Evaluate the limit with exponents using L'Hôpital's rule or series expansion Evaluate the limit$$\lim_{x\to 0}\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}
-\sqrt{ab}}{x}$$ It is known that $a>0,b>0$
My Attempt:
I could only fathom that  $$\lim_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$$

just use the l-H hospital for it,a tip for that is you can write the $((a^x+b^x)/2)^{1/x}$ to this inform: $e^{[\ln((a^x+b^x)/2)]/x}$,so you can use chain rule to differeitiate it,here you are,the answer leaves for you
