Question Number 46563 by Necxx last updated on 28/Oct/18
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{a}^{{x}} +{b}^{{x}} +{c}^{{x}} }{\mathrm{3}}\right)^{\mathrm{1}/{x}} \\ $$
Answered by ajfour last updated on 28/Oct/18
![lim_(x→0) (1+((a^x −1+b^x −1+c^x −1)/3))^(1/x) =lim_(x→0) {[1+f(a,b,c,x)]^(1/(f(a,b,c,x))) }^((f(a,b,c,x))/x) = e^(((ln a+ln b+ln c)/3) ) = (abc)^(1/3) .](https://www.tinkutara.com/question/Q46564.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{{a}^{{x}} −\mathrm{1}+{b}^{{x}} −\mathrm{1}+{c}^{{x}} −\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{1}/{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left[\mathrm{1}+{f}\left({a},{b},{c},{x}\right)\right]^{\frac{\mathrm{1}}{{f}\left({a},{b},{c},{x}\right)}} \right\}^{\frac{{f}\left({a},{b},{c},{x}\right)}{{x}}} \\ $$$$=\:{e}^{\frac{\mathrm{ln}\:{a}+\mathrm{ln}\:{b}+\mathrm{ln}\:{c}}{\mathrm{3}}\:} =\:\:\left(\boldsymbol{{abc}}\right)^{\mathrm{1}/\mathrm{3}} \:. \\ $$