<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] The indetermination is of the form $1^{+\infty}$, suggesting the application of fundamental limit:
$$\lim_{w\rightarrow 0^+}(1+w)^{\frac{1}{w}}=e. $$
Remembering that
$$\lim_{x\rightarrow 0}\frac{a^x-1}{x}=\ln a, $$
we have
$$\lim_{x\rightarrow 0^+} \left(\frac{3^x+5^x}{2} \right)^{\frac{1}{x}}=\lim_{x\rightarrow 0^+} \left(1+\frac{3^x+2^x}{2}-1 \right)^{\frac{1}{x}}=\lim_{x\rightarrow 0^+} \left( 1+\frac{3^x+5^x-2}{2}\right)^{\frac{2}{3^x+5^x-2}\cdot \frac{3^x+5^x-2}{2x}}=$$
$$=e^{\lim_{x\rightarrow 0^+} \frac{1}{2}\left(\frac{3^x-1}{x}+\frac{5^x-1}{x} \right)} =e^{\frac{1}{2}(\ln 3+\ln 5)}=e^{(\ln 3\cdot 5)^{\frac{1}{2}}}=\sqrt{15}.$$
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