Continuity and Differentiability — Class 12 Maths Solution

At x $=$ 0 :
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \cfrac{{|x|}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \cfrac{{ - x}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} ( - 1) = - 1$

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

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

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

So, the only point of discontinuity of f is 0.