Discontinuities of Real Functions
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:
\[\underset{x \rightarrow x_{0}}{lim} f ( x ) = f ( x_{0} )\]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 $x^{2} + 2 x + 1$, which represents a parabola:
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:
\[\underset{x \rightarrow x_{0}}{lim} f ( x ) = ℓ \in \mathbb{R}\]Moreover, at least one of the following conditions is satisfied:
- $f ( x_{0} )$ is undefined.
- $f ( x_{0} ) \neq ℓ$.
In such cases, the discontinuity may be removed by redefining the function at a single point as follows: \(g ( x ) = \{ ℓ & \text{if} x = x_{0} \\ f ( x ) & \text{if} x \neq x_{0}\)
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:
\[\underset{x \rightarrow 1}{lim} \frac{x^{2} - 1}{x - 1} = 2\]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: \(g ( x ) = \{ 2 & \text{if} x = 1 \\ f ( x ) & \text{if} x \neq 1\)
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:
\[\underset{x \rightarrow x_{0}^{-}}{lim} f ( x ) & = ℓ_{1} \in \mathbb{R} \\ \underset{x \rightarrow x_{0}^{+}}{lim} f ( x ) & = ℓ_{2} \in \mathbb{R} \\ ℓ_{1} & \neq ℓ_{2}\]In this case, the limit $\underset{x \rightarrow x_{0}}{lim} f ( x )$ 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 $x = 1.$
\[f ( x ) = \{ 0 & \text{if} x < 1 \\ 2 & \text{if} x \geq 1\]
For values of $x$ approaching $1$ from the left, the function remains constant at $0$. Therefore:
\[\underset{x \rightarrow 1^{-}}{lim} f ( x ) = 0\]For values of $x$ approaching 1 from the right, the function remains constantly equal to $2$, and therefore the limit is: \(\underset{x \rightarrow 1^{+}}{lim} f ( x ) = 2\)
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 $\underset{x \rightarrow 1}{lim} f ( x )$ 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:
\[\underset{x \rightarrow x_{0}^{-}}{lim} f ( x ) & = \pm \infty \\ \underset{x \rightarrow x_{0}^{+}}{lim} f ( x ) & = \pm \infty\]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 $x \rightarrow 2^{-}$, the denominator $x - 2$ becomes negative and approaches zero, causing the function to decrease without bound. Therefore we have:
\[\underset{x \rightarrow 2^{-}}{lim} f ( x ) = - \infty\]As $x \rightarrow 2^{+}$, the denominator is positive and approaches zero, which causes the function to increase without bound. The limit is:
\[\underset{x \rightarrow 2^{+}}{lim} f ( x ) = + \infty\]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:
\(f^{'} ( x_{0} ) = \underset{x \rightarrow x_{0}}{lim} \frac{f ( x ) - f ( x_{0} )}{x - x_{0}}\) \(\underset{x \rightarrow x_{0}}{lim} f ( x ) = f ( x_{0} )\)
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.
\[\underset{x \rightarrow x_{0}^{-}}{lim} \frac{f ( x ) - f ( x_{0} )}{x - x_{0}} \neq \underset{x \rightarrow x_{0}^{+}}{lim} \frac{f ( x ) - f ( x_{0} )}{x - x_{0}}\]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 $x = 0$:
\[f ( x ) = sin ( \frac{1}{x} )\]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.
Selected references
- Harvard University N. Masson. Discontinuities and Monotonic Functions
- Harvard University O. Knill. Introduction to Functions and Calculus – Continuity
- UC Berkeley A. Vizeff. Lecture 5: Continuity and Discontinuities