$\lim_{x\to0}\frac{e^x-1-x}{x^2}$ 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&#39;Hopital&#39;s rule nor any series expansion? do not fully address my question.

A challenging limit problem for the level of student who knows that:
$$\begin{align*}
\lim\limits_{x\to +\infty} e^x&amp;=+\infty\tag1\\
\lim\limits_{x\to -\infty} e^x&amp;=0\tag2\\
\lim\limits_{x\to +\infty} \frac{e^x}{x^n}&amp;=+\infty\tag3\\
\lim\limits_{x\to -\infty} x^ne^x&amp;=0\tag4\\
\lim\limits_{x\to 0} \frac{e^x-1}{x}&amp;=1\tag5
\end{align*}$$
$$
\lim\limits_{x\to 0}\dfrac{e^x-1-x}{x^2} = \lim\limits_{x\to 0}\left(\dfrac{1}{e^x+1+x}\right)\lim\limits_{x\to 0}\dfrac{e^{2x}-(1+x)^2}{x^2} = \dfrac{1}{2}\lim\limits_{x\to 0}\left(4\dfrac{e^{2x}-1-2x}{4x^2}-1\right) = 2\lim\limits_{x\to 0}\left(\dfrac{e^{2x}-1-2x}{4x^2}\right)-\dfrac{1}{2} = 2\lim\limits_{x\to 0}\left(\dfrac{e^x-1-x}{x^2}\right)-\dfrac{1}{2}.
$$

Therefore,


  $$\lim\limits_{x\to 0}\dfrac{e^x-1-x}{x^2} = \dfrac{1}{2}\,.$$




This can be extended in a natural way. Define $s_{n}(x):=\sum\limits_{k=0}^{n-1}\dfrac{x^k}{k!}$ and observe the following:
$$
\begin{eqnarray*}
\lim_{x\to0}\dfrac{e^x-s_{n}(x)}{x^n} &amp;=&amp; \lim_{x\to0}\left(\dfrac{1}{\sum\limits_{r=0}^{n-1}e^{rx}(s_n(x))^{n-1-r}}\right)\lim_{x\to0}\dfrac{e^{nx}-(s_n(x))^n}{x^n} \\ &amp;&amp;\ \\ \ \\ &amp;=&amp; \dfrac{1}{n}\left(\lim_{x\to0}\dfrac{e^{nx}-s_n(nx)-\frac{n^{n-1}-1}{(n-1)!}x^n-o(x^n)}{x^n}\right)\\ &amp;&amp;\ \\ \ \\&amp;=&amp; 
n^{n-1}\lim_{x\to0}\left(\dfrac{e^{nx}-s_n(nx)}{(nx)^n}\right)-\frac{n^{n-1}-1}{n!}\,, 
\end{eqnarray*}
$$

where, upon expansion, the polynomial $(s_n(x))^n = s_n(nx) + \frac{n^{n-1}-1}{(n-1)!}x^n + o(x^n)$. Consequently,


  $$
\lim_{x\to0}\dfrac{e^x-s_n(x)}{x^n} = \dfrac{1}{n!}\,.
$$




The foregoing arguments require the existence of the primary limits being sought after. Given that our hypothetical pupil is not armed with sophisticated $\epsilon, \delta$ analysis techniques, it is difficult to see how he might proceed in showing existence. Indeed, how does this pupil even know what a limit is, let alone the algebraic laws that follow by definition? Nevertheless, with these concerns in mind, we might at least allow our student one more atom of information: the interchange of limits for the particular class of functions we are interested in here. Thus, our pupil proceeds, somewhat mechanistically, as follows:

$$
\begin{eqnarray*}
\lim\limits_{x\to 0}\frac{e^{x}-s_n(x)}{x^n} &amp;=&amp; \lim\limits_{x\to 0}\lim\limits_{y\to 0}\frac{\left(e^{y}(\frac{e^{x}-1}{x})-\frac{1}{x}\Big(s_n(x+y)-s_n(y)\Big)\right)}{\Big(\sum\limits_{r=0}^{n-1}(x+y)^r y^{n-1-r}\Big)} \\ &amp;&amp;\ \\&amp;=&amp; \lim\limits_{y\to 0}\lim\limits_{x\to 0}\frac{\left(e^{y}(\frac{e^{x}-1}{x})-\frac{1}{x}\Big(s_n(x+y)-s_n(y)\Big)\right)}{\Big(\sum\limits_{r=0}^{n-1}(x+y)^r y^{n-1-r}\Big)} \\ &amp;&amp;\ \\&amp;=&amp; \frac{1}{n}\lim\limits_{y\to 0}\frac{\left(e^{y}-s_{n-1}(y)\right)}{y^{n-1}}\,.
\end{eqnarray*}$$

So, by successive application of such interchanges, our pupil deduces


  $$
\lim\limits_{x\to 0}\frac{e^{x}-s_n(x)}{x^n} = \frac{1}{n!}\,.
$$
$$e^x=1+x+\frac{1}{2}x^2+o(x^2)_{x\to0}\\
\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{1+x+\frac{1}{2}x^2+o(x^2)-1-x}{x^2}=
\lim_{x\to0}\frac{\frac{1}{2}x^2+o(x^2)}{x^2}=\frac{1}{2}+0
$$
($f=o(x^2)\Leftrightarrow\lim_{x\to0}\frac{f(x)}{x^2}=0$)