Question

$\underset{x\to 0}{lim}\frac{sin3x}{sin5x}=$

• A. -2
• B. $\frac{3}{5}$
• C. $\frac{5}{3}$
• D. 15

Answer 1

The correct option is B $\frac{3}{5}$

We have,$\underset{x\to 0}{lim}\frac{sin3x}{sin5x}$
This is an indeterminate limit of the form $\frac{0}{0}$

Applying L'Hospital's Rule,

$\underset{x\to 0}{lim}\frac{sin3x}{sin5x}=\underset{x\to 0}{lim}\frac{cos3x\frac{d}{dx}3x}{cos5x\frac{d}{dx}3x}$

$=\underset{x\to 0}{lim}\frac{3cos3x}{5cos5x}$

Substituting x=0, the limit evaluates to $\frac{3}{5}$