Question

What is the value of $\underset{x\to 0}{\lim }\frac{x\sin 5x}{{\sin }^{2}4x}$

• A. 0
• B. $\frac{5}{4}$
• C. $\frac{5}{16}$
• D. $\frac{25}{4}$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{x\sin 5x}{{\sin }^{2}4x}$
[multiply denominator and numerator with x]
We get,
$\underset{x\to 0}{\lim }\frac{x^2\sin 5x}{x{\sin }^{2}4x}=\underset{x\to 0}{\lim }\frac{\sin 5x}{x}.\frac{x^2}{{\sin }^{2}4x}$
Rearranging to bring a standard form, we get,
$=\underset{x\to 0}{\lim }\frac{5\sin 5x}{5x}.\frac{(4x{)}^2}{16{\sin }^{2}4x}$
$=\frac{5}{16}\left(\underset{x\to 0}{\lim }\frac{\sin 5x}{5x}\right).\frac{1}{\underset{x\to 0}{\lim }{\left(\frac{\sin 4x}{4x}\right)}^2}=\frac{5}{16}$