Compute $\lim \limits_{n \to \infty}\left(\frac {\sqrt[n]a + \sqrt[n]b}2\right)^{n} ~~~ (a, b>0)$ 
$$\lim_{n \to \infty}\left(\frac {\sqrt[n]a + \sqrt[n]b}2\right)^{n}  ~~~ (a, b>0)$$

I extended its domain and applied L'Hopital's rule to get the answer $\sqrt{ab}$.
However, is it possible to avoid using L'Hopital's rule here? (I tried for a few times, but failed.)
Or any hint maybe?
 A: You can as well compute
$$
\lim_{x\to0^+}\left(\frac{a^x+b^x}{2}\right)^{\!1/x}
$$
Take the logarithm, so the limit to compute is
$$
\lim_{x\to0^+}\frac{\log(a^x+b^x)-\log 2}{x}
$$
which is the derivative at $0$ of $f(x)=\log(a^x+b^x)$. No l'Hôpital needed, just the derivative, which is
$$
f'(x)=\frac{a^x\log a+b^x\log b}{a^x+b^x}
$$
and, clearly, $f'(0)=\frac{1}{2}(\log a+\log b)=\log\sqrt{ab}$. Therefore your given limit is $\sqrt{ab}$.
I find the function
$$
m(x;a,b)=\begin{cases}
\left(\dfrac{a^x+b^x}{2}\right)^{\!1/x} & x\ne 0 \\[4px]
\sqrt{ab} & x=0
\end{cases}
$$
very interesting, because $m(0;a,b)$ is the geometric mean, $f(-1;a,b)$ is the harmonic mean, $m(1;a,b)$ is the arithmetic mean, $m(2;a,b)$ the quadratic mean and so on. Also
$$
\lim_{x\to-\infty}m(x;a,b)=\min\{a,b\},\qquad\lim_{x\to\infty}m(x;a,b)=\max\{a,b\}
$$
The function (of $x$) is increasing (prove it). This provides all mean related inequalities.
It is not difficult to extend this to any number of parameters:
$$
m(x;a_1,a_2,\dots,a_n)=\begin{cases}
\left(\dfrac{a_1^x+a_2^x+\dots+a_n^x}{n}\right)^{\!1/x} & x\ne 0 \\[4px]
\sqrt[n]{a_1a_2\dots a_n} & x=0
\end{cases}
$$
A: If $a=b$, then it is trivial.So assume without loss of generality. $x = \frac ab>1.$
Then it is the same as: 
$$\left(\dfrac{\sqrt[n]{x}+1}{2}\right)^n\to\sqrt{x}$$.For simplicity, denote $\sqrt[2n]{x} = y >1.$ Then, we have $$\left(\dfrac{y^2+1}{2}\right)^n - y^n = y^n\left[\left(1+\dfrac{(y-1)^2}{2y}\right)^n-1\right].$$
However, notice that $y^n=\sqrt{x},\quad y\to 1,\quad\dfrac{(y-1)^2}{2y} <(\sqrt[2n]{1+(x-1)}-1)^2 <\frac{(x-1)^2}{4n^2} $ as $n\to 0.$
$$\left(\dfrac{y^2+1}{2}\right)^n - y^n = y^n\left[\left(1+\dfrac{(y-1)^2}{2y}\right)^n-1\right] < \sqrt{x} \left[\left(1+\frac{(x-1)^2}{4n^2}\right)^n-1\right] = $$
$$ = \sqrt{x}\left[\left(\left(1+\frac{(x-1)^2}{4n^2}\right)^{n^2}\right)^{\frac 1n}-1\right] < \sqrt{x}(e^{(x-1)^2/4n}-1)\to 0$$
as $n\to 0.$
Here, we used the fact that $(1+a/m)^m<e^a.$ 
