$\lim_{x\rightarrow 0}((a^{x}+b^{x})/2)^{1/x}$ gives $1^{\infty }$ form so we apply standard form
$e^{\lim_{x\rightarrow 0}((f(x)-1)*(g(x)))}$
here in this question f(x) =$((a^{x}+b^{x})/2)$ and g(x)=1/x
so we apply standard form $e^{\lim_{x\rightarrow 0}((a^{x}+b^{x})/2)-1)*1/x}$
now we calculate for$lim_{x\rightarrow 0}((a^{x}+b^{x})/2)-1)*1/x$ which is 0/0 form now we apply l hospital rule that gives
(ax lna+bxlnb)/2 =1/2*(lnab)=ln (ab)1/2
$\therefore e^{ln_{e}\sqrt{ab}}=\sqrt{ab}$