# Calculus

## Limits and Continuity of Functions # Discontinuous Functions – Page 2

• ### 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 $$x = 0$$. As $$\sin x$$ 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 $$f\left( x \right) = \begin{cases} 1 – {x^2}, & x \lt 0 \\ x +2, &x \ge 0 \end{cases}$$ 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 one-sided limits at $$x = 0.$$

${\lim\limits_{x \to 0 – 0} f\left( x \right) = \lim\limits_{x \to 0 – 0} \left( {1 – {x^2}} \right) = 1,\;\;\;}\kern-0.3pt {\lim\limits_{x \to 0 + 0} f\left( x \right) = \lim\limits_{x \to 0 – 0} \left( {x + 2} \right) = 2.}$

Thus, the function has a discontinuity of the first kind at $$x = 0.$$ The finite jump at this point is

${\Delta y = \lim\limits_{x \to 0 + 0} f\left( x \right) – \lim\limits_{x \to 0 – 0} f\left( x \right) }\kern0pt{= 2 – 1 = 1.}$

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 one-sided limits:

${\lim\limits_{x \to 0 – 0} \arctan \frac{1}{x} = \arctan \left( { – \infty } \right) = – \frac{\pi }{2},\;\;\;}\kern-0.3pt {\lim\limits_{x \to 0 + 0} \arctan \frac{1}{x} = \arctan \left( { + \infty } \right) = \frac{\pi }{2}.}$

As seen, the function has a discontinuity point of the first kind at $$x = 0$$ (Figure $$5$$). 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:

${\lim\limits_{x \to – \frac{5}{2} – 0} \frac{{\left| {2x + 5} \right|}}{{2x + 5}} } = {\lim\limits_{x \to – \frac{5}{2} – 0} \frac{{ – \left( {2x + 5} \right)}}{{2x + 5}} = – 1,\;\;\;}\kern-0.3pt {\text{if}\;\;x \lt – \frac{5}{2};}$

${\lim\limits_{x \to – \frac{5}{2} + 0} \frac{{\left| {2x + 5} \right|}}{{2x + 5}} } = {\lim\limits_{x \to – \frac{5}{2} + 0} \frac{{ \left( {2x + 5} \right)}}{{2x + 5}} = 1,\;\;\;}\kern-0.3pt {\text{if}\;\;x \ge – \frac{5}{2}.}$

As the values of the one-sided 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 in Figure $$6.$$