Q. For the function $f(x) = \frac{\log_{e}{(1 + x) + \log_{e}{(1 - x)}}}{x}$ to be continuous at $= 0$, the value of $f(0)$ is
A) -1
B) 0
C) -2
D) 2
Explanation
(b)
Since, the function $f(x)$ is continuous
$\therefore\ \ \ f(0) =$RHL $f(x) =$LHL$f(x)$
Now, RHL $f(X) = \lim_{h \rightarrow 0}\frac{\log{(1 + 0 + h) + \log{(1 - 0 - h)}}}{0 + h}$
$= \lim_{h \rightarrow 0}\frac{\log{(1 + h) + \log{(1 - h)}}}{h}$
$= \lim_{h \rightarrow 0}\ \frac{\frac{1}{1 + h} - \frac{1}{1 - h}}{1} = 0$
[by L ‘Hospital’s rule]
$\therefore\ \ \ f(0) =$RHL $f(x) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.