Question

Evaluate the following limits:

$\underset{x\to 0}{\lim }\frac{2\sin x-\sin 2x}{x^3}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{2\sin x-\sin 2x}{x^3} \\ =\underset{x\to 0}{\lim }\frac{2\sin x-2\sin x\cos x}{x^3} \\ =\underset{x\to 0}{\lim }\frac{2\sin x\left(1-\cos x\right)}{x^3} \\ =\underset{x\to 0}{\lim }\frac{2\sin x\times 2{\sin }^2{\displaystyle \frac{x}{2}}}{x^3} \end{gathered}$$
$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{\sin x\times {\sin }^2{\displaystyle \frac{x}{2}}}{x\times {\displaystyle \frac{x^2}{4}}} \\ =\underset{x\to 0}{\lim }\frac{\sin x}{x}\times {\left(\underset{x\to 0}{\lim }\frac{\sin {\displaystyle \frac{x}{2}}}{{\displaystyle \frac{x}{2}}}\right)}^2 \\ =1\times 1\left(\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1\right) \\ =1 \end{gathered}$$