<s>[INST] Limit of $\frac{\frac{1}{e}(1+x)^{1/x}-1+\frac{x}{2}}{x^2}$ when $x\to0$ 
Find the limit of $\dfrac{\frac{1}{e}(1+x)^{1/x}-1+\frac{x}{2}}{x^2}$ when $x\to0$.

I tried applying L'Hospital rule, but it is not working here. How should I solve this?
 [/INST] With the definition of $a^b$ we have for the numerator 
\begin{align*}
(1+x)^\frac{1}{x} \cdot \exp(-1) -1+\frac{x}{2}&= \exp\left(\frac{1}{x} \cdot \ln(1+x) \right) \cdot \exp(-1) -1+\frac{x}{2}\\
&=1-\frac{x}{2}+\frac{11 x^2}{24}-1+\frac{x}{2} +\mathcal{O}(x^3)\\
&= \frac{11x^2}{24} +\mathcal{O}(x^3)
\end{align*}
Now the limit is equal to 
$$\lim_{x\to 0} \frac{\frac{11}{24} x^2}{x^2}=\frac{11}{24}$$
Maybe some words how to derive they taylor series without calculating to much. 
We have 
$$\log(1+x)= \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} = x-\frac{x^2}{2} \pm \dots$$ 
and 
$$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2}+\dots$$. 
So we have 
\begin{align*}
\exp( \log(1+x)\cdot x^{-1} -1)&= 1+ \log(1+x)\cdot x^{-1} -1 + \frac{(\log(1+x)\cdot x^{-1} -1)^2}{2} +\mathcal{O}((\log(1+x)\cdot x^{-1} -1)^3)\\
&= 1-\frac{x}{2} +\frac{x^2}{3} +  \left(\frac{-x}{2}+\frac{x^2}{3}\right)^2 \cdot \frac{1}{2}+ \mathcal{O}(x^3)
\end{align*}
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