I want to calculate the limit of: $\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $ I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$
or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be greatly appreciated.

Notice, $$\lim_{x\to 0}\left(\frac{2^x+8^x}{2}\right)^{1/x}$$
$$=\lim_{x\to 0}\left(\frac{2^x+2^{3x}}{2}\right)^{1/x}$$
$$=\lim_{x\to 0}\exp\frac{1}{x}\ln\left(\frac{2^x+2^{3x}}{2}\right)$$
Applying L'hospital's rule for $\frac 00$ form $$=\lim_{x\to 0}\exp\frac{\frac{d}{dx}\ln\left(\frac{2^x+ 2^{3x}}{2}\right)}{\frac{d}{dx}(x)}$$
$$=\lim_{x\to 0}\exp\frac{\frac{1}{\left(\frac{2^x+2^{3x}}{2}\right)}\left(\frac{2^x\ln 2+3\cdot 2^{3x}\ln 2}{2}\right)}{1}$$
$$=\exp\frac{\left(\frac{2^0\ln 2+3\cdot 2^{0}\ln 2}{2}\right)}{\left(\frac{2^0+2^{0}}{2}\right)}=e^{2\ln 2}=2^2=\color{red}{4}$$
