# Discontinuous Functions

If *f* (*x*) is not continuous at *x* = *a*, then *f* (*x*) 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

### 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:

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

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

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