Examine the following functions for continuity :

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

(b)$f(x) = \cfrac{1}{{x - 5}},x \ne 5$

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

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

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

Let a be a real number, then

$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} (a + h) - 5 = a - 5$

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} (a - h) - 5 = a - 5$

Also, $f(a) = a - 5$
therefore, $\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a)$

Hence we can say that the given function $f(x) = (x - 5)$ is continuous.

(b) $f(x) = \cfrac{1}{{x - 5}}$
Let a be a real number, then

$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} \cfrac{1}{{a + h - 5}} = \cfrac{1}{{a - 5}}$

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} \cfrac{1}{{a - h - 5}} = \cfrac{1}{{a - 5}}$

Also, $f(a) = \cfrac{1}{{a - 5}}$
therefore, $\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a)$

Hence we can say that the given function $f(x) = \cfrac{1}{{x - 5}}$ is continuous at all point except at $x = 5$ .

(c)$f(x) = \cfrac{{{x^2} - 25}}{{x + 5}} = \cfrac{{(x + 5)(x - 5)}}{{(x + 5)}} = x - 5$
Let 'a' be a real number, then

$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} (a + h) - 5 = a - 5$

and $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} (a - h) - 5 = a - 5$

Also, $f(a) = a - 5$
therefore, $\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a)$

Hence we can say that the given function $f(x) = x - 5$ is continuous at every point of its domain.

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

$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} |a + h - 5| = |a - 5| = a - 5$

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} |a - h - 5| = |a - 5| = a - 5$

Also,$f(a) = |a - 5| = a - 5$
therefore, $\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a)$

Hence we can say that the given function $f(x) = |x - 5|$ is continuous.