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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.

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  • $\begingroup$ nice...........+1 $\endgroup$ – Bhaskara-III Dec 19 '15 at 13:05
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First, consider the limit of the logarithm of your function:

$\lim_{x \to 0} \frac{\ln \left( \frac{2^x + 8^x}{2} \right)}{x}$

This is in the indeterminate form $\frac{0}{0}$, so try L'Hopital's rule.

$\lim_{x \to 0} \frac{2}{2^x + 8^x} \left( \frac{2^x \ln 2 + 8^x \cdot 3 \ln 2}{2} \right) = \ln 4$

Using continuity of the exponential function, you get the original limit is 4.

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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 $$=\lim_{x\to 0}\exp\frac{\frac{d}{dx}\ln\left(\frac{2^x+ 2^{3x}}{2}\right)}{\frac{d}{dx}(x)}$$

$$=\lim_{x\to 0}\exp\frac{\frac{1}{\left(\frac{2^x+2^{3x}}{2}\right)}\left(\frac{2^x\ln 2+3\cdot 2^{3x}\ln 2}{2}\right)}{1}$$

$$=\exp\frac{\left(\frac{2^0\ln 2+3\cdot 2^{0}\ln 2}{2}\right)}{\left(\frac{2^0+2^{0}}{2}\right)}=e^{2\ln 2}=2^2=\color{red}{4}$$

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More generally, for $a>0$ and $b>0$, $$ \lim_{x\to0}\frac{1}{x}\log\frac{a^x+b^x}{2}= \lim_{x\to0}\frac{\log(a^x+b^x)-\log 2}{x} $$ is the derivative at $0$ of the function $f(x)=\log(a^x+b^x)$. Since $$ f'(x)=\frac{a^x\log a+b^x\log b}{a^x+b^x} $$ we have $$ f'(0)=\frac{\log a+\log b}{2}=\log\sqrt{ab} $$ Thus $$ \lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{1/x}= e^{\log\sqrt{ab}}=\sqrt{ab} $$

You might enjoy proving that $$ \lim_{x\to\infty}\left(\frac{a^x+b^x}{2}\right)^{1/x}=\max(a,b) $$

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Let $f(x) = \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $. For $x \to 0$, We have $$\begin{align*} \log f(x) &= \frac{1}{x} \log \left[ \frac{1}{2}\left(e^{x \log 2} + e^{x \log 8}\right) \right] \\ &= \frac{1}{x}\log\left(1 + x \log 4 + o(x)\right) \\ &= \frac{1}{x}\left(x \log 4 + o(x)\right) \\ &= \log 4 + o(1). \end{align*}$$ Thus $f(x) \to 4$.

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This solution use series expansion. Near $0$ we have $$2^x=1+x\log2 +o(x)$$ $$8^x=1+x\log8+o(x)$$ Then applying $\log$ to your limit $$\frac{1}{x}\log \Big( \frac{2^x+8^x}{2}\Big)=\frac{1}{x}\log \Big( 1+x\frac{\log8+\log2}{2}+o(x)\Big)$$ and using $\log(1+x)$ expansion this is eq. to $$\frac{1}{x} x\frac{\log8+\log2}{2}+o(x)$$ The goes to $\frac{\log8+\log2}{2}$ when $x$ goes to $0$, hence the original limit is $$e^{\frac{\log8+\log2}{2}}=4$$

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