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).


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!
