Calculus

Limits and Continuity of Functions

Limits and Continuity Logo

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.

example 1 of continuous function
Fig 1. Continuous function.
example 1 of discontinuous function
Fig 2. Discontinuous function.
example 2 of continuous function
Fig 3. Continuous function.
example 2 of discontinuous function
Fig 4. Discontinuous function.

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.

Page 1
Problem 1
Page 2
Problems 2-5