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 or tap 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.