Discuss the continuity of the function f, where f is defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{3,if}&{0 \le x \le 1}\\{4,if}&{1 < x < 3}\\{5,if}&{3 \le x \le 10}\end{array}} \right.$ }}
Discuss the continuity of the function f, where f is defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{3,if}&{0 \le x \le 1}\\{4,if}&{1 < x < 3}\\{5,if}&{3 \le x \le 10}\end{array}} \right.$ }}
Official Solution
At x $=$ 1 :
L.H.L.$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} 3 = 3$ and
R.H.L.$= \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} 4 = 4$
therefore, $L.H.L. \ne R.H.L.$ at $x = 1.$
At x $=$ 3 :
L.H.L.$= \mathop {\lim }\limits_{x \to {3^ - }} f(x) = = \mathop {\lim }\limits_{x \to {3^ - }} 4 = 4$ and
R.H.L.$= = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} 5 = 5$
L.H.L.$\ne$ R.H.L. at x $=$ 3.
Hence we can say that function is not continuous at x $=$ 1 and x $=$ 3.
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