Question

Let $f \left(x\right)=\frac{ln\left(1+ax\right)-ln\left(1-bx\right)}{x}$, $x \ne 0$.
If $f(x)$ is continuous at $x = 0$, then $f(0) =$

• A. $a - b$
• B. $a + b$
• C. $b - a$
• D. $ln\,a + ln\, b$

Answer 1

Answer: (B)

$\displaystyle \lim_{x \to 0} f (x)$
$=\displaystyle \lim_{x \to 0}$ $\frac{ln\left(1+ax\right)-ln\left(1-bx\right)}{x}$ $(\frac{0}{0}$ form$)$
$=\displaystyle \lim_{x \to 0}$ $\left[\frac{a}{1+ax}+\frac{b}{1-bx}\right]$ ($L'$ Hospital Rule)
$=a+b$.