<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] If the limit $L=\lim_{n\rightarrow a} f(x)^{g(x)}$,
is of the form $1^{\infty}$ maning $\lim_{x\rightarrow a}f(x)=1, \lim_{x\rightarrow a} g(x) = \infty.$  Then $L=\exp[\lim_{x\rightarrow a}(f(x)-1)g(x)]$. So in this case
$$L= \exp \left [\lim_{x\rightarrow 0} \frac{1}{x} \left( \frac{3^x + 5^x}{2}-1 \right)\right]=
\exp \left [ \frac{1}{2} \left (\lim_{x\rightarrow 0} \frac {3^x-1}{x} + \lim_{x\rightarrow 0} \frac{5^x-1}{x} \right ) \right]=\exp[\frac{1}{2}(\ln 3 + \ln 5)] =\sqrt{15}.$$
</s>