Q:

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!

A:

Write $f(x) = \frac{1}{x}\log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,
\begin{align*}
\lim_{x\to0} \frac{e^{f(x)} - e + \frac{e}{2}x}{x^2}
&= \lim_{x\to0} \frac{e^{f(x)}f'(x) + \frac{e}{2}}{2x} \\
&= \lim_{x\to0} \frac{e^{f(x)}f''(x) + e^{f(x)}f'(x)^2}{2} \\
&= \frac{e}{2}f''(0) + \frac{e}{2}f'(0)^2.
\end{align*}
Since $ f(x) = 1 - \frac{1}{2}x + \frac{1}{3}x^2 + \cdots $ near $0$, it follows that $f'(0) = -\frac{1}{2}$ and $f''(0) = \frac{2}{3}$. Therefore the limit equals
$$ \frac{e}{2}\cdot\frac{2}{3} + \frac{e}{2}\left(-\frac{1}{2}\right)^2
= \frac{11}{24}e. $$

A:

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.

