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Question

# Use series to evaluate the following limit.$lim x→0 sin x-x/x^3$

Solution

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\begin{align*} \sin x = \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dotsb \end{align*}

Substitute the series for $\sin x$ and simplify

\begin{align*} \lim\limits_{x \to 0} \dfrac{\sin x - x}{x^3} &= \lim\limits_{x \to 0} \dfrac{ \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dotsb \right) - x}{x^3}\\ \\ &= \lim\limits_{x \to 0} \dfrac{ - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dotsb }{x^3}\\ \\ &= \lim\limits_{x \to 0} \left( - \dfrac{1}{3!} + \dfrac{x^2}{5!} - \dfrac{x^4}{7!} + \dotsb \right)\\ \\ &= - \dfrac{1}{3!} + 0 - 0 + \dotsb\\ \\ &= -\dfrac{1}{6} \end{align*}

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