Showing that $\lim_{{x\to 0}}(1+\sin{x})^{\frac{1}{x}} = e$ How do you calculate $\displaystyle \lim_{{x\to 0}}(1+\sin{x})^{\frac{1}{x}}$? I got it from here. It says L'Hopital, but I can't figure out how to apply it as I don't have a denominator. I also tried to rewrite the limit using trig identities:
$\displaystyle \lim_{{x\to 0}}(1+\sin{x})^{\frac{1}{x}} = \lim_{x \to 0} 2^{\frac{1}{x}}\sin^{\frac{2}{x}}\left(\frac{\pi}{4}+\frac{x}{2}\right) = ?$

my answer: 
$\lim_{x\to 0}\left(1+\sin x\right)^{1/x}$
$=\lim_{x\to 0}(\left(1+\sin x\right)^{\frac{1}{\sin x}})^{\frac{\sin x}{x}}$
Note: $\lim_{x\to 0}\frac{\sin x}{x}=1$, so i get
$=(e)^1=e$
