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?
 A: If we try with
$$
\lim_{x\to 0} \frac{\log(a^x+b^x)-\log 2}{x}
$$
and apply l'Hôpital's theorem, we get
$$
\lim_{x\to 0}\frac{a^x\log a+b^x\log b}{a^x+b^x}=\frac{\log a+\log b}{2}=
\log\sqrt{ab}.
$$
It's just the derivative of $x\mapsto (a^x+b^x)/2$ at $0$, of course.
However,
$$
\lim_{x\to0}\frac{\log(1+x)}{x}=1
$$
so that
$$
\lim_{x\to 0} \frac{\log\dfrac{a^x+b^x}{2}}{x}=
\lim_{x\to 0} \frac{\log\dfrac{a^x+b^x}{2}}{\dfrac{a^x+b^x}{2}-1}
              \frac{\dfrac{a^x+b^x}{2}-1}{x}
$$
and the limit of the first factor is $1$. I don't think it's a real simplification.

It may be worth noting that the function
$$
\mu_{a,b}(x)=\begin{cases}
\left(\dfrac{a^x+b^x}{2}\right)^{1/x} & \text{if $x\ne0$}\\[2ex]
\sqrt{ab} & \text{if $x=0$}
\end{cases}
$$
for $a,b>0$ is quite interesting, because it's increasing, $\mu_{a,b}(-1)$ is the harmonic mean, $\mu_{a,b}(0)$ is the geometric mean, $\mu_{a,b}(1)$ is the arithmetic mean and
$$
\lim_{x\to-\infty}\mu_{a,b}(x)=\min(a,b),\qquad
\lim_{x\to\infty}\mu_{a,b}(x)=\max(a,b).
$$
A: 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 $\frac{\ln ab}{2}$.
