112 views
The function

$f(x) = \begin{cases} 0 & \text{if x is rational} \\ x& \text{if x is irrational} \end{cases}$

is not continuous anywhere on the real line.
asked in Calculus | 112 views

False ;

It is continuous at x=0 since |f(x0)-0|<a for all a , whenever |x-x0|<a.