Question
Discuss the continuity of the function $f$ defined by $f(x)=\left\{\begin{array}{l} {x+2, \text { if } x \leq 1} \\ {x-2, \text { if } x>1} \end{array}\right.$
✓
Answer
The function $f$ is defined at all points of the real line.
Case $1$: If $c < 1,$ then $f(c) = c + 2$.
Now, $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} (x + 2) = c + 2$
Thus $,f$ is continuous at all real numbers less than $1$.
Case $2$: If $c > 1,$ then $f(c) = c - 2$ .
Now, $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} (x - 2) = c - 2 = f (c)$
Thus $, f$ is continuous at all points $x > 1.$
Case $3$ : If $c = 1,$ then 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
$\mathop {\lim }\limits_{x \to 1^+} f(x) = \mathop {\lim }\limits_{x \to c^+} (x - 2) = 1 - 2 = -1$
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 $f.$
Case $1$: If $c < 1,$ then $f(c) = c + 2$.
Now, $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} (x + 2) = c + 2$
Thus $,f$ is continuous at all real numbers less than $1$.
Case $2$: If $c > 1,$ then $f(c) = c - 2$ .
Now, $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} (x - 2) = c - 2 = f (c)$
Thus $, f$ is continuous at all points $x > 1.$
Case $3$ : If $c = 1,$ then 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
$\mathop {\lim }\limits_{x \to 1^+} f(x) = \mathop {\lim }\limits_{x \to c^+} (x - 2) = 1 - 2 = -1$
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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