Find the limit as x goes to 0 of the given expression

Evaluate the limit.

$\lim_{x \to 0} \frac{x (e^x + 1) - 2 (e^x -1)}{x^3}.$

From page 287 we will use the following expansion for $e^x$ as $x \to 0$ ,

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3).$

Therefore, we compute the limit as

$\begin{align*} \lim_{x \to 0} &\frac{x (e^x + 1) - 2(e^x - 1)}{x^3} \\[9pt] &= \lim_{x \to 0} \frac{x \left( 2 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3) \right) - 2 \left( x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3)\right)}{x^3} \\[9pt] &= \lim_{x \to 0} \frac{ 2x + x^2 + \frac{x^3}{2} + \frac{x^4}{6} + o(x^4) - 2x - x^2 - \frac{x^3}{3} + o(x^3)}{x^3} \\[9pt] &= \lim_{x \to 0} \frac{\frac{1}{6}x^3 + o(x^3)}{x^3} \\[9pt] &= \frac{1}{6}. \end{align*}$

(In the second to last line, all of the $x^4$ terms got absorbed into the $o(x^3)$ since any function $f(x)$ such that $\frac{f(x)}{x^4} \to 0$ as $x \to 0$ will certainly also obey $\frac{f(x)}{x^3} \to 0$ as $x \to 0$ . Again, this is why little $o$ -notation can be very convenient.)