Evaluating a limit. What makes the equality right? I'm reading a proof of a limit calculation. The limit is:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x}$$
where $a,b>0$.
The aother claims that:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left( \lim\limits_{x\to 0}\frac{\frac{a^x+b^x}{2} - 1}{x} \right)$$
How come?
Update:
Of course, 
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left(\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^x+b^x}{2} \right)}{x} \right)$$
But how to proceed to reach the auther's expression?

First use simple fact, that $\displaystyle\lim_{y \to 0}\frac{\ln(1+y)}{y}=1$, so:
$$\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{y}=\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{\frac{a^x+b^x}{2}-1} \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{y}=\\=1 \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}$$
Now $\lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}=\lim_{x \to 0}\frac{1}{2}\frac{a^x-1}{x}+\lim_{x \to 0}\frac{1}{2}\frac{b^x-1}{x}$
But $a^x=e^{x \ln a }$, so $\lim_{x \to 0}\frac{1}{2}\frac{a^x-1}{x}=\lim_{x \to 0}\ln a\frac{1}{2}\frac{e^{\ln a x}-1}{x \ln a}=\frac{1}{2}\ln a$. The same with second limit. Finally the result is $\frac{\ln ab}{2}$.
