Let $f$ be the real valued function on $[0, \infty)$ defined by $f(x) = \begin{cases} x^{\frac{2}{3}}\log x& \text {for x > 0} \\ 0& \text{if x=0 } \end{cases}$ Then $f$ is discontinuous at $x = 0$ $f$ is continuous on $[0, \infty)$, but not ... $f$ is uniformly continuous on $[0, \infty)$ $f$ is not uniformly continuous on $[0, \infty)$, but uniformly continuous on $(0, \infty)$.

asked
Dec 10, 2015
in Calculus
makhdoom ghaya
105 views