Limit as x goes to zero $y(x)=(\frac{a^x+b^x}{2})^{1/x}$ for $a,b>0$

Question

I did this two different ways and got max{a,b} and 1. I have seen on the internet things like sqrt(ab). I recently had this problem on an exam so I am very curious...

Answer 1

Hint: write $$e^{\frac{\ln\left(\frac{a^x+b^x}{2}\right)}{x}}$$ and use L'Hospital. The result should be $$\sqrt{ab}$$

Answer 2

Assuming $a,b >0$,$$y= \left(\frac{a^x+b^x}{2}\right)^{\frac 1x}\implies \log(y)={\frac 1x}\log\left(\frac{a^x+b^x}{2}\right)$$ Use Taylor built at $x=0$ to get $$\log\left(\frac{a^x+b^x}{2}\right)=\frac{1}{2} x (\log (a)+\log (b))+\frac{1}{8} x^2 \left(\log ^2(a)+\log ^2(b)-2 \log (a) \log (b)\right)+O\left(x^3\right)$$ that is to say $$\log\left(\frac{a^x+b^x}{2}\right)=\frac{1}{2} \log (ab)\,x+\frac{1}{8}\log ^2\left(\frac{a}{b}\right)\,x^2+O\left(x^3\right)$$ $$\log(y)= \log (\sqrt{ab})+\frac{1}{8}\log ^2\left(\frac{a}{b}\right)\,x+O\left(x^2\right)$$ making, reusing Taylor for $y=e^{\log(y)}$ $$y=\sqrt{ab} \left(1+ \frac{1}{8}\log ^2\left(\frac{a}{b}\right)\,x\right)+O\left(x^2\right)$$ which shows the limit and also how it is approached.

Answer 3

Note that

$$\left(\frac{a^x+b^x}{2}\right)^{\frac 1x}=e^{{\frac 1x}\log\left(\frac{a^x+b^x}{2}\right)}=e^{{\frac 1x}\log\left(1+\frac{a^x+b^x-2}{2}\right)}= e^{{\frac{a^x+b^x-2}{2x}}\log\left(1+\frac{a^x+b^x-2}{2}\right)^{\frac2{a^x+b^x-2}}}\to \sqrt {a b}$$

indeed

$$\log\left(1+\frac{a^x+b^x-2}{2}\right)^{\frac2{a^x+b^x-2}}\to \log e =1$$

$$e^{{\frac{a^x+b^x-2}{2x}}}\to \sqrt {ab}$$

indeed by definition of derivative or by standard limit of exponentials

$$\frac{a^x+b^x-2}{2x}=\frac12\frac{a^x-1}{x}+\frac12\frac{b^x-1}{x}\to \frac12 \log a+ \frac12 \log b = \log \sqrt{ab}$$