Question

$\underset{x\to 0}{lim}{\left[\frac{2^x+2^{2x}+2^{3x}}{3}\right]}^{\frac{1}{x}}$ is equal to

Answer 1

This is $1^{\infty }\text{form}$
$=e^{\underset{x\to 0}{lim}\frac{1}{x}[\frac{2^x+2^{2x}+2^{3x}}{3}-1]}$
$=e^{\underset{x\to 0}{lim}\frac{1}{x}\left[\frac{2^x+2^{2x}+2^{3x}-3}{3}\right]}$
$=e^{\underset{x\to 0}{lim}(\left[\frac{2^x-1}{x}\right]+2\left[\frac{2^{2x}-1}{2x}\right]+3\left[\frac{2^{3x}-1}{3x}\right]).\frac{1}{3}}$
$=e^{\underset{x\to 0}{lim}\left[\frac{\ln 2+2\ln 2+3\ln 2}{3}\right]}$
$=e^{\underset{x\to 0}{lim}\left[\frac{2\ln 2}{1}\right]}$
$=e^{\left(2\ln 2\right)}$
$=e^{\left(\ln 4\right)}=4$