That is correct. It cannot be factored in that way.
[URL="[QUOTE=HallsofIvy;794322]First obvious step is to factor out $\displaystyle \frac{1}{2}$. I would then factor out $\displaystyle a^x$ so this becomes $\displaystyle \frac{a}{2}(1+ (b/a)^x)^{1/x}$. Now, do you know the limit definition of "e"?[/QUOTE]"
Look at this link.
http://www.wolframalpha.com/input/?i...9%2F%28x%29%29
$\displaystyle \begin{align*}&\lim_{x\to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0}e^{\left(\ln\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\right)}\\=& \lim_{x\to 0} e^{\left(\frac{1}{x}\ln{\left(a^x + b^x\right)} - \frac{1}{x}\ln{2}\right)}\\ & \\=& \lim_{x\to 0} \frac{ \left( e^{\frac{1}{x}} \right) \left(e^{\ln{\left(a^x + b^x \right)}}\right)}{\left( e^{\frac{1}{x}}\right) \left(e^{\ln{2}}\right)}\\& \\=& \lim_{x\to 0} \frac{\left(a^x + b^x\right)}{2}\\=& \frac{(1+1)}{2} \\=& 1........\text{[Answer]}\end{align*}$
Hope it helps.
This is a bit rough and ready, but I think that the result is okay.
Suppose that $\displaystyle b>a,$ then
$\displaystyle \left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=b\left(\frac{(a/b)^{x}+1}{2}\right)^{1/x}\rightarrow b,$ as $\displaystyle x \rightarrow \infty,$
(since $\displaystyle (a/b)^{x}\rightarrow 0,$ and the resulting $\displaystyle (1/2)^{1/x}\rightarrow 1).$
Notice btw that if $\displaystyle a=b$ then
$\displaystyle \left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=a,$ (or $\displaystyle b$).
Please look at this page.
NOTE $\displaystyle (a/b)^{x}\not\rightarrow 0,$ you have $\displaystyle (a/b)^{x}\rightarrow 1,$
On the wolfram page l'Hospitial's rule gives $\displaystyle \exp\left(\frac{\ln(ab)}{2}\right)$
$\displaystyle \begin{align*}&\lim_{x\to 0} \left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0} e^{\ln{\left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}}}\\=& \lim_{x\to 0} e^{\frac{\ln{\left(a^x + b^x\right)}-\ln(2)}{x}}\end{align*}$
Now using I'Hopital's rule and property of limit this last line becomes:
$\displaystyle \begin{align*}&e^{\lim_{x\to 0} \frac{a^x \ln{(a)} + b^x \ln{(b)}}{a^x + b^x}}\\=& e^{\lim_{x\to 0} \frac{a^0 \ln{(a)} + b^0 \ln{(b)}}{a^0 + b^0}}\\=& e^{\left(\frac{\ln{(ab)}}{2}\right)}\\=& ab - e^2\text{.......[Answer]}\end{align*}$
$\displaystyle \exp{\left(\frac{\ln{(ab)}}{2}\right)}=\exp\left( \ln{(\sqrt{ab}) \right)}=\sqrt{ab}$
@x3bnm: Please make sure that you understand the process before posting.
Did you look at either of the Wolfram webpages links that I posted in replies #5 & #9?
If not, why not?
Hi Plato
Regarding your comment to my post #8, your post #9, somewhat sheepishly I have to admit that, (have just noticed that), I was calculating the wrong limit, $\displaystyle x\rightarrow \infty$ rather than $\displaystyle x\rightarrow 0.$
Apologies to all.