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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^xln3+5^xln5}{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.

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Note, you can check numerical questions such as this using wolframalpha.com –  Ozera Mar 20 at 19:07

2 Answers 2

up vote 2 down vote accepted

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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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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