TITLE: I want to calculate the limit of: $\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $
QUESTION [8 upvotes]: I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$
or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be greatly appreciated.

REPLY [3 votes]: More generally, for $a>0$ and $b>0$,
$$
\lim_{x\to0}\frac{1}{x}\log\frac{a^x+b^x}{2}=
\lim_{x\to0}\frac{\log(a^x+b^x)-\log 2}{x}
$$
is the derivative at $0$ of the function $f(x)=\log(a^x+b^x)$. Since
$$
f'(x)=\frac{a^x\log a+b^x\log b}{a^x+b^x}
$$
we have
$$
f'(0)=\frac{\log a+\log b}{2}=\log\sqrt{ab}
$$
Thus
$$
\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{1/x}=
e^{\log\sqrt{ab}}=\sqrt{ab}
$$
You might enjoy proving that
$$
\lim_{x\to\infty}\left(\frac{a^x+b^x}{2}\right)^{1/x}=\max(a,b)
$$