Question

If the function $f\left(x\right)=\{\begin{array}{c}\frac{x\tan 2x}{\sin 3x.\sin 5x},forx\ne 0\\ k,forx=0\end{array}$, is continuous at $x=0$, then $f\left(0\right)$:

• A. $\frac{2}{13}$
• B. $\frac{2}{17}$
• C. $\frac{2}{11}$
• D. $\frac{2}{15}$

Answer 1

The correct option is D $\frac{2}{15}$

For $f$ to be continuous, we must have,

$f\left(0\right)=\underset{x\to 0}{lim}\frac{x\tan 2x}{\sin 3x.\sin 5x}$

$\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\frac{2\left(\frac{\tan 2x}{2x}\right)}{15\left(\frac{\sin 3x}{3x}\right)\left(\frac{\sin 5x}{5x}\right)}=\frac{2}{15}=k$