Limit with indeterminate form

Question

What is $${\lim_{x\rightarrow 0}} \frac{(1+x)^{1/x}-e+\frac{ex}2}{x^2}$$ Here's what I tried $$\text {ATTEMPT}$$ We know that ${\lim_{h\rightarrow 0}} (1+h)^{1/h}=e$ so the $e.-e$ get cancelled out now we want ${\lim_{x\rightarrow 0}} \frac{ex/2}{x^2}$ which is $\infty$ but it isn't in any option given. So please tell me any approximation or any better way to evaluate it.

Answer 1

$$(1+x)^{\frac{1}{x}}=e^{\ln((1+x)^{\frac{1}{x}})}=e^{1-\frac{x}{2}+\frac{x^2}{3}...}$$ $$=e.e^{-\frac{x}{2}+\frac{x^2}{3}...}$$ $$=e\Big[1+\Big(-\frac{x}{2}+\frac{x^2}{3}...\Big)+\Big(-\frac{x}{2}+\frac{x^2}{3}...\Big)^2 +...\Big]$$ $$=e\Big[1-\frac{x}{2}+\frac{11x^2}{24}...\Big]$$

$$\therefore \lim_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}=\frac{11e}{24}$$