Question
Examine the function for continuity. f(x) = |x – 5|
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Answer
The given function is $f(x)=|x-5|=\left\{\begin{array}{l} {5-x \text { , if } x<5} \\ {x-5, \text { if } x \geq 5} \end{array}\right.$
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e. , k < 5, or k = 5 or k >5
Now,
Case I: k<5
Then, f(k) = 5 – k
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (5 - x)$ = 5 – k = f(k)
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at all real number less than 5.
Case II: k = 5
Then, f(k) = f(5) = 5 – 5 = 0
$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to 5} (5 - x)$= 5 – 5 = 0
$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to 5} (x - 5)$ = 5 – 5 = 0
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at x = 5.
Case III: k > 5
Then, f(k) = k – 5
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (x - 5) = k - 5 = f(k)$
Thus, $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at all real number greater than 5.
Therefore, f is a continuous function.
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e. , k < 5, or k = 5 or k >5
Now,
Case I: k<5
Then, f(k) = 5 – k
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (5 - x)$ = 5 – k = f(k)
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at all real number less than 5.
Case II: k = 5
Then, f(k) = f(5) = 5 – 5 = 0
$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to 5} (5 - x)$= 5 – 5 = 0
$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to 5} (x - 5)$ = 5 – 5 = 0
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at x = 5.
Case III: k > 5
Then, f(k) = k – 5
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (x - 5) = k - 5 = f(k)$
Thus, $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at all real number greater than 5.
Therefore, f is a continuous function.
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