How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion?

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*}

Let us call our limit $\ell$.
I was considering the following identity

$$ 4\frac{e^{2x}-1-2x}{(2x)^2}-2\frac{e^x-1-x}{x^2}=\left(\frac{e^x-1}{x}\right)^2\quad\forall x\ne0 $$

If $\mathbf{\ell}$ exists and is not infinite, taking the limit of the above identity we have

$$ 4\ell-2\ell=1\implies\ell=\frac{1}{2} $$

but I am not able to prove the bold part above (if at all possible, in a simple way).

I thought it might be useful to present a way forward that relies on an integral representation of the numerator along with the mean-value theorem for integrals. To that end, we now proceed.

Note that we can write the numerator as

$$\begin{align} e^x-x-1&=\int_0^x \int_0^t e^s \,ds\,dt\\\\ &=\int_0^x \int_s^x e^s\,dt\,ds\\\\ &=\int_0^x (x-s)e^s\,ds \end{align}$$

Next, we apply the Mean-Value-Theorem for integrals to reveal

$$\begin{align} e^x-x-1&=e^{s^*}\int_0^x(x-s)\,ds\\\\ &=\frac12 x^2e^{s^*} \end{align}$$

for some value of $s^*\in (0,x)$.

Finally, exploiting the continuity of the exponential function yields the coveted limit

$$\begin{align} \lim_{x\to 0}\frac{e^x-x-1}{x^2}&=\lim_{x\to 0}\frac{\frac12 x^2e^{s^*}}{x^2}\\\\ &=\frac12 \end{align}$$

as expected!