A limit with sine

#limits#taylor

Statement

Compute
$\lim_{x\to 0}\frac{\sin x - x}{x^3}.$

Hint

Use the Taylor expansion of $\sin$ at $0$ up to order $3$ .

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As $x\to0$ , $\sin x = x-\dfrac{x^3}{6}+o(x^3)$ .

Hence $\dfrac{\sin x-x}{x^3}=-\dfrac16+o(1)$ , and the limit is $-\dfrac16$ .