Limits Continuity And Differentiability Question 73
Question: $ f(x)= \begin{cases} \frac{x}{2x^{2}+| x |,}x\ne 0 \\ 1.x=0 \\ \end{cases} . $ . Then $ f(x) $ is
Options:
A) continuous but non-differentiable at x=0
B) differentiable at x=0
C) discontinuous at x=0
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(0+0)=\underset{h\to 0}{\mathop{\lim }}f(h)=\underset{h\to 0}{\mathop{\lim }}\frac{h}{2h^{2}+h}=\underset{h\to 0}{\mathop{\lim }}\frac{1}{2h+1}=1 $ And $ f(0-0)=\underset{h\to 0}{\mathop{\lim }}f(-h)=\underset{h\to 0}{\mathop{\lim }}\frac{-h}{2h^{2}+| -h |} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{-h}{2h^{2}+h}=\underset{h\to 0}{\mathop{\lim }}\frac{-1}{2h+1}=-1 $
BETA
