Question

Examine the following functions for continuity.

(i) $f\left(x\right)=x-5$ (ii) $f\left(x\right)=$ $\frac{1}{x-5}$, $x\ne 5$

(iii) $f\left(x\right)=\frac{x^2-25}{x+5},x\ne -5$ (iv) $f\left(x\right)=|x-5|$

Answer 1

(i)

Given function is polynomial of degree one,
and we know every polynomial function is continuous for all $x\in R$
Hence $f$ is continuous.

(ii)

The given function is $f\left(x\right)=\frac{1}{x-5},x\ne 5$
For any real number $k\ne 5$, we obtain

$\underset{x\to k}{lim}f\left(x\right)=\underset{x\to k}{lim}\frac{1}{x-5}=\frac{1}{k-5}$

Also, $f\left(k\right)=\frac{1}{k-5}$ (As k $\ne$ 5)
$\therefore$ $\underset{x\to k}{lim}f\left(x\right)=f\left(k\right)$
Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(iii)

The given function is $f\left(x\right)=\frac{x^2-25}{x+5},x\ne 5$
For any real number $x\ne -5,$ we obtain

$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}\frac{(x+5)(x-5)}{x+5}=x-5$

$\therefore$ $\underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Hence,

$f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(iv)

The given function is $f\left(x\right)=|x-5|=\{\begin{array}{c}5-x,\text{if}x<5\\ x-5,\text{if}x\ge 5\end{array}$
The function is defined at all points of the real line.
Let $c$ be a point on a real line. Then, $c<5$ or $c=5$ or $c>5$
Case I : $c<5$
Then, $f\left(c\right)=5-c$
$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(5-x)=5-c$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all real numbers less than $5$.
Case II : $c=5$
Then $f\left(c\right)=f\left(5\right)=(5-5)=0$
Then, $\underset{x\to 5}{lim}f\left(x\right)=\underset{x\to 5}{lim}(5-x)=(5-5)=0$
$\underset{x\to 5}{lim}f\left(x\right)=\underset{x\to 5}{lim}(x-5)=0$
$\therefore \underset{x\to 5}{lim}f\left(x\right)=\underset{x\to 5}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at $x=5$
Case III : $c>5$
Then, $f\left(c\right)=f\left(5\right)=c-5$
$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(x-5)=c-5$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
therefore, $f$ is continuous at all real numbers greater than $5$.
Hence, $f$ is continuous at every real number and therefore, it is a continuous function.