The value of $\underset{x→0}{\lim}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$ is

$\frac{11e}{24}$

$\underset{x→0}{\lim}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$

$(1+x)^{1/x}$ can be expanded as $y=(1+x)^{1/x}$

$\log y=\frac{1}{x}\log (1+x)=\frac{1}{x}\left[x-\frac{x^2}{2}+\frac{x^3}{3}\right]$ [Higher power is neglected as → 0]

$⇒y=e^{1-\frac{x}{2}+\frac{x^2}{3}}=e[e^{-\frac{x}{2}+\frac{x^2}{3}}]$

Now from the expansion of exponential function

$y=e\left[1+\left(\frac{-x}{2}+\frac{x^2}{3}\right)+\frac{1}{2!}\left[\frac{x^2}{4}+\frac{x^4}{9}-\frac{2x^3}{6}\right]+......\right]$

Thus, $y=e-\frac{ex}{2}+ex^2[\frac{1}{3}+\frac{1}{8}]$+ higher powers

Therefore, $\underset{x→0}{\lim}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\underset{x→0}{\lim}\frac{e-\frac{ex}{2}+x^2(\frac{11}{24})e-e+\frac{ex}{2}}{x^2}=\frac{11e}{24}$