Without using Taylor expansion or L'Hospital rule evaluate the limit: 
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$


I give the two alternate ways so that no one gives them again in their answer:
Using L'Hospital:
$$L=\lim_{x\to0}\frac{e^x-x-1}{3x^2}=\lim_{x\to0}\frac{e^x-1}{6x}=\lim_{x\to0}\frac{e^x}{6}=\frac16$$
Using Taylor:
$$L=\lim_{x\to0}\frac{\color{red}{(1+x+x^2/2+x^3/6+O(x^4))}-1-x-x^2/2}{x^3}=\frac16$$
Using nothing(not infact):
$$L=\lim_{x\to0}\frac{e^x-1-x^2/2}{x^3}\tag{sorry I omitted the x}\\
=\lim_{x\to0}\frac{e^{2x}-1-2x^2}{8x^3}\\
8L=\lim_{x\to0}\frac{e^{2x}-1-2x^2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{2x}-e^x-3x^2/2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{8x^3}\\
56L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{x^3}\\
49L=\lim_{x\to0}\frac{e^{4x}-2e^{2x}+e^x-9x^2/2}{x^3}=?$$
 A: 
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$


If Limit exists:
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}\\
L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{8x^3}\\
8L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{2x}-e^x-x-3x^2/2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-2x-6x^2}{8x^3}\\
28L=\lim_{x\to0}\frac{\frac12e^{4x}-\frac12e^{2x}-x-3x^2}{x^3}\\
21L=\lim_{x\to0}\frac{\frac12e^{4x}-\frac32e^{2x}+e^x-\frac32x^2}{x^3}\\
21L=\lim_{x\to0}\frac{\frac12e^{8x}-\frac32e^{4x}+e^{2x}-6x^2}{8x^3}\\
168L=\lim_{x\to0}\frac{\frac12e^{8x}-\frac32e^{4x}+e^{2x}-6x^2}{x^3}\\
42L=\lim_{x\to0}\frac{\frac18e^{8x}-\frac38e^{4x}+\frac14e^{2x}-\frac32x^2}{x^3}\\
21L=\lim_{x\to0}\frac{\frac18e^{8x}-\frac78e^{4x}+\frac74e^{2x}-e^x}{x^3}\\
21L=\lim_{x\to0}\frac{\frac18e^x(e^x-1)^3(10e^x+6e^{2x}+3e^{3x}+e^{4x}+8)}{x^3}\\
21L=\frac1828\\
\huge L=\frac16
$$
A: I will deal with the case $x \to 0^{+}$ and leave the $x \to 0^{-}$ case for OP and other readers. Let us use the formula $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ We then have $$\begin{aligned}L &= \lim_{x \to 0^{+}}\dfrac{e^{x} - 1  - x - \dfrac{x^{2}}{2}}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\lim_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} - 1  - x - \dfrac{x^{2}}{2}}{x^{3}}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}\dfrac{\left(1 + \dfrac{x}{n}\right)^{n} - 1  - x - \dfrac{x^{2}}{2}}{x^{3}}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}\frac{1}{x^{3}}\left\{-\frac{x^{2}}{2n} + \dfrac{\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)}{3!}x^{3} + \cdots\right\}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}\left\{-\frac{1}{2nx} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)\left(1 - \dfrac{3}{n}\right)}{4!}x + \cdots\right\}\\
&= \lim_{x \to 0^{+}}\lim_{n \to \infty}f(x, n) - \frac{1}{2nx}\end{aligned}$$ where $f(x, n)$ is a finite sum with number of terms dependent on $n$. By using Monotone convergence theorem we can show that the sum $f(x, n)$ tends to a limit (as $n\to\infty$) dependent on $x$ say $f(x)$ and we can also show that $f(x) = (1/6) + o(x)$. The term $-1/2nx$ tends to $0$. It follows that $L = \lim_{x \to 0^{+}}f(x) = 1/6$.
A: Your answers are wrong because you omit the $-x\;$ term. And in your L'Hospital answer
$e^x - x$ is not $0$ for $x=0!\,$ Your function goes to $+\infty$ for $x \rightarrow 0.$
