Use L'Hôpital's rule to evaluate this limit $$\lim_{x\to0} \left(\frac{a^x + b^x}{2}\right)^{1/x}  = (ab)^{1/2}$$
I tried taking logarithm to both sides, but got stuck 
I got to
e^ (lim (a^x loga + b^x logb)/(a^x + b^x)
 A: Hint : Write
$$
f(x) = \left ( \frac{a^x + b^x}{2} \right )^{1/x}
$$
and consider
$$
\lim_{x\to 0} \ln f(x)
$$
A: Note that $ \ln \left( \left( \dfrac {a^x+b^x}{2} \right)^{1/x} \right) = \dfrac {1}{x} \cdot \ln \left( \dfrac {a^x+b^x}{2} \right) $.
Now, consider $ \lim \left( \dfrac {\ln \left( \dfrac {a^x+b^x}{2} \right)}{x} \right) $. 
The derivative of the numerator is $ \dfrac {a^x \ln a + b^x \ln b}{a^x+b^x} $. 
The derivative of the denominator is $ 1 $. 
Now, our original limit is just $$ \exp \left( \lim_{x \to 0} \left( \dfrac {a^x \ln a + b^x \ln b}{a^x+b^x} \right) \right). $$
Can you do it from here?
A: If I am not missing anything, here is a solution without LH rule, probably you will not see anywhere:
$$\lim _{ x\to 0 } \left( \frac { a^{ x }+b^{ x } }{ 2 }  \right) ^{ 1/x }=\lim _{ x\to 0 } \left( \frac { { \left( a^{ x/2 }-b^{ x/2 } \right)  }^{ 2 }+{ 2\left( ab \right)  }^{ x/2 } }{ 2 }  \right) ^{ 1/x }$$$$=\lim _{ x\to 0 } \left( \frac { { \left( a^{ x/2 }-b^{ x/2 } \right)  }^{ 2 } }{ 2 } +{ \left( ab \right)  }^{ x/2 } \right) ^{ 1/x }\\ =\lim _{ x\to 0 } \left( 0+{ \left( ab \right)  }^{ x/2 } \right) ^{ 1/x }={ \left( ab \right)  }^{ 1/2 }$$
