# Examine the following functions for continuity.

Question:

(a) $f(x)=x-5$

(b) $f(x)=\frac{1}{x-5}, x \neq 5$

(c) $f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$

(d) $f(x)=|x-5|$

Solution:

(a) The given function is

It is evident that f is defined at every real number k and its value at k is k − 5.

It is also observed that, $\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k}(x-5)=k-5=f(k)$

$\therefore \lim _{x \rightarrow k} f(x)=f(k)$

Hence, f is continuous at every real number and therefore, it is a continuous function.

(b) The given function is $f(x)=\frac{1}{x-5}, x \neq 5$

For any real number $k \neq 5$, we obtain

$\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} \frac{1}{x-5}=\frac{1}{k-5}$

Also, $f(k)=\frac{1}{k-5} \quad($ As $k \neq 5)$

$\therefore \lim _{x \rightarrow k} f(x)=f(k)$

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

(c) The given function is $f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$

For any real number $c \neq-5$, we obtain

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c} \frac{x^{2}-25}{x+5}=\lim _{x \rightarrow c} \frac{(x+5)(x-5)}{x+5}=\lim _{x \rightarrow c}(x-5)=(c-5)$

Also, $f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) \quad($ as $c \neq-5)$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

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

(d) The given function is $f(x)=|x-5|=\left\{\begin{array}{l}5-x, \text { if } x<5 \\ x-5, \text { if } x \geq 5\end{array}\right.$

This function f 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 l: $c<5$

Then, $f(c)=5-c$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(5-x)=5-c$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, f is continuous at all real numbers less than 5.

Case II : $c=5$

Then, $f(c)=f(5)=(5-5)=0$

$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5}(5-x)=(5-5)=0$

$\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(x-5)=0$

$\therefore \lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c)$

Thereforeis continuous at x = 5

Case III: $c>5$

Then, $f(c)=f(5)=c-5$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x-5)=c-5$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

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.