Examine the following function for continuity: $f(x) = |x - 5|$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
The given function is $f(x) = |x - 5| = \begin{cases} 5 - x, & \text{if } x < 5 \\ x - 5, & \text{if } x \ge 5 \end{cases}$.
This function $f$ is defined for all real numbers.
Let $c$ be any real number. Then $c < 5$,$c = 5$,or $c > 5$.
Case $I$: $c < 5$.
Then $f(c) = 5 - c$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (5 - x) = 5 - c$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $c < 5$.
Case $II$: $c = 5$.
Then $f(5) = 5 - 5 = 0$.
Left-hand limit: $\lim_{x \to 5^-} f(x) = \lim_{x \to 5} (5 - x) = 5 - 5 = 0$.
Right-hand limit: $\lim_{x \to 5^+} f(x) = \lim_{x \to 5} (x - 5) = 5 - 5 = 0$.
Since $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$,$f$ is continuous at $x = 5$.
Case $III$: $c > 5$.
Then $f(c) = c - 5$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $c > 5$.
Conclusion: Since $f$ is continuous at all real numbers,it is a continuous function.

Explore More

Similar Questions

Which of the following function$(s)$ not defined at $x = 0$ has/have a removable discontinuity at $x = 0$?

Given that $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $a = $

$f(x) = \begin{cases} 3x - 8 & \text{if } x \leq 5 \\ 2k & \text{if } x > 5 \end{cases}$ is continuous,find $k$.

If $f(x) = \frac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}, x \neq 0$ is continuous at $x=0$, then the value of $f(0)$ is

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo