When Function Values Have No Finite Bound
Some discontinuities involve two finite one-sided limits that disagree. In other cases, function values do not approach any finite number at all: as the input approaches a particular point, the outputs grow beyond every fixed bound, either positively or negatively. This behavior is called an infinite discontinuity.
The direction of approach matters. A function may grow without bound on one side of a point and decrease without bound on the other. To recognize the behavior correctly, examine the signs of the function values on each side, not just the fact that a denominator becomes small.
One-Sided Limits Equal to Infinity
Let \(f\) be defined for inputs immediately to the right of \(a\), though \(f(a)\) itself need not be defined. The statement that \(f(x)\) tends to positive infinity as \(x\) approaches \(a\) from the right means that \(f(x)\) eventually exceeds every positive bound.
These symbols do not mean that positive or negative infinity is a real number that the function reaches. They describe unbounded behavior. For example, \(\lim_{x\to a^+}f(x)=+\infty\) says that however large a positive threshold \(M\) is chosen, all sufficiently nearby values on the right exceed it.
An infinite discontinuity at \(a\) occurs when at least one one-sided limit is \(+\infty\) or \(-\infty\). This can happen even if the function is assigned a finite value at \(a\). The value \(f(a)\), when defined, records the output at one input; an infinite limit describes outputs at nearby inputs.
First Example: A Reciprocal Changes Sign Across Its Centre
Worked Example: The two sides of a reciprocal function
Consider \(f(x)=\frac{1}{x-3}\) for \(x\ne 3\). Determine its behavior from the left and from the right as \(x\) approaches \(3\).
If \(x<3\), then \(x-3<0\), so \(f(x)\) is negative. As \(x\) gets closer to \(3\) from the left, \(x-3\) is negative and close to zero. For instance, \(f(2.9)=\frac{1}{-0.1}=-10\), and \(f(2.99)=\frac{1}{-0.01}=-100\). These values decrease without bound.
If \(x>3\), then \(x-3>0\), so \(f(x)\) is positive. For instance, \(f(3.1)=\frac{1}{0.1}=10\), and \(f(3.01)=\frac{1}{0.01}=100\). These values grow without bound.
The definition verifies both conclusions. Given any \(M>0\), choose \(\delta=\frac{1}{M}\). If \(0<3-x<\delta\), then \(0<3-x<\frac{1}{M}\), so \(\frac{1}{3-x}>M\). Since \(f(x)=-\frac{1}{3-x}\) on the left, \(f(x)<-M\). If \(0<x-3<\delta\), then \(\frac{1}{x-3}>M\), so \(f(x)>M\). Therefore the left-hand limit is \(-\infty\), and the right-hand limit is \(+\infty\).
Check: At \(x=2.99\), the function value is \(-100\), which is less than \(-10\); at \(x=3.01\), the value is \(100\), which exceeds \(10\). These substitutions agree with the signs and directions established above.
The signs follow from the denominator: it is negative to the left of \(3\) and positive to the right. A denominator close to zero does not by itself tell you whether the quotient is large positive or large negative.
Second Example: Squaring Makes Both Sides Positive
Worked Example: A squared reciprocal
Let \(g(x)=\frac{1}{(x+2)^2}\) for \(x\ne -2\). Determine the one-sided behavior as \(x\) approaches \(-2\).
For every \(x\ne -2\), the squared denominator is positive, so \(g(x)>0\) on both sides. When \(x=-1.9\), \(g(x)=\frac{1}{(0.1)^2}=100\). When \(x=-2.1\), \(g(x)=\frac{1}{(-0.1)^2}=100\). Moving closer to \(-2\) makes the squared distance smaller and the reciprocal larger.
To verify unbounded growth, let \(M>0\) and choose \(\delta=\frac{1}{\sqrt{M}}\). Whenever \(0<|x+2|<\delta\), we have \((x+2)^2<\frac{1}{M}\). Both sides are positive, so taking reciprocals gives \(\frac{1}{(x+2)^2}>M\). This condition applies whether \(x<-2\) or \(x>-2\).
Check: For \(M=100\), the choice is \(\delta=\frac{1}{10}=0.1\). At either \(x=-1.95\) or \(x=-2.05\), the distance to \(-2\) is \(0.05<0.1\), and \(g(x)=\frac{1}{(0.05)^2}=400>100\). Thus both one-sided limits equal positive infinity.
Third Example: The Value at the Point Does Not Control Nearby Values
Worked Example: A defined point value with unbounded limits
Define \(h\) by \(h(x)=-\frac{1}{x-1}\) when \(x\ne1\), and set \(h(1)=6\). Find the one-sided limits at \(1\) and decide whether \(h\) is continuous there.
When \(x<1\), \(x-1<0\), so \(-\frac{1}{x-1}\) is positive. At \(x=0.9\), \(h(x)=-\frac{1}{-0.1}=10\); at \(x=0.99\), \(h(x)=-\frac{1}{-0.01}=100\). Thus the values grow without bound from the left.
When \(x>1\), \(x-1>0\), so \(h(x)\) is negative. At \(x=1.1\), \(h(x)=-\frac{1}{0.1}=-10\); at \(x=1.01\), \(h(x)=-\frac{1}{0.01}=-100\). Thus the values decrease without bound from the right.
For a definition-based check, take any \(M>0\) and set \(\delta=\frac{1}{M}\). If \(0<1-x<\delta\), then \(0<1-x<\frac{1}{M}\), so \(\frac{1}{1-x}>M\). Since \(h(x)=\frac{1}{1-x}\) to the left, \(h(x)>M\). If \(0<x-1<\delta\), then \(\frac{1}{x-1}>M\) and \(h(x)=-\frac{1}{x-1}<-M\).
Check: The substitutions \(x=0.99\) and \(x=1.01\) give \(100\) and \(-100\), respectively, consistent with the left and right behavior. The assigned value \(h(1)=6\) is finite, but it does not change either limit. Therefore \(h\) is not continuous at \(1\).
Why an Infinite Limit Rules Out Continuity
Continuity at a point requires the function values near that point to remain close to its value there. More precisely, if \(f(a)\) is defined, continuity at \(a\) means that for every \(\varepsilon>0\), there exists \(\delta>0\) such that \(|x-a|<\delta\) implies \(|f(x)-f(a)|<\varepsilon\), for inputs in the domain.
Here is why an infinite one-sided limit contradicts that condition. Suppose \(f(a)\) is finite and \(\lim_{x\to a^+}f(x)=+\infty\). If \(f\) were continuous at \(a\), applying continuity with \(\varepsilon=1\) would give a \(\delta_0>0\) such that \(|x-a|<\delta_0\) implies \(|f(x)-f(a)|<1\). The triangle inequality then gives \(|f(x)|\le |f(a)|+|f(x)-f(a)|<|f(a)|+1\).
But the infinite-limit definition, with \(M=|f(a)|+1\), gives a \(\delta_1>0\) such that \(0<x-a<\delta_1\) implies \(f(x)>|f(a)|+1\). Choose \(x\) with \(0<x-a<\min(\delta_0,\delta_1)\), which is possible because the function is defined immediately to the right of \(a\). Then \(|x-a|<\delta_0\), so continuity would give \(|f(x)|<|f(a)|+1\), while the infinite-limit condition gives \(f(x)>|f(a)|+1\). This contradiction proves that \(f\) is not continuous at \(a\).
If \(a\) is not in the function’s domain, continuity at \(a\) is not defined for that function. The unbounded one-sided behavior is still present, but there is no point value to compare with nearby values.
Reading the Behavior Correctly
A vertical line \(x=a\) is called a vertical asymptote when the function has an infinite limit from at least one side of \(a\). The one-sided limits need not agree: one side may tend to positive infinity while the other tends to negative infinity. The asymptote describes the unbounded behavior near the line; it does not say that the function attains an infinite value.
When analyzing a reciprocal expression near a point, use this sequence of checks:
Determine whether the input is less than or greater than the point being approached.
Find the sign of the denominator, and account for any negative sign or even power in the formula.
As the denominator’s magnitude shrinks toward zero, a reciprocal’s magnitude grows without bound.
State the left-hand and right-hand limits, including the sign of each infinite limit.
A Common Sign Error
For \(f(x)=\frac{1}{x-3}\), a common incorrect calculation is: “As \(x\to3^-\), the denominator approaches zero, so \(f(x)\to+\infty\).” The conclusion ignores the sign of the denominator. On the left, \(x-3<0\), so the quotient is negative. For example, \(x=2.9\) gives \(f(x)=\frac{1}{-0.1}=-10\), not \(+10\). The corrected result is \(\lim_{x\to3^-}\frac{1}{x-3}=-\infty\); from the right, the denominator is positive and the limit is \(+\infty\).
Another point to keep distinct is approaching a finite input versus allowing the input itself to grow without bound. Here the input approaches a fixed number such as \(3\), while the function values become unbounded. The signs and one-sided behavior are the evidence for an infinite discontinuity.
Summary
An infinite discontinuity occurs when function values exceed every finite bound in magnitude on at least one side of a point. The sign determines whether the one-sided limit is \(+\infty\) or \(-\infty\). A finite assigned value at the point cannot remove unbounded nearby behavior, and a function with an infinite one-sided limit at a point where it is defined is not continuous there.
Check Your Understanding
For each question, pay attention to which side of the point is being considered and to the sign of the function values.
- For \(f(x)=\frac{1}{x+4}\), determine the left-hand and right-hand limits as \(x\) approaches \(-4\).
- Why do both one-sided limits of \(g(x)=\frac{1}{(x-5)^2}\) equal positive infinity?
- If a function has value \(f(2)=3\) but \(\lim_{x\to2^+}f(x)=+\infty\), can it be continuous at \(2\)? Explain.
- State the definition of \(\lim_{x\to a^-}f(x)=-\infty\) using \(M\) and \(\delta\).
- For \(h(x)=-\frac{1}{x+1}\), determine the signs of the one-sided limits as \(x\) approaches \(-1\).
