Question

What is the value of $\displaystyle\lim_{x \to 0} \frac{x \, \sin \, 5x}{\sin^2 \, 4x}$

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

Answer 1

Answer: (C)

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