Let f(x)={xsin(1x)x≠00x=0f(x) = \begin{cases} x \sin(\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}f(x)={xsin(x1)0x=0x=0. What is the nature of fff at x=0x=0x=0?
Continuous and differentiable.
Continuous but not differentiable.
Jump discontinuity.
Essential singularity.