Limits Continuity And Differentiability Question 100
Question: If $ f(x)= \begin{cases} & \frac{x\log \cos x}{\log (1+x^{2})},x\ne 0 \\ & 0,x=0 \\ \end{cases} .then\text{f(x)is} $ is
Options:
A) Continuous as well as differentiable at x = 0
B) Continuous but not differentiable at x = 0
C) Differentiable but not continuous at x = 0
D) Neither continuous nor differentiable at x = 0
Show Answer
Answer:
Correct Answer: A
Solution:
We have, $ Lf’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\frac{-h\log \cosh }{-h\log (1+h^{2})} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\log \cosh }{\log (1+h^{2})}( \frac{0}{0}form ) $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{-\tan h}{2h/(1+h^{2})}=-1/2 $
$ Rf’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\frac{h\log \cosh }{h\log (1+h^{2})} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\log \cosh }{\log (1+h^{2})}( \frac{0}{0}form ) $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{-tanh}{2h/(1+h^{2})}=\frac{-1}{2} $
Since $ Lf’(0)=Rf’(0), $ therefore $ f(x) $ is differentiable at $ x=0 $
Since differentiability
$ \Rightarrow $ continuity, therefore f(x) is continuous at x = 0.
BETA
