^{\frac{1}{\cos h -1}})^{\frac{\cos h -1}{h}}(\left(\tan{h}+1\right)^{\frac{1}{\tan h}})^\frac{\tan h}{h}\\
&=&e^{\lim_{h\to 0}{\frac{\cos h -1}{h}}}e^{\lim_{h\to 0}\frac{\sin h}{h}\cdot \lim_{h\to 0}\frac{1}{\cos h}}\\
&=&e^0e^{1\cdot 1}\\
&=&e\\
\end{array}$$
A: Set $1/x=2h,$ 
$$F=\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x=\lim_{h\to0}(1+\sin2h+\cos2h-1)^{1/2h}$$
Now using $\sin2h=2\sin h\cos h,\cos2h=1-2\sin^2h$ $$\sin2h+\cos2h-1=2\sin h(\cos h-\sin h)$$
$$F=\left[\lim_{h\to0}\{1+2\sin h(\cos h-\sin h)\}^{1/2\sin h(\cos h-\sin h)}\right]^{\lim_{h\to0}\frac{\sin h(\cos h-\sin h)}h}=e^1$$
