<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] $$\lim_{x\to0} \dfrac{e^x-1-x}{x^2}= \lim_{x\to0} \dfrac{e^{-x}-1+x}{x^2}= \lim_{x\to0} \dfrac{e^x+e^{-x}-2}{2x^2}=\displaystyle \dfrac{1}{2}  \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-e^{\frac{-x}{2}}}{x}\Bigg)^2$$
$$=\displaystyle \dfrac{1}{2} \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-1}{x}-\dfrac{e^{\frac{-x}{2}}-1}{x}\Bigg)^2=\dfrac{1}{2} \Bigg(\dfrac{1}{2}-\dfrac{-1}{2}\Bigg)^2=\dfrac{1}{2}$$
</s>