Question
Find all the points of discontinuity of the function f defined by $f(x)=\left\{\begin{array}{cc} {x+2,} & {\text { if } x<1} \\ {0,} & {\text { if } x=1} \\ {x-2,} & {\text { if } x>1} \end{array}\right.$
✓
Answer
f is continuous at all real numbers such that x $\neq$ 1. The left-hand limit of f at x = 1 is
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} $(x + 2) = 1 + 2 = 3
The right hand limit of f at x = 1 is
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} $(x - 2) = 1 - 2 = -1
Since the left and right-hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f.
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} $(x + 2) = 1 + 2 = 3
The right hand limit of f at x = 1 is
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} $(x - 2) = 1 - 2 = -1
Since the left and right-hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f.
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