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
Figure 1.
Example 1 of discontinuous function
Figure 2.
Example 2 of continuous function
Figure 3.
Example 2 of discontinuous function
Figure 4.

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

Further there may be the following two options:

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 \[f\left( x \right) = {3^{\frac{x}{{1 - {x^2}}}}}.\]

Example 2

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

Example 1.

Investigate continuity of the function \[f\left( x \right) = {3^{\frac{x}{{1 - {x^2}}}}}.\]

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:

\[\lim\limits_{x \to - 1 - 0} {3^{\frac{x}{{1 - {x^2}}}}} = 3^{\frac{{ - 1}}{{ - 0}}} = 3^\infty = \infty ,\;\;\lim\limits_{x \to - 1 + 0} {3^{\frac{x}{{1 - {x^2}}}} = 3^{\frac{{ - 1}}{{ + 0}}}} = 3^{ - \infty } = \frac{1}{{{3^\infty }}} = 0.\]

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

\[\lim\limits_{x \to 1 - 0} {3^{\frac{x}{{1 - {x^2}}}} = 3^{\frac{{ 1}}{{ +0}}}} = 3^\infty = \infty ,\;\; \lim\limits_{x \to 1 + 0} {3^{\frac{x}{{1 - {x^2}}}} = 3^{\frac{{ 1}}{{ -0}}}} = 3^{ - \infty } = \frac{1}{{{3^\infty }}} = 0.\]

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

Example 2.

Show that the function \[f\left( x \right) = {\frac{{\sin x}}{x}}\] 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 \(f\left( x \right) = {\frac{{\sin x}}{x}}\) is also continuous for all \(x\) except the point \(x = 0.\)

Since \(\lim\limits_{x \to 0} {\frac{{\sin x}}{x}} = 1,\) the function has a removable discontinuity at this point. We can construct the new function

\[{f_1}\left( x \right) = \begin{cases} \frac {\sin x}{x}, & x \ne 0 \\1, &x = 0 \end{cases},\]

which is continuous at every real \(x.\)

See more problems on Page 2.

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