Question

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

Answer 1

Given limit of the form $1^{\infty }$, as we put $x=0$.

let $y=\underset{x\to }{lim}{\left(\frac{2^x+2^{2x}+23x}{3}\right)}^{1/x}$ [$1^{\infty }$ from indeterminate form of limit]

Taking natural logarithm both sides:-

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

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

$\Rightarrow \log =\underset{x\to 0}{lim}\frac{\ln (2^x+2^{2x}+2^{3x})-\ln 3}{x}\left[\frac{0}{0}formindetiominateform\right]$

Applying,

Hospital's rule:-

$\Rightarrow \log =\underset{x\to 0}{lim}\frac{\frac{d}{dx}\left[\ln \right(2^x+2^{2x}+2^{3x})-\ln 3]}{\frac{d}{dx}\left(x\right)}$

$\Rightarrow \log =\underset{x\to 0}{lim}[\frac{[2^x\ln 2+{2.2}^x.\ln 2+{3.2}^x.\ln 2]-0}{(2^x+2^{2x}+2^{3x})}÷1]$

$\Rightarrow \log =\underset{x\to 0}{lim}\frac{2^2[\ln 2+\ln 2^2+\ln 2^3]}{2^x+2^{2x}+2^{3x}}$

Applying times $x=0$:-

$\log =\frac{2^0\left[\ln \right(\mathrm{2.4.8}\left)\right]}{2^0+2^{2.0}+2^{3.0}}$

$\log =\frac{1}{3}\ln 64$

$\log =\ln {64}^{1/3}$

$\log =\ln 4$

Taking antilogarthim both sides:-

$y=4$