Question

If $f(x)=\frac{xtan2x}{\sin 3x\sin 5x}$ is continuous at $x=0$, then $f(0)$ is equal to:

• A. $\frac{1}{15}$
• B. $\frac{15}{2}$
• C. $\frac{2}{15}$
• D. none of these

Answer 1

Answer: (C)

Given function is :
$f(x)=\frac{xtan2x}{sin3xsin5x}$
If $f(x)$ is continuous at $x=0$, then limit
$x\to 0$ should exist and $f(0)=\underset{x\to 0}{\lim }f(x)$
So, $f(0)=\underset{x\to 0}{\lim }\frac{xtan2x}{sin3xsin5x}$
Multiplying Nr and Dr by 3,we get
$\Rightarrow f(0)=\underset{x\to 0}{\lim }\frac{1}{3}\left(\frac{3x}{sin3x}\right)\frac{tan2x}{sin5x}$
$\Rightarrow f(0)=\frac{1}{3}\underset{x\to 0}{\lim }\left(\frac{3x}{sin3x}\right)\underset{x\to 0}{\lim }\frac{tan2x}{sin5x}$
$=\frac{1}{3}\underset{x\to 0}{\lim }\frac{tan2x}{sin5x}[\because \underset{x\to 0}{\lim }\frac{x}{sinx}=1]$
$\Rightarrow f(0)=\frac{1}{3}\underset{x\to 0}{\lim }\frac{tan2x}{2x}\times 2x\frac{5x}{sin5x}.\frac{1}{5x}$
$=\frac{1}{3}.\underset{x\to 0}{\lim }.\frac{tan2x}{2x}.\underset{x\to 0}{\lim }.\frac{5x}{sin5x}.\frac{2x}{5x}$
$\Rightarrow f(0)=\frac{1}{3}.1.1.\frac{2}{5}=\frac{2}{15}$