Question

Find
$\underset{x\to 0}{lim}{\left(\frac{1^x+2^x+3^x}{3}\right)}^{\frac{1}{x}}$

Answer 1

Putting $x=0$,

${\left(\frac{1^o+2^o+3^o}{3}\right)}^{\frac{1}{0}}={\left(\frac{1+1+1}{3}\right)}^{\infty }=1^{\infty }$

This limit is of the form $1^{\infty }$.

So, using $\underset{x\to 0}{lim}\left[f\right(x){]}^{g\left(x\right)}=e^{\underset{x\to 0}{lim}(\frac{1^x+2^x+3^x}{3}-1).\left(\frac{1}{x}\right)}$

$=e^{\underset{x\to 0}{lim}\left(\frac{1^x+2^x+3^x-3}{3x}\right)}=|\frac{0}{0}$ form

$=e^{\underset{x\to 0}{lim}\left(\frac{1^x\log 1+2^x\log 2+3^x\log 3}{3x}\right)}$

$\because$ using L-Hopital's Rule: $\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

if $\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}$ form, at $x=0$

$=e^{(\frac{\log 2}{3}+\frac{\log 3}{3})}$

$=\left(e^{\frac{1}{3}\log 2}\right)\left(e^{\frac{1}{3}\log 3}\right)$

$=\left(e^{\log 2^{\frac{1}{3}}}\right)\left(e^{\log 3^{\frac{1}{3}}}\right)$

$=2^{\frac{1}{3}}.3^{\frac{1}{3}}=6^{\frac{1}{3}}$