Q116 / 180 · COMEDK 2018 · Mathematics

The function $f(x)=\left(\frac{\log _{e}(1+a x)-\log _{e}(1-b x)}{x}\right)$ 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
    $\frac{a+b}{2}$
  • Correct answer
    B
    $a+b$
  • C
    $\log _{e}(a b)$
  • D
    $a-b$

Solution

We have, $f(x)=\frac{\log _{e}(1+a x)-\log _{e}(1-b x)}{x}$ $=\lim _{x \rightarrow 0} \frac{\log _{e}(1+a x)-\log _{e}(1-b x)}{x} \quad\left[\frac{0}{0}\right.$ form $]$ Applying L'Hospital rule, $$ \lim _{x \rightarrow 0} \frac{\frac{a}{1+a x}+\frac{b}{1-b x}}{1}=\frac{\frac{a}{1}+\frac{b}{1}}{1}=a+b $$

Hence, the correct answer is option: 'B'.

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