Question

$\underset{x\to 0}{lim}\frac{\sin 3x}{\sin 5x}$ is equal to

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

Answer 1

The correct option is C

$\frac{3}{5}$

We have, $\underset{x\to 0}{lim}\frac{\sin 3x}{\sin 5x}$

At $x=0$, the value of the expression is in determinate form of $\frac{0}{0}$
Applying L'Hospital Rule
Takin derivative of Numerator and Denominator,
$\underset{x\to 0}{lim}\frac{\frac{d}{dx}\sin 3x}{\frac{d}{dx}\sin 5x}$
$\Rightarrow \underset{x\to 0}{lim}\frac{\cos 3x\frac{d}{dx}3x}{\cos 5x\frac{d}{dx}5x}$
$\underset{x\to 0}{lim}\frac{3\cos 3x}{5\cos 5x}$

Putting $x=0$, we get
The value of limit $=\frac{3\cos 0}{5\cos 0}=\frac{3}{5}$