Evaluating $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$ without expansions in limits 
Evaluate $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e  + \frac{1}{2}ex}{x^2}$

One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{\log_e (1+x)^{1/x}} = e^{\frac{1}{x} (x-\frac{x^2}{2} -\frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $\frac{11e}{24}$ as an answer.
Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!

If I may suggest, the problem of
$$y=\frac{(1+x)^{\frac1 x} - e  + \frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{\frac1 x}\implies \log(a)= {\frac1 x}\log(1+x)$$
$$ \log(a)={\frac1 x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+O\left(x^4\right)$$ Now, continuing with Taylor
$$a=e^{\log(a)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}-\frac{7 e x^3}{16}+O\left(x^4\right)$$
$$y=\frac{\frac{11 e x^2}{24}-\frac{7 e x^3}{16}+O\left(x^4\right) }{x^2}=\frac{11 e}{24}-\frac{7 e x}{16}+O\left(x^2\right)$$ which gives not only the limit but also how it is approached.
