Question

Evaluate:
$li{m}_{x\to 0}(\frac{a^x+b^x+c^x}{3}{)}^{1/x}.$

Answer 1

${lim}_{x⟶0}{\left(\frac{a^x+b^x+c^x}{3}\right)}^{\frac{1}{x}}={lim}_{x⟶0}{\left[\left(\frac{a^x+b^x+c^x+3-3}{3}\right)\right]}^{\frac{1}{x}}={lim}_{x⟶0}{\left[\frac{\frac{x(a^x-1)}{x}+\frac{x(b^x-1)}{x}+\frac{x(c^x-1)}{x}+9}{3}\right]}^{\frac{1}{x}}\because ({lim}_{x⟶0}\frac{(a^x-1)}{a}={\log }_{e}a)\therefore {lim}_{x⟶0}{\left(\frac{a^x+b^x+c^x}{3}\right)}^{\frac{1}{x}}={lim}_{x⟶0}{\left[\frac{x\ln a+x\ln b+x\ln c+3}{3}\right]}^{\frac{1}{x}}={lim}_{x⟶0}{[\frac{x\ln \left(abc\right)}{3}+1]}^{\frac{1}{x}}={lim}_{x⟶0}{[1+\frac{x\ln \left(abc\right)}{3}]}^{\frac{1}{x}\frac{3}{\ln \left(abc\right)}\frac{\ln \left(abc\right)}{3}}={lim}_{x⟶0}{[1+\frac{x\ln \left(abc\right)}{3}]}^{\frac{3}{x\ln \left(abc\right)}\frac{\ln \left(abc\right)}{3}}=e^{\frac{\ln \left(abc\right)}{3}}(\because {lim}_{x⟶0}{+(1+x)}^{\frac{1}{x}}=e)=e^{{\ln \left(abc\right)}^{\frac{1}{3}}}={\left(abc\right)}^{\frac{1}{3}}$