If \( f(x)=\begin{cases}\displaystyle\frac{\log_{e}(1+ax)-\log_{e}(1-bx)}{x}, & x\ne 0\\ k, & x=0\end{cases} \) and \( f(x) \) is continuous at \( x=0 \), then the value of \( k \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( a+b \)
For continuity at 0, k must equal the limit:
\( \displaystyle k=\lim_{x\to 0}\frac{\log_{e}(1+ax)-\log_{e}(1-bx)}{x} \)
Using the standard limit \( \frac{\log(1+u)}{u}\to 1 \) as \( u\to 0 \):
\( \displaystyle k=a\lim_{x\to 0}\frac{\log_{e}(1+ax)}{ax}+b\lim_{x\to 0}\frac{\log_{e}(1-bx)}{-bx} \)
\( \displaystyle =a+b \)
Shortcut: The form is 0/0, so apply L’Hôpital’s rule:
\( \displaystyle \lim_{x\to 0}\left(\frac{a}{1+ax}+\frac{b}{1-bx}\right)=a+b \)