Question

$\displaystyle \lim_{x \to 0}$ $\frac{\left(1-\cos\,2x\right)\sin\,5x}{x^{2}\,\sin\,3x}$ equals

• A. $\frac{10}{3}$
• B. $\frac{3}{10}$
• C. $\frac{6}{5}$
• D. $\frac{5}{6}$

Answer 1

Answer: (A)

We have, $\displaystyle\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^{2} \sin 3 x}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^{2}} \displaystyle\lim _{x \rightarrow 0} \frac{\sin 5 x}{\sin 3 x}$
$=\displaystyle\lim _{x \rightarrow 0} \cdot \frac{2 \sin ^{2} x}{x^{2}} \displaystyle\lim _{x \rightarrow 0} \frac{\sin ^{2} 5 x}{5 x} \frac{3 x}{\sin 3 x} \cdot \frac{5 x}{3 x}$
$=\displaystyle\lim _{x \rightarrow 0} 2\left(\frac{\sin x}{x}\right) \cdot \frac{5}{3}$
$=\frac{2 \times 5}{3}=\frac{10}{3}$