Solve limit using the concept of equivalent functions How to solve this limit $$\lim_{x\rightarrow 0}{\frac{(2+x)^x-2^x}{x^2}}$$ using the concept of equivalent functions?
For example, if $x\rightarrow 0 $ function $\sin x$ is equivalent to $x$, $\ln(1+x)\sim x$,  $a^x-1 \sim x \ln a$, etc.

we have $$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}=\lim_{x\to 0}2^x\lim_{x\to 0}{\frac{(1+\frac{x}{2})^x-1}{x^2}}=\lim_{x\to 0}{\frac{e^{x\ln(1+\frac{x}{2})}-1}{x^2}}$$ therefore we have that$$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}\sim\lim_{x\to 0}{\frac{{x\ln(1+\frac{x}{2})}}{x^2}}=\frac{1}{2}$$ 
