# 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. Figures $$1 – 4$$ show the graphs of four functions, two of which are continuous at $$x =a$$ and two are not.

### 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.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Investigate continuity of the function $$f\left( x \right) = {3^{\large\frac{x}{{1 – {x^2}}}\normalsize}}.$$

### Example 2

Show that the function $$f\left( x \right) = {\large\frac{{\sin x}}{x}\normalsize}$$ has a removable discontinuity at $$x = 0.$$

### 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.

### Example 4

Find the points of discontinuity of the function $$f\left( x \right) = \arctan {\large\frac{1}{x}\normalsize}$$ if they exist.

### 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.

### 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:

${\lim\limits_{x \to – 1 – 0} {3^{\large\frac{x}{{1 – {x^2}}}\normalsize}} = {3^{\large\frac{{ – 1}}{{ – 0}}\normalsize}} }={ {3^\infty } = \infty ,\;\;\;}\kern-0.3pt {\lim\limits_{x \to – 1 + 0} {3^{\large\frac{x}{{1 – {x^2}}}\normalsize}} = {3^{\large\frac{{ – 1}}{{ + 0}}\normalsize}} }={ {3^{ – \infty }} = \frac{1}{{{3^\infty }}} = 0.}$

Since the left-side limit at $$x = -1$$ is infinity, we have an essential discontinuity at this point.

${\lim\limits_{x \to 1 – 0} {3^{\large\frac{x}{{1 – {x^2}}}\normalsize}} = {3^{\large\frac{{ 1}}{{ +0}}\normalsize}} }={ {3^\infty } = \infty ,\;\;\;}\kern-0.3pt {\lim\limits_{x \to 1 + 0} {3^{\large\frac{x}{{1 – {x^2}}}\normalsize}} = {3^{\large\frac{{ 1}}{{ -0}}\normalsize}} }={ {3^{ – \infty }} = \frac{1}{{{3^\infty }}} = 0.}$

Similarly, the right-side limit at $$x = 1$$ is infinity. Hence, here we also have an essential discontinuity.

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Problem 1
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Problems 2-5