Is the function f defined by $f(x) = \left\{ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x,}&{if}&{x \le 1}\\{5,}&{if}&{x > 1}\end{array}} \right.$ continuous at x $=$ 0 ? At x $=$ 1? At x $=$ 2 ? }

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

Also, $f(0) = 0$

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.

(ii) At x $=$ 1, .
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (x) = 1$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} 5 = 5$

therefore, $\mathop {\lim }\limits_{x \to {1^ + }} f(x) \ne \mathop {\lim }\limits_{x \to {1^ - }} f(x)$

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

(iii) At x $=$ 2,
$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} 5 = 5$
$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} 5 = 5$

Also, f(2) $=$ 5

Thus $\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2)$

Hence we can say that f(x) is continuous at x $=$ 2.