Limits Continuity And Differentiability Question 100
If $ f(x)= \begin{cases} \frac{x\log \cos x}{\log (1+x^{2})}, & x\ne 0 \ 0, & x=0 \end{cases} .\text{then }f(x)\text{ 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 (0-h)}{-h\log (1+h^{2})} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\log \cosh (h)}{\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}{h\log (1+h^{2})} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\log \cosh (h)}{\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 $ if $ f(x) $ is continuous at $ x=0 $ Since differentiability is a local property $ \Rightarrow $ continuity, therefore f(x) is continuous at x = 0.
BETA
