$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{2x,if}&{x < 0}\\{0,if}&{0 \le x \le 1}\\{4x,if}&{x > 1}\end{array}} \right.$ }}

At x $=$ 0 :
L.H.L. $=$ $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} 2(0) = 0$

R.H.L.$= \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} 0 = 0$

Also, $f(0) = 0$
therefore, L.H.L. $=$ R.H.L $=$ f (x)

So, f (x) is continuous at x $=$ 0.
At x $=$ 1 :

L.H.L.$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} 0 = 0$
and $R.H.L. = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} 4(1) = 4$

Also, f (1) $=$ 0
therefore, L.H.L.$\ne$R.H.L.

Hence we can say that f(x) is discontinuous at x $=$ 1.