Question

Let $f\left(x\right) = \frac{\log\left(1+ex\right)-\log\left(1-x\right)}{x} , x\ne0 $ . Then $f$ is continuous at $x = 0$ if $f(0)$ =

• A. $e - 1$
• B. $ \log (e +1)$
• C. $ \log (e - 1)$
• D. $ (e +1)$

Answer 1

The Correct Option is D

$f\left(x\right) = \frac{\log\left(1+ex\right)-\log\left(1-x\right)}{x}$
It is continuous at $x = 0$
$\therefore \:\:\: \displaystyle\lim_{x\to0^{-}} f\left(x\right) = \displaystyle\lim _{x\to 0^{+}} f\left(x\right) =f\left(0\right)$
$ \displaystyle\lim _{x\to 0^{-}} \frac{\log\left(1+ex\right) -\log \left(1-x\right) }{x} $
$=\displaystyle\lim _{x\to 0^{-}} \frac{\log \left(1+e\left(0-h\right)\right) -\log \left(1-\left(0 -h\right)\right)}{0-h} $
$= \displaystyle\lim _{x\to 0^{-}}\frac{\log \left(1-eh\right)-\log \left(1+h\right)}{-h}$
It is continuous at
$ \frac{=\displaystyle\lim _{x\to 0^{-}} \left(\frac{1}{1-eh} \right)\left(-e\right) -\left(\frac{1}{1+h}\right)}{-1}$
$ = \frac{\frac{-e}{1}-1}{-1} \Rightarrow e+1$
$ \therefore \:\:\: f(0) = e + 1$