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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 left-hand limit $$\lim\limits_{x \to a - 0} f\left( x \right)$$ and right-hand limit $$\lim\limits_{x \to a + 0} f\left( x \right)$$;
• These one-sided limits are finite.
Further there may be the following two options:
• The right-hand limit and the left-hand 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 right-hand limit and the left-hand 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 one-sided 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 one-sided 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 left-side 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 right-side 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 one-sided limits at $$x = 0$$. ${\lim\limits_{x \to 0 - 0} f\left( x \right) = \lim\limits_{x \to 0 - 0} \left( {1 - {x^2}} \right) = 1,}\;\;\; {\lim\limits_{x \to 0 + 0} f\left( x \right) = \lim\limits_{x \to 0 - 0} \left( {x + 2} \right) = 2.}$ 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 one-sided 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 $$\pi.$$
 Fig.2 Fig.3
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 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 on Figure $$3.$$

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