Question

The functions $f\left(x\right)=\left(\frac{{\log }_e(1+ax)-{\log }_e(1-bx)}{x}\right)$ is undefined at $x=0$. The value which should be assigned to $f$ at $x=0$ so that it is continuous at $x=0$ is

• A. $\frac{a+b}{2}$
• B. $a+b$
• C. ${\log }_e(a,b)$
• D. $a-b$

Answer 1

Answer: (B)

$a+b$

To make the function continuous at x=0, it should be assigned the value $\underset{x\to 0}{lim}\left(x\right)$ at x=0

On writing the given function in form

$\Rightarrow \underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{a{\log }_e(1+ax)}{ax}-\underset{x\to 0}{lim}\frac{b{\log }_e(1-bx)}{bx}$
$\Rightarrow \underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{a{\log }_e(1+ax)}{ax}+\underset{x\to 0}{lim}\frac{b{\log }_e(1-bx)}{-bx}$

We know that $\underset{h\to 0}{lim}\frac{{\log }_e(1+h)}{h}$ equals 1

$\therefore$ on solving, we get $f\left(0\right)=a+b$