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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.)

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