limit

$\begin{aligned} L & = \lim_{x \to 0} \frac {(1+x)^\frac 1x-e + \frac 12 ex}{x^2} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies} \\ & = \lim_{x \to 0} \frac {\left(\color{#D61F06}\frac 1{x(1+x)} - \frac {\ln(1+x)}{x^2} \right)(1+x)^\frac 1x+ \frac 12 e}{2x} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. } x \\ & = \lim_{x \to 0} \frac {\left(\color{#D61F06}\frac 1x (1-x+x^2-\cdots) - \frac 1{x^2}\left(x - \frac {x^2}2 + \frac {x^3}3 - \cdots \right) \right)(1+x)^\frac 1x+ \frac 12 e}{2x} & \small \color{#D61F06} \text{Using Maclaurin series} \\ & = \lim_{x \to 0} \frac {\left(-\frac 12 + \frac 23 x - \frac 34 x^2 + \cdots \right)(1+x)^\frac 1x+ \frac 12 e}{2x} & \small \color{#3D99F6} \text{A 0/0 case again} \\ & = \lim_{x \to 0} \frac {\left(\frac 23 - \frac 32 x + \frac {12}5 x^2 - \cdots \right)(1+x)^\frac 1x +\left(\frac 1{x(1+x)} - \frac {\ln(1+x)}{x^2} \right)^2(1+x)^\frac 1x}2 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. } x \\ & = \frac {\frac 23 e + \frac 14 e}2 = \frac {11e}{24} \approx \boxed{1.246} \end{aligned}$

References:

• L'Hôpital's rule
• Maclaurin series

If $f$ is twice differentiable near $0$ , then by applying L'Hospital's rule twice, $\begin{aligned} \lim_{x \to 0} \frac{e^{f(x)} - (1 + f'(0)x)e^{f(0)}}{x^2} &= \lim_{x \to 0} \frac{f'(x)e^{f(x)} - f'(0)e^{f(0)}}{2x} \\ &= \lim_{x \to 0} \frac{(f''(x) + f'(x)^2)e^{f(x)}}{2} \\ &= \frac{f''(0) + f'(0)^2}{2} e^{f(0)}. \end{aligned}$ Now plug $f(x) = \frac{\log(1+x)}{x} = 1 - \frac{1}{2}x + \frac{1}{3}x^2 + \cdots$ , so that $f(0) = 1$ , $f'(0) = -\frac{1}{2}$ , and $f''(0) = \frac{2}{3}$ . Then this limit becomes $\lim_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{x}{2}e}{x^2} = \frac{11}{24}e.$

The answer is not 0.458.Answer is 1.246. (1+x)^{1/x}=e[1-x/2+11x^{2}/24 . . . .]

use expansion of $(1+x)^(1/x)$ = $e(1-(x/2)+(11x^(2))/24 ....$