Question

The function $f\left(x\right) = \left(\frac{\log_{e}\left(1+ax\right) - \log_{e}\left(1-bx\right)}{x}\right)$ 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. $\log_e (ab)$
• D. $a - b$

Answer 1

Answer: (B)

$a + b$

Answer 2

$f\left(x\right) = \left(\frac{\log_{e}\left(1+ax\right) - \log_{e}\left(1-bx\right)}{x}\right)$
For $f(x)$ to be continuous at $x = 0$
$\lim_{x\to0^{-}} f\left(x\right) = \lim _{x\to 0^{+}} f\left(x\right) =f\left(0\right)$
$\therefore \:\:\: \lim _{x\to 0^{-}} \left(\frac{\log\left(1+ax\right)-\log \left(1-bx\right)}{x}\right)$
$= \lim _{h\to 0} \frac{\log \left(1-ah\right)-\log \left(1+bh\right)}{-h}$ ($\frac{0}{0}$ from)
Applying L' Hospital Rule, we get
$\lim_{h\to0} \frac{\frac{-a}{1-ah} - \frac{b}{1+bh}}{-1} = \lim _{h\to 0} \frac{a}{1-ah} +\frac{b}{1+bh} = a+b$
$\therefore \:\:\: f\left(0\right) =a+b$ for continuous at $x=0$