<s>[INST] $\lim_{x\to0}\frac{e^x-1-x}{x^2}$ using only rules of algebra of limits. I would like to solve that limit  solved using only rules of algebra of limits.
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
All the answers in How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? do not fully address my question.
A challenging limit problem for the level of student who knows that:
$$\begin{align*}
\lim\limits_{x\to +\infty} e^x&=+\infty\tag1\\
\lim\limits_{x\to -\infty} e^x&=0\tag2\\
\lim\limits_{x\to +\infty} \frac{e^x}{x^n}&=+\infty\tag3\\
\lim\limits_{x\to -\infty} x^ne^x&=0\tag4\\
\lim\limits_{x\to 0} \frac{e^x-1}{x}&=1\tag5
\end{align*}$$
 [/INST] $$\left(\frac{e^x-1}x\right)^2=\frac{e^{2x}-2e^x+1}{x^2}=2^2\frac{e^{2x}-1-2x}{(2x)^2}-2\frac{e^x-1-x}{x^2}.$$
Then, taking the limit,
$$1=4L-2L.$$

UPDATE: the same approach can be used for the next order,
$$\left(\frac{e^x-1}x\right)^3=\frac{e^{3x}-3e^{2x}+3e^x-1}{x^3}=3^3f(3x)-3\cdot2^3(2x)+3f(x),$$
where $f(x)=\dfrac{e^x-1-x-\dfrac{x^2}2}{x^3},$
and
$$1=27L-24L+3L.$$
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