Limit of $\lim_{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ with a > 0 and b > 0 Question:

$\lim_{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ with $a>0$ and $b>0$.$a$ and $b$ are constant.

 A: Hint
Without loss of generality, assume $b\ge a$ and use $$b^x\le a^x+b^x\le 2\cdot b^x$$
A: Now that I have solved the problem, I think I should post the answers and thank everyone for their help.

$\lim _{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$
$=\lim _{x\to -\infty }\left(\frac{1}{2}\right)^{\frac{1}{x}}\cdot \left(a^x+b^x\right)^{\frac{1}{x}}$
$=\lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}$
$\lim _{x\to -\infty }\left(max\left(a^x,b^x\right)\right)^{\frac{1}{x}}\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:-\infty \:}\left(2\cdot max\left(a^x,b^x\right)\right)^{\frac{1}{x}}$

when $x\to -\infty$
,$max\left(a^x,b^x\right)=min\left(a,b\right)^x$
So

$\lim _{x\to -\infty }\left(min\left(a,b\right)^x\right)^{\frac{1}{x}}\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:-\infty \:}\left(2\cdot min\left(a,b\right)^x\right)^{\frac{1}{x}}\:$
$\lim _{x\to -\infty }min\left(a,b\right)\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:-\infty \:}2^{\frac{1}{x}}\cdot min\left(a,b\right)$

Conclusion:

$min\left(a,b\right)\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le min\left(a,b\right)$

By the way, I found the following method is also true when $x\to +\infty$.

$\lim _{x\to +\infty }\left(max\left(a^x,b^x\right)\right)^{\frac{1}{x}}\le \lim _{x\to +\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:+\infty \:}\left(2\cdot max\left(a^x,b^x\right)\right)^{\frac{1}{x}}$

