Infinite Discontinuities

When Function Values Grow Without Bound

At a jump discontinuity, the function approaches two different finite values from opposite sides. Another possibility is that the values do not approach any finite number because their magnitude grows without bound as the input approaches a point. This behavior is described by limits equal to positive or negative infinity. Such a limit is not a real number; it records a precise kind of unbounded behavior near the point.

Throughout this tutorial, let \(f:I\to\mathbb{R}\), where \(I\) is an interval, and let \(a\) be an interior point of \(I\). Thus inputs in \(I\) can approach \(a\) from either side. The value \(f(a)\) is finite because \(f\) is real-valued, but it does not determine what happens to \(f(x)\) as \(x\) approaches \(a\) through values different from \(a\).

For example, a right-hand limit of \(+\infty\) means that no matter how large a positive threshold \(M\) is chosen, all sufficiently close inputs to the right of \(a\) have output greater than \(M\). A function that merely takes some very large values near \(a\) need not have this property: the definition requires the inequality to hold for every sufficiently close input on that side.

Worked Examples: Checking the Direction and Sign

A Sequential Test for Infinite Limits

The Sequential Criterion for Continuity from “Proof of the Sequential Criterion” concerns finite-valued limits. There is a parallel test for limits equal to infinity: every sequence approaching the point from the relevant side must have function values tending to the specified infinity. The test is useful because it can establish a limit using arbitrary approaching sequences, or disprove one by producing a single sequence that fails.

Proof. Suppose first that \(\lim_{x\to a^+}f(x)=+\infty\), and let \((x_n)\) be any sequence in \(I\) with \(x_n>a\) and \(x_n\to a\). To show that \(f(x_n)\to+\infty\), fix \(M>0\). By the definition of the right-hand infinite limit, there is a \(\delta>0\) such that \(a<x<a+\delta\) and \(x\in I\) imply \(f(x)>M\). Since \(x_n\to a\), there is an integer \(N\) such that \(n\geq N\) implies \(|x_n-a|<\delta\). Together with \(x_n>a\), this gives \(a<x_n<a+\delta\), and therefore \(f(x_n)>M\) for every \(n\geq N\). This holds for every \(M>0\), so \(f(x_n)\to+\infty\).

Conversely, suppose the right-hand limit is not \(+\infty\). Negating its definition, there is an \(M>0\) such that for every \(\delta>0\), some \(x\in I\) satisfies \(a<x<a+\delta\) and \(f(x)\leq M\). For each positive integer \(n\), apply this statement with \(\delta=1/n\) to choose \(x_n\in I\) with \(a<x_n<a+1/n\) and \(f(x_n)\leq M\). The inequalities \(0<x_n-a<1/n\) show that \(x_n\to a\) from the right. But \(f(x_n)\leq M\) for every \(n\), so the sequence \(f(x_n)\) does not tend to \(+\infty\). This contradicts the proposed sequence property and proves the converse. The left-hand cases follow by the same argument, using \(a-1/n<x_n<a\) for the sequence construction; the cases with limit \(-\infty\) follow by applying the corresponding arguments to \(-f\). \(\square\)

A failed sequence test is often especially direct. For instance, to rule out a right-hand limit of \(+\infty\), it is enough to find one sequence approaching from the right whose function values stay bounded above, or that tend instead to \(-\infty\). By contrast, checking just one sequence whose outputs tend to \(+\infty\) cannot establish the limit: the definition requires the behavior to hold for all sufficiently close inputs on that side.

Infinite Limits Force Local Unboundedness

An infinite limit entails more than discontinuity: the function is unbounded on every sufficiently small punctured neighborhood on the relevant side. This provides a useful way to rule out an infinite limit. If a function is bounded on even one such one-sided neighborhood, it cannot tend to either \(+\infty\) or \(-\infty\) there.

Proof. First suppose the right-hand limit is \(+\infty\). Let \(\delta_0>0\) be any neighborhood width. By the definition of this limit, using the threshold \(M=1\), there is a \(\delta>0\) such that \(a<x<a+\delta\) and \(x\in I\) imply \(f(x)>1\). Since \(a\) is interior to \(I\), there is a point \(x\in I\) with \(a<x<a+\min\{\delta,\delta_0\}\). In fact, every such point has \(f(x)>1\), and for any threshold \(M>0\) the same argument with that \(M\) gives values greater than \(M\) arbitrarily close to \(a\) on the right. Thus \(f\) is unbounded on the right-hand punctured neighborhood of width \(\delta_0\). If the limit is \(-\infty\), the definition gives values less than \(-M\) arbitrarily close on that side, so the function is again unbounded. The proof for a left-hand limit is identical with inputs to the left.

Now suppose there were a finite two-sided punctured limit \(L\). By its definition, using \(\varepsilon=1\), there would be a \(\rho>0\) such that \(0<|x-a|<\rho\) and \(x\in I\) imply \(|f(x)-L|<1\). The triangle inequality then gives \(|f(x)|\leq |L|+1\) throughout that punctured neighborhood. This contradicts the unboundedness just proved on the side with an infinite limit. Hence no finite two-sided punctured limit exists. The Continuity as a Limit theorem from earlier in this course says that continuity at \(a\) requires the finite punctured limit to equal \(f(a)\). Since no such finite limit exists, \(f\) is not continuous at \(a\). \(\square\)

Unboundedness Alone Is Not Enough

The converse of the local-unboundedness conclusion is false. A function may be unbounded in every neighborhood of a point without tending to \(+\infty\) or \(-\infty\) from either side. For example, define \(q(x)=\sin(1/x)/x\) for \(x\ne0\), and give \(q(0)\) any finite value. For positive integers \(n\), set

Both sequences are positive and tend to \(0\). Since \(\sin(\pi/2+2\pi n)=1\), we have \(q(u_n)=1/u_n=\pi/2+2\pi n\to+\infty\). Since \(\sin(3\pi/2+2\pi n)=-1\), we have \(q(v_n)=-1/v_n=-(3\pi/2+2\pi n)\to-\infty\). The Sequential Criterion shows that \(q\) does not tend to \(+\infty\) from the right, because the second sequence has outputs tending to \(-\infty\); it does not tend to \(-\infty\) from the right either, because the first sequence has outputs tending to \(+\infty\). This is unbounded behavior, but not a one-sided infinite limit.

This distinction is important when classifying a discontinuity. To identify an infinite discontinuity, verify that outputs eventually exceed every positive threshold or eventually fall below every negative threshold on at least one side. Finding arbitrarily large outputs is not enough. The next topic, oscillatory discontinuities, develops further examples in which nearby function values fail to settle into a single limiting behavior.