Mathematical Reflections 2017, Issue 2 - Problem U406

Problem:
Evaluate $$\lim_{x \to 0} \dfrac{\cos(n+1)x\cdot \sin nx-n\sin x}{x^3}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{rcl} \cos(n+1)x&=&1-\dfrac{1}{2}(n+1)^2x^2+o(x^3) \\  \sin nx&=&nx-\dfrac{1}{6}n^3x^3+o(x^3) \\ n\sin x&=&nx-\dfrac{1}{6}nx^3+o(x^3). \end{array}$$
So, $$\dfrac{\left(1-\frac{1}{2}(n+1)^2x^2+o(x^3)\right)\left(nx-\frac{1}{6}n^3x^3+o(x^3)\right)-\left(nx-\frac{1}{6}nx^3+o(x^3)\right)}{x^3}=-\dfrac{\frac{x^3}{3}n(n+1)(2n+1)+o(x^3)}{x^3}.$$ So, $$\lim_{x \to 0} \dfrac{\cos(n+1)x\cdot \sin nx-n\sin x}{x^3}=-\dfrac{n(n+1)(2n+1)}{3}.$$