# 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 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 $$x.$$