


Discontinuous Functions


If \(f\left( x \right)\) is not continuous at \(x = a\),
then \(f\left( x \right)\) is said to be discontinuous at this point.
Figure \(1\) shows the graphs of four functions, two of which are continuous at \(x =a\)
and two are not.

 
Continuous at \(x = a\). 
 Discontinuous at \(x = a\). 

 
Continuous at \(x = a\). 
 Discontinuous at \(x = a\). 
Figure 1. 
Classification of Discontinuity Points
All discontinuity points are divided into discontinuities of the first and second kind.
The function \(f\left( x \right)\) has a discontinuity of the first kind
at \(x = a\) if
 There exist lefthand limit \(\lim\limits_{x \to a  0} f\left( x \right)\) and righthand limit \(\lim\limits_{x \to a + 0} f\left( x \right)\);
 These onesided limits are finite.
Further there may be the following two options:
 The righthand limit and the lefthand limit are equal to each other:
\[\lim\limits_{x \to a  0} f\left( x \right) = \lim\limits_{x \to a + 0} f\left( x \right).\]
Such a point is called a removable discontinuity.
 The righthand limit and the lefthand limit are unequal:
\[\lim\limits_{x \to a  0} f\left( x \right) \ne \lim\limits_{x \to a + 0} f\left( x \right).\]
In this case the function \(f\left( x \right)\) has a jump discontinuity.
The function \(f\left( x \right)\) is said to have a discontinuity of the second kind
(or a nonremovable or essential discontinuity)
at \(x = a\), if at least one of the onesided limits either does not exist or is infinite.

Example 1


Investigate continuity of the function \(f\left( x \right) = {3^{\large\frac{x}{{1  {x^2}}}\normalsize}}.\)
Solution.
The given function is not defined at \(x = 1\) and \(x = 1\).
Hence, this function has discontinuities at \(x = \pm 1\). To determine the type of the discontinuities,
we find the onesided limits:
\[
{\lim\limits_{x \to  1  0} {3^{\large\frac{x}{{1  {x^2}}}\normalsize}} = {3^{\large\frac{{  1}}{{  0}}\normalsize}} = {3^\infty } = \infty ,\;\;}
{\lim\limits_{x \to  1 + 0} {3^{\large\frac{x}{{1  {x^2}}}\normalsize}} = {3^{\large\frac{{  1}}{{ + 0}}\normalsize}} = {3^{  \infty }} = \frac{1}{{{3^\infty }}} = 0.}
\]
Since the leftside limit at \(x = 1\) is infinity, we have an essential discontinuity at this point.
\[
{\lim\limits_{x \to 1  0} {3^{\large\frac{x}{{1  {x^2}}}\normalsize}} = {3^{\large\frac{{ 1}}{{ +0}}\normalsize}} = {3^\infty } = \infty ,\;\;}
{\lim\limits_{x \to 1 + 0} {3^{\large\frac{x}{{1  {x^2}}}\normalsize}} = {3^{\large\frac{{ 1}}{{ 0}}\normalsize}} = {3^{  \infty }} = \frac{1}{{{3^\infty }}} = 0.}
\]
Similarly, the rightside limit at \(x = 1\) is infinity.
Hence, here we also have an essential discontinuity.

Example 2


Show that the function \(f\left( x \right) = {\large\frac{{\sin x}}{x}\normalsize}\) has a removable
discontinuity at \(x = 0\).
Solution.
Obviously, the function is not defined at \(x = 0\). Since \(\sin x\)
is continuous at every \(x\), then the initial function \(f\left( x \right) = {\large\frac{{\sin x}}{x}\normalsize}\) is also continuous for all
\(x\) except the point \(x = 0\).
Since \(\lim\limits_{x \to 0} {\large\frac{{\sin x}}{x}\normalsize} = 1\), the function has a removable discontinuity at this point. We can construct
the new function
\[
{f_1}\left( x \right) =
\begin{cases}
\large\frac {\sin x}{x}\normalsize, & x \ne 0 \\
1, &x = 0
\end{cases}
,\]
which is continuous at every real \(x.\)

Example 3


Find the points of discontinuity of the function \(
f\left( x \right) =
\begin{cases}
1  {x^2}, & x < 0 \\
x +2, &x \ge 0
\end{cases}
,\,\) if they exist.
Solution.
The function exists for all \(x,\) however it is defined by two different functions and, therefore, is not elementary.
We investigate "behavior" of the function near to the point \(x = 0\) where its analytic expression changes.
Calculate onesided limits at \(x = 0\).
\[
{\lim\limits_{x \to 0  0} f\left( x \right) = \lim\limits_{x \to 0  0} \left( {1  {x^2}} \right) = 1,}\;\;\;
{\lim\limits_{x \to 0 + 0} f\left( x \right) = \lim\limits_{x \to 0  0} \left( {x + 2} \right) = 2.}
\]
Thus, the function has a discontinuity of the first kind at \(x = 0\).
The finite jump at this point is
\[
{\Delta y = \lim\limits_{x \to 0 + 0} f\left( x \right)  \lim\limits_{x \to 0  0} f\left( x \right) }
{= 2  1 = 1.}
\]
The function is continuous for all other \(x,\) because both the functions defined from the left and from the right of the point \(x = 0\)
are elementary functions without any discontinuities.

Example 4


Find the points of discontinuity of the function \(f\left( x \right) = \arctan {\large\frac{1}{x}\normalsize}\) if they exist.
Solution.
This elementary function is defined for all \(x\) except \(x = 0\), where it has a discontinuity.
Find onesided limits at
\[
{\lim\limits_{x \to 0  0} \arctan \frac{1}{x} = \arctan \left( {  \infty } \right) =  \frac{\pi }{2},}\;\;
{\lim\limits_{x \to 0 + 0} \arctan \frac{1}{x} = \arctan \left( { + \infty } \right) = \frac{\pi }{2}.}
\]
As can be seen, the function has a discontinuity point of the first kind at \(x = 0\) (Figure \(2\)). The jump of the function is equal to \(\pi.\)

Example 5


Find the points of discontinuity of the function \(f\left( x \right) = {\large\frac{{\left {2x + 5} \right}}{{2x + 5}}\normalsize}\) if they exist.
Solution.
The function is defined and continuous for all \(x\) except \(x =  {\large\frac{5}{2}\normalsize}\), where it has a discontinuity.
Investigate this discontinuity point:
\[
{\lim\limits_{x \to  \frac{5}{2}  0} \frac{{\left {2x + 5} \right}}{{2x + 5}} }
= {\lim\limits_{x \to  \frac{5}{2}  0} \frac{{  \left( {2x + 5} \right)}}{{2x + 5}} =  1,}\;\;
{\text{if}\;\;x <  \frac{5}{2};}
\]
\[
{\lim\limits_{x \to  \frac{5}{2} + 0} \frac{{\left {2x + 5} \right}}{{2x + 5}} }
= {\lim\limits_{x \to  \frac{5}{2} + 0} \frac{{ \left( {2x + 5} \right)}}{{2x + 5}} = 1,}\;\;
{\text{if}\;\;x \ge  \frac{5}{2}.}
\]
Since the values of the onesided limits are finite, then there's a discontinuity of the first kind at the point \(x =  {\large\frac{5}{2}\normalsize}.\)
The graph of the function is sketched on Figure \(3.\)



