Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 22

Determine the $\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3} = \frac{e^0 - 1 - 0 - \frac{(0)^2}{2}}{0^3} = \frac{1-1-0-\frac{0^2}{2}}{0^3} = \frac{0}{0} \text{ Indeterminate}$


Thus, by applying L'Hospital's Rule...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3} &= \lim_{x \to 0}\frac{e^x - 0 - 1 - \frac{2x}{2}}{3x^2}\\
\\
&= \lim_{x \to 0} \frac{e^x - 1 - x}{3x^2}
\end{aligned}
\end{equation}
$

If we evaluate the limit, we will still get an indeterminate form. So, by applying L'Hospital's Rule again...
$\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x}{3x^2} = \lim_{x \to 0} \frac{e^x - 1}{6x}$

The limit is still indeterminate form so, again, by Applying L'Hospital's Rule for the third time.

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{e^x - 1}{6x} &= \lim_{x \to 0} \frac{e^x}{6}\\
\\
&= \frac{e^0}{6}\\
\\
&= \frac{1}{6}
\end{aligned}
\end{equation}
$