Solveeit Logo
Login

Question

Mathematics Question on Continuity and differentiability

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

A

e−1e - 1

B

log⁡(e+1) \log (e +1)

C

log⁡(e−1) \log (e - 1)

D

(e+1) (e +1)

Answer

(e+1) (e +1)

Explanation

Solution

f(x)=log⁡(1+ex)−log⁡(1−x)xf\left(x\right) = \frac{\log\left(1+ex\right)-\log\left(1-x\right)}{x}
It is continuous at x=0x = 0
∴   lim⁡x→0−f(x)=lim⁡x→0+f(x)=f(0)\therefore \:\:\: \displaystyle\lim_{x\to0^{-}} f\left(x\right) = \displaystyle\lim _{x\to 0^{+}} f\left(x\right) =f\left(0\right)
lim⁡x→0−log⁡(1+ex)−log⁡(1−x)x\displaystyle\lim _{x\to 0^{-}} \frac{\log\left(1+ex\right) -\log \left(1-x\right) }{x}
=lim⁡x→0−log⁡(1+e(0−h))−log⁡(1−(0−h))0−h=\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}
=lim⁡x→0−log⁡(1−eh)−log⁡(1+h)−h= \displaystyle\lim _{x\to 0^{-}}\frac{\log \left(1-eh\right)-\log \left(1+h\right)}{-h}
It is continuous at
=lim⁡x→0−(11−eh)(−e)−(11+h)−1\frac{=\displaystyle\lim _{x\to 0^{-}} \left(\frac{1}{1-eh} \right)\left(-e\right) -\left(\frac{1}{1+h}\right)}{-1}
=−e1−1−1⇒e+1= \frac{\frac{-e}{1}-1}{-1} \Rightarrow e+1
∴   f(0)=e+1\therefore \:\:\: f(0) = e + 1