$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\cfrac{x}{{|x|}},if}&{x < 0}\\{ - 1,if}&{x \ge 0}\end{array}} \right.$ }

At x $=$ 0 :
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \cfrac{x}{{|x|}} = \mathop {\lim }\limits_{x \to {0^ - }} \cfrac{x}{{ - x}}$

$\mathop {\lim }\limits_{x \to {0^ - }} ( - 1) = - 1$
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} ( - 1) = - 1$

Thus, $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$

Hence we can say that $f(x)$ is continuous at x $=$ 0.

So, f(x) has no point of discontinuity.