Question

$\underset{x\to 0}{\lim }\frac{\left(1-\cos 2x\right)\sin 5x}{x^2\sin 3x}=$

(a) 10/3
(b) 3/10
(c) 6/5
(d) 5/6

Answer 1

(a) 10/3

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\left(1-\cos 2x\right)\sin 5x}{x^2\sin 3x} \\ =\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x\sin 5x}{x^2\sin 3x} \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{2{\sin }^{2}x}{x^2}}\left({\displaystyle \frac{\sin 5x}{5x}}\right)\times 5}{\left({\displaystyle \frac{\sin 3x}{3x}}\right)3} \\ =\frac{10\times 1^2\times 1}{1\times 3} \\ =\frac{10}{3} \end{gathered}$$