A) Has no limit
B) Is discontinuous
C) Is continuous but not differentiable
D) Is differentiable
Correct Answer: B
Solution :
[b] For \[x\ne 0,\] we have, \[f(x)=x+\frac{x/1+x}{1-\frac{1}{1+x}}=x+\frac{x/1+x}{x/1+x}=x+1\] For \[x=0,f(x)=0\]. Thus, \[f(x)=\left\{ \begin{matrix} x+1, & x\ne 0 \\ 0, & x=0 \\ \end{matrix} \right.\] Clearly, \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=1\ne f(0)\]. So, \[f(x)\] is discontinuous and hence not differentiable at \[x=0\].
You need to login to perform this action.
You will be redirected in
3 sec
