Discontinuities of Real Functions

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Introduction

Continuity is a property of a function in which small variations in the input result in correspondingly small variations in the output within the neighbourhood of a given point. If this local stability does not hold, the function is considered discontinuous. Discontinuities are typically classified into three distinct types:

  • A removable discontinuity occurs when the limit exists and is finite, but the function is either undefined at the point or its value does not equal the limit.
  • A jump discontinuity is present when both the left-hand and right-hand limits exist and are finite, but these limits are not equal.
  • An infinite discontinuity occurs when at least one of the one-sided limits is infinite, causing the function to diverge near the point rather than approach a finite value.

A discontinuity at $x_0$ can occur in exactly one of the three mutually exclusive ways described above. A point cannot simultaneously exhibit more than one type of discontinuity.

Each of these types will be examined in detail in the following sections. In this discussion, $f$ denotes a real-valued function, and $x_0$ represents a point in its domain or a point at which the function may fail to be defined.

Recall of continuity

A function $f$ is continuous at $x_0$ if the limit as $x$ approaches $x_0$ exists, is finite, and coincides with the value of the function at that point. This condition is expressed by the following limit:

\lim_{x \to x_0} f(x) = f(x_0)

Polynomials constitute a fundamental class of elementary continuous functions. These functions represent smooth curves in the plane and exhibit no points of discontinuity. Below is the graph of the quadratic function $x^2 + 2x + 1$, which represents a parabola:

The graph of a second-degree polynomial is a continuous parabola, with no jumps or interruptions.

A discontinuity at $x_0$ arises whenever this equality does not hold, and the specific way in which the condition fails determines the type of discontinuity.

Intuitively, a function is continuous if its graph can be drawn in the plane without any interruptions, breaks, or sudden jumps.

Removable discontinuity

A removable discontinuity arises when a function possesses a well-defined finite limit at $x_0$, yet the function’s value at that point is either undefined or does not coincide with the limit. Formally, a function $f$ has a removable discontinuity at $x_0$ if the following limit exists and is finite:

\lim_{x \to x_0} f(x) = \ell \in \mathbb{R}

Moreover, at least one of the following conditions is satisfied:

  • $f(x_0)$ is undefined.
  • $f(x_0) \neq \ell$.

In such cases, the discontinuity may be removed by redefining the function at a single point as follows:

g(x) = \begin{cases} \ell & \text{if } x = x_0 \\[6pt] f(x) & \text{if } x \ne x_0 \end{cases}

With this definition, the function $g$ becomes continuous at $x_0$. The term “removable” refers to the fact that the discontinuity can be resolved in this manner.

Removable discontinuities typically occur in rational functions containing cancellable factors, resulting in a hole in the graph. They can also be present in piecewise-defined functions or in functions where the value at a single point has been modified, provided the limit at that point exists and is finite.

Example 1

Consider the function defined by the following rational expression, which is undefined at $x = 1$:

f(x) = \frac{x^2 - 1}{x - 1}

Factoring the numerator demonstrates that the expression simplifies for all values of $x$ except $1$ since $x=1$ would cancel the denominator and make the function undefined.

x^2 - 1 = (x - 1)(x + 1)

For all $x \neq 1$, the function is equivalent to a linear function:

f(x) = x + 1

Although the function is undefined at $x = 1$, the limit as $x$ approaches $1$ exists and is finite:

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2

This demonstrates that $x = 1$ is a removable discontinuity, as the graph corresponds to the straight line $y = x + 1$ with a single missing point at$(1,2) .$

Redefining the function at that point by assigning it the value of the limit eliminates the discontinuity:

g(x) = \begin{cases} 2 & \text{if } x = 1 \\[6pt] f(x) & \text{if } x \ne 1 \end{cases}
With this modification, the function is continuous at $x = 1$.

Jump discontinuity

A jump discontinuity arises when both the left-hand and right-hand limits at $x_0$ exist and are finite, yet these limits are not equal. Formally, $f$ has a jump discontinuity at $x_0$ if:

\begin{align} \lim_{x \to x_0^-} f(x) &= \ell_1 \in \mathbb{R} \\[6pt] \lim_{x \to x_0^+} f(x) &= \ell_2 \in \mathbb{R} \\[6pt] \ell_1 &\neq \ell_2 \end{align}

In this case, the limit $\lim_{x \to x_0} f(x)$ does not exist. The function approaches two distinct finite values depending on the direction of approach. Unlike a removable discontinuity, this type cannot be resolved by redefining the function at a single point, as the discrepancy is inherent to the local behavior.

Example 2

To analyse the jump discontinuity, consider the following simple function, which exhibits a discontinuity at the point $x = 1.$

f(x) = \begin{cases} 0 & \text{if } x < 1 \\[6pt] 2 & \text{if } x \ge 1 \end{cases}

For values of $x$ approaching $1$ from the left, the function remains constant at $0$. Therefore:

\lim_{x \to 1^-} f(x) = 0

For values of $x$ approaching 1 from the right, the function remains constantly equal to $2$, and therefore the limit is:

\lim_{x \to 1^+} f(x) = 2

Both one-sided limits exist and are finite but they are not equal. Since $0 \neq 2$, it follows that the two one-sided limits do not coincide, and consequently, the limit $\lim_{x \to 1} f(x)$ does not exist. The graph of the function shows a vertical jump at $x = 1$, transitioning from $0$ to $2$.

This discontinuity cannot be removed by redefining the function at $x = 1$, as the difference between the two limiting values indicates a break in the local behaviour of the function.

Infinite discontinuity

An infinite discontinuity occurs when a function diverges as $x$ approaches $x_0$, with at least one of the one-sided limits being infinite. Formally, a function $f$ exhibits an infinite discontinuity at $x_0$ if at least one of the following conditions is satisfied:

\begin{align} \lim_{x \to x_0^-} f(x) &= \pm \infty \\[6pt] \lim_{x \to x_0^+} f(x) &= \pm \infty \end{align}

In such cases, the function does not approach any finite value as $x$ nears $x_0$. The graph typically displays a vertical asymptote. This discontinuity reflects unbounded growth rather than a finite discontinuity.

Example 3

For example, consider the following function:

f(x) = \frac{1}{x - 2}

The behaviour of this function near $x_0 = 2$ is analysed as follows. As $x \to 2^-$, the denominator $x - 2$ becomes negative and approaches zero, causing the function to decrease without bound. Therefore we have:

\lim_{x \to 2^-} f(x) = -\infty

As $x \to 2^+$, the denominator is positive and approaches zero, which causes the function to increase without bound. The limit is:

\lim_{x \to 2^+} f(x) = +\infty

At least one of the one-sided limits is infinite, and they diverge with opposite signs. Consequently, the function exhibits an infinite discontinuity at $x = 2$. The graph displays a vertical asymptote at the line $x = 2$, and this divergence indicates unbounded growth rather than a finite jump or a removable discontinuity.

Discontinuity, continuity and differentiability

It is instructive to establish a precise link between the notions of discontinuity and differentiability. We know that if a function f is differentiable at a point $x_0$, it must also be continuous at that point. The existence of the derivative ensures that the function satisfies the condition of continuity:

f’(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}
\lim_{x \to x_0} f(x) = f(x_0)

Therefore, if a function exhibits a discontinuity of the type just described at $x_0$, meaning the limit does not exist or does not equal the function’s value the derivative at that point does not exist.

However, the converse is not true. A function can be continuous at $x_0$ yet not differentiable there. This situation arises when the one-sided derivatives exist but differ, when at least one is infinite, or when one of the limits diverges.

\lim_{x \to x_0^-} \frac{f(x) - f(x_0)}{x - x_0} \neq \lim_{x \to x_0^+} \frac{f(x) - f(x_0)}{x - x_0}

A common example is the absolute value function, which is continuous at $x = 0$ but not differentiable there, resulting in a corner on its graph. In summary, every discontinuity implies non-differentiability, whereas not every point of non-differentiability is associated with a discontinuity.

A particular case: essential discontinuity

An additional category, known as essential discontinuity, is sometimes recognised but not universally adopted as a formal classification. This type arises when the limit does not exist and cannot be described as infinite. Unlike a jump discontinuity, where both one-sided limits exist but are unequal, or an infinite discontinuity, where the function diverges in a particular direction, an essential discontinuity reflects fundamentally irregular behaviour that cannot be reduced to simpler forms.

A classic example is the following function, which exhibits an essential discontinuity at $x = 0$:

f(x) = \sin\left(\frac{1}{x}\right)

As $x$ approaches $0$, the argument $1/x$ grows without bound, causing the function to oscillate between $-1$ and $1$ with increasing frequency. Neither one-sided limit exists, and no value can be assigned to $f(0)$ that would restore any form of continuity.

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