Question

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

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

Answer 1

Answer: (A)

Using, $1-cos2x=2{sin}^{2}x$,
The expression transforms to,
$li{m}_{x\to 0}\frac{2{sin}^{2}xsin5x}{x^{2}sin3x}$
Rewriting the expression in a different form,
$li{m}_{x\to 0}\frac{2{sin}^{2}x}{x^2}\times \frac{sin5x}{5x}\times \frac{3x}{sin3x}\times \frac{5}{3}$
Therefore, the limit to the expression is $\frac{10}{3}$
Hence, option 'A' is correct.