Q.If $f(x)=\begin{cases} \frac{x^2-1}{|x-1|} & x\ne1 \ 0 & x=1 \end{cases}$ , then $f$ is

continuous at

x=1

discontinuous at

x=1

not differentiable at

x=1

both (a) and (c)

The correct solution involves applying the fundamental concept to derive the final value step by step...

Already a member? Login here

Q.If f(x)={x2−1∣x−1∣x≠1 0x=1f(x)=\begin{cases} \frac{x^2-1}{|x-1|} & x\ne1 \ 0 & x=1 \end{cases}f(x)={∣x−1∣x2−1​​x=1 0​x=1​, then fff is