<s>[INST] Evaluate the limit $\lim\limits_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\frac1x}$ Evaluate
$$
\displaystyle\lim_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
$$

And actually I have my answer and just need someone to verify this for me since I haven't done something like this for a long time.

First, to deal with the pesky $1/x$, I take the natural log inside the limit:
\begin{align}
\lim_{x\to0+}\ln\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
&= \lim_{x\to0+}\frac{1}{x}\ln\left(\frac{3^x+5^x}{2}\right)\\
&= \lim_{x\to0+}\frac{\ln(3^x+5^x)-\ln2}{x}\\
&= \lim_{x\to0+}\frac{3^x\ln3+5^x\ln5}{3^x+5^x}......L'Hopital's \;Rule\\
&=\frac{\ln3+\ln5}{2}\\
&=\frac{1}{2}\ln3+\frac{1}{2}\ln5
\end{align}
And since what we calculated was the limit the of the natural log, the final answer would be $\displaystyle e^{\frac{1}{2}ln3+\frac{1}{2}ln5}=e^{\sqrt{3}+\sqrt{5}}$. Please tell me if I did this correctly, thanks.
 [/INST] Everything is good until you try to simplify $e^{\frac{1}{2}\ln 3 +\frac{1}{2}\ln 5}$. By the rules of logarithms, $\frac{1}{2}\ln 3 + \frac{1}{2}\ln 5 = \ln \sqrt{3} + \ln \sqrt{5} = \ln (\sqrt{3}\cdot \sqrt{5}) = \ln \sqrt{15}$. So the final answer is $e^{\ln \sqrt{15}} = \sqrt{15}$.
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