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?

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 $(1+a/m)^m<e^a.$ 
