How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? Is it possible to determine the limit
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
without using l'Hopital's rule nor any series expansion?
For example, suppose you are a student that has not studied derivative yet (and so not even Taylor formula and Taylor series).

Define $f(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$.  One possibility is to take $f(x)$ as the definition of $e^x$.  Since the OP has suggested a different definition, I will show they agree.
If $x=\frac{p}{q}$ is rational, then
\begin{eqnarray*}
f(x)&=&\lim_{n\to\infty}\left(1+\frac{p}{qn}\right)^n\\
&=&\lim_{n\to\infty}\left(1+\frac{p}{q(pn)}\right)^{pn}\\
&=&\lim_{n\to\infty}\left(\left(1+\frac{1}{qn}\right)^n\right)^p\\
&=&\lim_{n\to\infty}\left(\left(1+\frac{1}{(qn)}\right)^{(qn)}\right)^{p/q}\\
&=&\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^{p/q}\\
&=&e^{p/q}
\end{eqnarray*}
Now, $f(x)$ is clearly non-decreasing, so
$$
\sup_{p/q\leq x}e^{p/q}\leq f(x)\leq \inf_{p/q\geq x}e^{p/q}
$$
It follows that $f(x)=e^x$.
Now, we have
\begin{eqnarray*}
\lim_{x\to0}\frac{e^x-1-x}{x^2}&=&\lim_{x\to0}\lim_{n\to\infty}\frac{\left(1+\frac{x}{n}\right)^n-1-x}{x^2}\\
&=&\lim_{x\to0}\lim_{n\to\infty}\frac{n-1}{2n}+\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-2}\\
&=&\frac{1}{2}+\lim_{x\to0}x\lim_{n\to\infty}\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-3}\\
\end{eqnarray*}
We want to show that the limit in the last line is 0.  We have $\frac{{n\choose k}}{n^k}\leq\frac{1}{k!}\leq 2^{-(k-3)}$, so we have
\begin{eqnarray*}
\left|\lim_{x\to0}x\lim_{n\to\infty}\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-3}\right|&\leq&\lim_{x\to0}|x|\lim_{n\to\infty}\sum_{k=3}^n \left(\frac{|x|}{2}\right)^{k-3}\\
&=&\lim_{x\to0}|x| \frac{1}{1-\frac{|x|}{2}}\\
&=&0
\end{eqnarray*}
