$\lim_{x\to0}\frac{e^x-1-x}{x^2}$ using only rules of algebra of limits.

$$ \lim\limits_{x\to 0}\dfrac{e^x-1-x}{x^2} = \lim\limits_{x\to 0}\left(\dfrac{1}{e^x+1+x}\right)\lim\limits_{x\to 0}\dfrac{e^{2x}-(1+x)^2}{x^2} = \dfrac{1}{2}\lim\limits_{x\to 0}\left(4\dfrac{e^{2x}-1-2x}{4x^2}-1\right) = 2\lim\limits_{x\to 0}\left(\dfrac{e^{2x}-1-2x}{4x^2}\right)-\dfrac{1}{2} = 2\lim\limits_{x\to 0}\left(\dfrac{e^x-1-x}{x^2}\right)-\dfrac{1}{2}. $$

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: $$ \begin{eqnarray*} \lim_{x\to0}\dfrac{e^x-s_{n}(x)}{x^n} &=& \lim_{x\to0}\left(\dfrac{1}{\sum\limits_{r=0}^{n-1}e^{rx}(s_n(x))^{n-1-r}}\right)\lim_{x\to0}\dfrac{e^{nx}-(s_n(x))^n}{x^n} \\ &&\ \\ \ \\ &=& \dfrac{1}{n}\left(\lim_{x\to0}\dfrac{e^{nx}-s_n(nx)-\frac{n^{n-1}-1}{(n-1)!}x^n-o(x^n)}{x^n}\right)\\ &&\ \\ \ \\&=& n^{n-1}\lim_{x\to0}\left(\dfrac{e^{nx}-s_n(nx)}{(nx)^n}\right)-\frac{n^{n-1}-1}{n!}\,, \end{eqnarray*} $$

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:

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

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

$$ \lim\limits_{x\to 0}\frac{e^{x}-s_n(x)}{x^n} = \frac{1}{n!}\,. $$

$$\left(\frac{e^x-1}x\right)^2=\frac{e^{2x}-2e^x+1}{x^2}=2^2\frac{e^{2x}-1-2x}{(2x)^2}-2\frac{e^x-1-x}{x^2}.$$

Then, taking the limit,

$$1=4L-2L.$$

UPDATE: the same approach can be used for the next order,

$$\left(\frac{e^x-1}x\right)^3=\frac{e^{3x}-3e^{2x}+3e^x-1}{x^3}=3^3f(3x)-3\cdot2^3(2x)+3f(x),$$ where $f(x)=\dfrac{e^x-1-x-\dfrac{x^2}2}{x^3},$ and $$1=27L-24L+3L.$$

In order to show that

$$\lim_{x\to0}{e^x-1-x\over x^2}={1\over2}$$

it suffices to show that

$$\lim_{x\to0}{1\over x^2}\int_0^x(e^u-1-u)\,du=0$$

Let $g(u)=e^u-1-u$. Since $g'(u)=e^u-1$ and $g''(u)=e^u$, we see that $g(u)$ has a global minimum at $u=0$, with $g(0)=0$. Consequently

$$0\le{1\over x^2}\int_0^x(e^u-1-u)\,du\le{1\over x}\int_0^x{e^u-1-u\over u}\,du\quad\text{for }x\not=0$$

By the Mean Value Theorem,

$$\lim_{x\to0}{1\over x}\int_0^x{e^u-1-u\over u}\,du=\lim_{u\to0}{e^u-1-u\over u}$$

But

$$\lim_{u\to0}{e^u-1-u\over u}=\lim_{u\to0}{g(u)-g(0)\over u-0}=g'(0)=0$$

Thus, by the Squeeze Theorem, $\lim_{x\to0}{1\over x^2}\int_0^x(e^u-1-u)\,du=0$.

Note that

$$e^x-1-x= \int_0^x (e^t-1)\, dt = \int_0^x \int_0^t e^s\,ds\, dt.$$

Now $1\le e^s \le e^x$ above. Thus $e^x-1-x$ lies between

$$\tag 1 \int_0^x \int_0^t 1 \,ds\, dt,\, \,\,\int_0^x \int_0^t e^x\,ds\, dt.$$

The first iterated integral in $(1)$ equals $x^2/2,$ the one on the right equals $e^x(x^2/2).$ So we have $e^x-1-x$ between $x^2/2$ and $e^x(x^2/2).$ Dividing by $x^2$ and using the squeeze theorem gives $1/2$ for the desired limit.

One could consider the more general limit:

$$L_n=\lim_{x\to0}x^{-n}(e^x-P_{n-1}(x))$$

where

$$P_n(x)=\sum_{k=0}^n\frac{x^k}{k!}$$

It may be easily seen that

$$e^x-P_n(x)=\int_0^xe^t-P_{n-1}(t)~\mathrm dt$$

And by induction,

$$e^x-P_{n-1}(x)=\int_0^x\int_0^{\sigma_n}\dots\int_0^{\sigma_2}e^{\sigma_1}~\mathrm d\sigma_1~\mathrm d\sigma_2\dots\mathrm d\sigma_n$$

By Cauchy's repeated integral formula, it can be seen that

$$e^x-P_{n-1}(x)=\frac1{(n-1)!}\int_0^x(x-t)^{n-1}e^t~\mathrm dt$$

And by substituting $t=xu$, this simplifies down to

$$e^x-P_{n-1}(x)=\frac{x^n}{(n-1)!}\int_0^1(1-u)^{n-1}e^{xu}~\mathrm du$$

Which so nicely cancels the $x^n$, and likewise, it is easy to see that

$$\int_0^1(1-u)^{n-1}e^{-|x|}~\mathrm du<\int_0^1(1-t)^{n-1}e^{xu}~\mathrm du<\int_0^1(1-u)^{n-1}e^{|x|}~\mathrm du$$

So by the squeeze theorem,

$$\begin{align}\lim_{x\to0}x^{-n}(e^x-P_{n-1}(x))&=\lim_{x\to0}\frac1{(n-1)!}\int_0^1(1-u)^{n-1}e^{xu}~\mathrm du\\&=\frac1{(n-1)!}\int_0^1(1-u)^{n-1}~\mathrm du\\&=\frac1{(n-1)!}\frac1n\\&=\frac1{n!}\end{align}$$

As expected.