What I've tried so far is to use the exponent and log functions:
$$\lim_{x \to 0}{(\sin(x) + 2^x)^\frac{\cos x}{\sin x}}= \lim_{x \to 0}e^ {\ln {{(\sin(x) + 2^x)^\frac{\cos x}{\sin x}}}}=\lim_{x \to 0}e^ {\frac{1}{\tan x}{\ln {{(\sin(x) + 2^x)}}}}$$.
From here I used the expansion for $\tan x$ but the denominator turned out to be zero. I also tried expanding $\sin x$ and $\cos x$ with the hope of simplifying $\frac{\cos x}{\sin x}$ to a constant term and a denominator without $x$ but I still have denominators with $x$.
Any hint on how to proceed is appreciated.
Best Answer
Take the logarithm and use standard first order Taylor expansions: $$ \lim_{x\to0} \frac{\log\bigl(\sin(x)+2^x\bigr)}{\tan(x)} =\lim_{x\to0} \frac{\log\bigl(\sin(x)+2^x\bigr)}{x+o(x)} =\lim_{x\to0} \frac{x+\log(2)x+o(x)}{x+o(x)} = 1+\log(2). $$ Then $$ \lim_{x\to0} \bigl(\sin(x)+2^x\bigr)^{\cot(x)} = e^{1+\log(2)} = 2e. $$
EDIT
Maybe it's important to clarify why $\log\bigl(\sin(x)+2^x\bigr)=x+\log(2)x+o(x)$. I'm using the following facts: