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).

Given the familiar limit ${e^t-1\over t}\to1$ as $t\to0$, we can argue as follows:
$${e^x-1-x\over x^2}={1\over x^2}\int_0^x(e^u-1)\,du={1\over x}\int_0^1(e^{xv}-1)\,dv=\int_0^1v\left(e^{xv}-1\over xv\right)\,dv$$
Now intuitively we have ${e^{xv}-1\over xv}={e^t-1\over t}\to1$ as $xv=t\to0$, hence
$$\int_0^1v\left(e^{xv}-1\over xv\right)\,dv\to\int_0^1v\cdot1\,dv={1\over2}v^2\Big|_0^1={1\over2}$$
but to be rigorous we need to justify the intuition of bringing the limit inside the integral. If you have a fancy enough theorem, you can cite it and be done, but let's do it from first principles: We need to show that for any $\epsilon\gt0$, there is a $\delta\gt0$ such that $0\lt|x|\lt\delta$ implies
$$\left|\int_0^1 v\left(e^{xv}-1\over xv\right)\,dv-{1\over2} \right|=\left|\int_0^1 v\left({e^{xv}-1\over xv}-1\right)\,dv\ \right|\le\epsilon$$
What we do know (from the familiar limit) is that for any $\epsilon\gt0$ there is a $\delta\gt0$ such that $0\lt|t|\lt\delta$ implies $\left|{e^t-1\over t}-1\right|\lt\epsilon$. Now if $0\lt|x|\lt\delta$ and $0\lt v\lt1$, then $0\lt|xv|\lt\delta$ as well, so for any $\epsilon\gt0$ we conclude there is a $\delta\gt0$ such that $0\lt|x|\lt\delta$ implies
$$\left|\int_0^1 v\left({e^{xv}-1\over xv}-1\right)\,dv\ \right|\le\int_0^1 v\left|{e^{xv}-1\over xv}-1\right|\,dv\
\le\int_0^1v\epsilon\,dv={\epsilon\over2}\lt\epsilon$$
and we're done.
Remark: The "familiar" limit ${e^t-1\over t}\to1$ as $t\to0$ is the definition of the derivative of the exponential function at $0$. The whole proof here is built around knowing that $(e^x)'=e^x$, in the form $\int_0^xe^u\,du=e^x-1$.
