Question

$\text{If}f\left(x\right)=\{\begin{array}{cc}\frac{\sin 3x}{e^{2x}-1};& x\ne 0\\ k-2;& x=0\end{array}\text{is continuous at}x=0,\text{then}k=$

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{7}{2}$

Answer 1

The correct option is D $\frac{7}{2}$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\sin 3x}{e^{2x}-1}=\underset{x\to 0}{lim}\frac{\frac{\sin 3x}{3x}\times 3x}{\frac{e^{2x}-1}{2x}\times 2x}=\frac{1\times 3x}{1\times 2x}=\frac{3}{2}$
Since, $f\left(x\right)$ is continuous at $x=0$.
$\therefore \underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)\frac{3}{2}=k-2\Rightarrow k=\frac{7}{2}$