Why Prove an Infinite Limit With Inequalities?
A graph may suggest that a function grows without bound as its input approaches a particular value. A limit calculation may suggest the same conclusion. But to verify an infinite one-sided limit from its definition, we must show that the function exceeds any chosen positive threshold (or falls below its negative), provided the input is sufficiently close to the target from the specified side.
For rational functions near a zero of the denominator, the proof has two separate tasks. First, control the sign: the function must be positive to tend to \(+\infty\), or negative to tend to \(-\infty\). Second, control its magnitude: the numerator must not become too small, while the denominator becomes small enough. Keeping these tasks separate makes the argument precise and helps prevent sign mistakes.
The number \(M\) is an arbitrary height that the function must eventually exceed in magnitude. The distance \(\delta\) may depend on \(M\). A proof must work for every \(M>0\), not just for one convenient value.
The Inequality Strategy
Suppose a rational function has a denominator that approaches zero at \(x=a\). A useful proof often finds a neighborhood of \(a\) in which the numerator has a fixed sign and stays at least a fixed positive distance from zero. At the same time, we bound the absolute value of the denominator above by a quantity that becomes small as \(x\) approaches \(a\).
For example, if in a chosen one-sided interval the function is positive and satisfies \(\displaystyle f(x)\geq \frac{c}{|x-a|}\) for some constant \(c>0\), then requiring \(\displaystyle 0<|x-a|<\frac{c}{M}\) ensures \(f(x)>M\). The exact bound depends on the function, but the structure is consistent: find a lower bound for the function and choose \(\delta\) to make that lower bound exceed \(M\).
Choose a small one-sided interval around \(a\) that avoids other denominator zeros and makes the sign factors clear.
Show that its magnitude is at least some fixed \(c>0\) throughout that interval.
Show that its magnitude is at most a quantity involving \(|x-a|\), and verify whether the quotient is positive or negative.
Make the resulting lower bound for the magnitude larger than the arbitrary threshold.
Worked Example: A Positive Infinite Limit From the Right
Worked Example: Proving a Limit at a Simple Denominator Zero
Prove from the definition that \(\displaystyle \lim_{x\to 3^+}\frac{2x+1}{x-3}=+\infty\).
Let \(M>0\) be arbitrary. We will show that a suitable choice of \(\delta\) makes the quotient greater than \(M\) whenever \(3<x<3+\delta\).
Choose \(\delta=\min\{1,7/M\}\). If \(0<x-3<\delta\), then \(3<x<4\), so \(2x+1>7\). Also, \(x-3\) is positive and \(x-3<\delta\leq 7/M\). Consequently,
The first inequality uses \(2x+1>7\) and \(x-3>0\). The second uses \(0<x-3\leq 7/M\), so its reciprocal is at least \(M/7\); because the original inequality is strict, the result is strictly greater than \(M\). Thus the defining condition holds for every \(M>0\), and the limit is \(+\infty\).
As a check, the function is positive on the interval used: its numerator is greater than \(7\), and its denominator is positive. The inequality also gives the required size, not merely the sign.
Worked Example: A Two-Sided Blow-Up
Worked Example: A Squared Denominator
Prove that \(\displaystyle \lim_{x\to 1^-}\frac{x+2}{(x-1)^2}=+\infty\).
Let \(M>0\). Choose \(\delta=\min\{1,\sqrt{2/M}\}\). If \(0<1-x<\delta\), then \(0<x<1\), and hence \(x+2>2\). The squared denominator is positive, and \((x-1)^2=(1-x)^2<\delta^2\leq 2/M\). Therefore,
Every \(M>0\) has a corresponding positive \(\delta\), so this proves the stated left-hand limit. The same bounds apply when \(0<x-1<\delta\), so they also prove \(\lim_{x\to 1^+}\frac{x+2}{(x-1)^2}=+\infty\). In this case both sides work because the numerator stays positive and the squared denominator is positive on either side.
For a final check, as \(x\) approaches \(1\) from either side, the numerator remains greater than \(2\), while the positive denominator can be made smaller than \(2/M\). The quotient is therefore greater than \(M\), as required.
Worked Example: Proving a Negative Infinite Limit
Worked Example: Controlling the Sign on the Left
Prove that \(\displaystyle \lim_{x\to 2^-}\frac{x+1}{2x-4}=-\infty\).
Let \(M>0\) be arbitrary. Choose \(\delta=\min\{1,1/M\}\). If \(0<2-x<\delta\), then \(1<x<2\), so \(x+1>2\). Rewrite the denominator as \(2x-4=-2(2-x)\). Since \(2-x>0\), the quotient is negative. Its magnitude satisfies
The last inequality follows from \(0<2-x<\delta\leq 1/M\). Because the quotient is negative and its magnitude is greater than \(M\), it follows that \(\frac{x+1}{2x-4}<-M\). This is the defining condition for the limit to equal \(-\infty\).
To check the sign explicitly, \(x+1>0\) and \(2x-4=-2(2-x)<0\) throughout the interval. Thus the quotient is indeed negative there, and the magnitude estimate proves it falls below every \(-M\).
A General Justification for the Method
The examples use the same underlying principle. Suppose a rational function can be written near \(a\) as
where \(m\) is a positive integer, \(P\) and \(S\) are continuous near \(a\), and \(P(a)\neq 0\) and \(S(a)\neq 0\). Since \(P\) and \(S\) are continuous and nonzero at \(a\), there is a sufficiently small interval around \(a\) on which neither changes sign. We can also choose positive constants \(p\) and \(s\) such that \(|P(x)|\geq p\) and \(|S(x)|\leq s\) throughout that interval.
On either one-sided part of this interval, the sign of \(f(x)\) is fixed: it is determined by the signs of \(P(x)\), \(S(x)\), and \((x-a)^m\). If that sign is positive, then
Given any \(M>0\), make the input close enough that \(\displaystyle |x-a|<\left(\frac{p}{sM}\right)^{1/m}\). Then \(\displaystyle \frac{p}{s|x-a|^m}>M\), so \(f(x)>M\). If the sign is negative instead, the same argument gives \(|f(x)|>M\), and therefore \(f(x)<-M\). This proves the relevant one-sided infinite limit directly from its definition whenever the sign on that side is known.
The argument also explains why a numerator that vanishes at \(a\) needs separate attention: the lower bound \(|P(x)|\geq p>0\) may no longer be available. A denominator zero by itself does not prove an infinite limit. The function’s sign and the relative sizes of numerator and denominator must still be checked.
A Common Sign Mistake
Consider \(\frac{x+1}{2x-4}\) as \(x\to 2^-\). A tempting but incorrect conclusion is that the limit is \(+\infty\), on the grounds that the denominator approaches zero and the numerator approaches a positive number. This ignores which side the denominator approaches zero from.
For \(x<2\), \(2x-4=2(x-2)<0\), while \(x+1>0\) near \(2\). The quotient is negative. The worked proof established the stronger statement that it is less than \(-M\) for every \(M>0\) when \(x\) is sufficiently close to \(2\) from the left. Thus the limit is \(-\infty\), not \(+\infty\).
Connection to Vertical Asymptotes
A vertical asymptote is associated with an infinite one-sided limit at a finite input. The inequality proofs here give a direct verification of such behavior: they show precisely that the function exceeds every positive bound, or falls below every negative bound, close enough to the relevant input. The important feature of the proof is not a sketch or a pattern in computed values; it is the quantified statement for every \(M>0\).
When proving a particular limit, keep the order of reasoning clear: choose a small interval, establish a fixed sign, derive a bound for the magnitude, and then select \(\delta\) in terms of \(M\). This order makes each condition visible and provides a check that the conclusion follows from the definition.
Check Your Understanding
For each question, identify the sign and the inequality needed to verify the proposed limit.
- For \(\displaystyle f(x)=\frac{x+4}{x-1}\), which one-sided limit at \(x=1\) is \(+\infty\), and which is \(-\infty\)?
- In proving \(\displaystyle \lim_{x\to 0^+}\frac{3}{x}=+\infty\), what choice of \(\delta\) works for an arbitrary \(M>0\)?
- Why does \((x-2)^2\) have the same sign on both sides of \(x=2\), and how can that help prove a limit equal to \(+\infty\)?
- What condition on the numerator allows a fixed positive lower bound for its magnitude near a denominator zero?
- In a proof of a limit equal to \(-\infty\), what two facts must be established about the function?
