Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\ln (1+3x)-\ln (1-2x)}{x},& x\ne 0\\ a,& x=0\end{array}$ is continuous at $x=0$, then the value of $a$ is

• A. $5$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A $5$

$f\left(x\right)$ to be continuous at $x=0$, we must have
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
$\Rightarrow a=\underset{x\to 0}{lim}\frac{\ln (1+3x)-\ln (1-2x)}{x}$
$\Rightarrow a=\underset{x\to 0}{lim}\frac{\ln (1+3x)}{3x}\cdot 3-\underset{x\to 0}{lim}\frac{\ln (1-2x)}{-2x}\cdot (-2)$

$\therefore a=3+2=5$