Without using L'Hôpital's Rule or the Taylor Series expansion for the sine function, I would like to show that the function
$$\frac{x - \sin{x}}{x^{3}}$$
has a limit at $0$. I understand that if it is assumed that the function has a limit at $0$, the limit must be $1/6$.
Wednesday, December 14, 2016
calculus - Convergence of $lim_{x to 0}frac{x - sin{x}}{x^3}$
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