Show that
Question:

Show that $f(x)=\left\{\begin{array}{rrr}\frac{|x-a|}{x-a}, & \text { when } & x \neq a \\ 1 & , \text { when } & x=a\end{array}\right.$

is discontinuous at x = a.

Solution:

The given function can be rewritten as:

$f(x)=\left\{\begin{array}{l}\frac{x-a}{x-a}, \text { when } x>a \\ \frac{a-x}{x-a}, \text { when } x<a \\ 1, \quad \text { when } x=a\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{c}1, \text { when } x>a \\ -1, \text { when } x<a \\ 1, \text { when } x=a\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{l}1, \text { when } x \geq a \\ -1, \text { when } x<a\end{array}\right.$

We observe

$(\mathrm{LHL}$ at $x=a)=\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a-h)=\lim _{h \rightarrow 0}(-1)=-1$

$(\mathrm{RHL}$ at $x=a)=\lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a+h)=\lim _{h \rightarrow 0}(1)=1$

$\therefore \lim _{x \rightarrow a^{-}} f(x) \neq \lim _{x \rightarrow a^{+}} f(x)$

Thus, f(x) is discontinuous at x = a.