Limit (sin x − x cos x) over x³

Problem

Compute $\lim_{x\to 0}\dfrac{\sin x - x\cos x}{x^3}$ with Taylor.

Solution

Expansions: $\sin x = x - \dfrac{x^3}{6} + o(x^4)$ and $\cos x = 1 - \dfrac{x^2}{2} + o(x^3)$, whence $x\cos x = x - \dfrac{x^3}{2} + o(x^4)$. Then $\sin x - x\cos x = \Bigl(x - \frac{x^3}{6}\Bigr) - \Bigl(x - \frac{x^3}{2}\Bigr) + o(x^3) = \frac{x^3}{2} - \frac{x^3}{6} + o(x^3) = \frac{x^3}{3} + o(x^3).$ Hence: $\frac{\sin x - x\cos x}{x^3} = \frac{1}{3} + o(1) \xrightarrow{x\to 0} \boxed{\frac{1}{3}}.$

Links

Topics: Derivatives
Concepts: Limit · Standard expansion
Methods: Limits via Taylor
Skills: Compute limits
Exercise type: Limit computation