Question

The values of $p$ and $q$ so that the function
$f\left(x\right)=\{\begin{array}{c}{(1+\left|\sin x\right|)}^{\frac{p}{\sin x}},\frac{-\pi }{6}<x<0\\ q,x=0\\ e^{\frac{\sin 2x}{\sin 3x}},0<x<\frac{\pi }{6}\end{array}$ is continuous at $x=0$ is

• A. $p=\frac{1}{3},q=e^{2/3}$
• B. $p=0,q=e^{2/3}$
• C. $p=\frac{2}{3},q=e^{-2/3}$
• D. $p=-\frac{2}{3},q=e^{2/3}$

Answer 1

The correct option is D $p=-\frac{2}{3},q=e^{2/3}$

Left hand limit :

$\Rightarrow \underset{x\to 0}{lim}(1+|\sin x|{)}^{\frac{p}{\sin x}}$

Above is $1^{\infty }$ form

$\Rightarrow e^{\underset{x\to 0}{lim}(1+|\sin x|-1)\times \frac{p}{\sin x}}$

$\Rightarrow e^{\underset{x\to 0}{lim}(-\sin x)\times \frac{p}{\sin x}}$

$\Rightarrow LHL=e^{-p}$

Right hand limit :

$\Rightarrow e^{\underset{x\to 0}{lim}\frac{\sin 2x}{\sin 3x}}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin 2x}{\sin 3x}\times \frac{3x}{2x}\times \frac{2}{3}$

$\Rightarrow \underset{x\to 0}{lim}\frac{3x}{\sin 3x}\times \frac{\sin 2x}{2x}\times \frac{2}{3}=\frac{2}{3}$

$RHL=e^{2/3}$

If $f\left(x\right)$ is continuous then, $LHL=RHL$

$q=e^{2/3}$ and $p=\frac{-2}{3}$