Why Signs Matter Near a Singular Input
A rational expression can become unbounded near a point where its denominator is zero. But knowing that its magnitude grows without bound does not, by itself, tell us whether the expression grows toward positive infinity or negative infinity.
To determine the direction, examine the signs of the numerator and denominator on each side of the input. A factor chart makes this systematic: it records the sign of each factor just to the left and just to the right, then combines those signs to find the sign of the quotient.
The left and right limits must be considered separately. A factor that is negative on one side of a zero may be positive on the other, so the quotient can head in different directions on the two sides.
Reading a Sign Chart
For a real number \(a\), a left-hand neighborhood consists of inputs \(x<a\) sufficiently close to \(a\); a right-hand neighborhood consists of inputs \(x>a\) sufficiently close to \(a\). A one-sided limit describes the behavior as \(x\) approaches \(a\) within one of those neighborhoods.
For a factor \(x-r\), its sign follows directly from the comparison between \(x\) and \(r\): it is negative when \(x<r\), zero when \(x=r\), and positive when \(x>r\). Any factor that is nonzero at \(a\) keeps its sign on a sufficiently small interval around \(a\), because it is continuous.
In this tutorial, we use the chart for rational functions \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials. The singular input \(a\) satisfies \(Q(a)=0\), while \(P(a)\ne 0\). The examples use a single denominator factor that vanishes at \(a\); the remaining factors keep fixed signs nearby.
A Justification for the Signed Infinite Limit
A sign chart identifies the eventual sign of the quotient. The following result explains why that sign, together with a denominator tending to zero, gives a signed infinite limit.
Here “sufficiently close” means that there is a positive distance from \(a\) within which the stated property holds, excluding \(a\) itself. We prove the result for either chosen side.
Since \(P\) is continuous and \(P(a)\ne0\), \(P(x)\) has the same sign as \(P(a)\) for all \(x\) sufficiently close to \(a\). For example, continuity allows us to require \(|P(x)-P(a)|<|P(a)|/2\), which prevents \(P(x)\) from reaching zero or changing sign.
The denominator is continuous and nonzero throughout a sufficiently small punctured interval on that side. If it took both positive and negative values there, the intermediate value theorem would force it to equal zero between those inputs. Thus it has one fixed sign on that interval.
Since \(Q(a)=0\) and \(Q\) is continuous, \(Q(x)\to0\). Also, sufficiently close to \(a\), \(|P(x)|>|P(a)|/2\). For any \(M>0\), take \(x\) close enough that \(|Q(x)|<|P(a)|/(2M)\). Then \(|P(x)/Q(x)|>M\).
The quotient has a fixed sign and magnitude greater than any prescribed \(M>0\) sufficiently close to \(a\). If its sign is positive it eventually exceeds every \(M\); if negative it is eventually less than \(-M\). These are precisely the meanings of limits \(+\infty\) and \(-\infty\), respectively.
The fixed-sign condition matters. For arbitrary functions, a denominator could approach zero while changing sign repeatedly, so an unbounded quotient need not tend to either signed infinity. For polynomials, the nonzero condition on the chosen punctured side and continuity ensure that the sign is fixed, as the theorem requires.
Worked Examples
Worked Example: Opposite Infinite Limits on Either Side
Find the one-sided limits of \(f(x)=\dfrac{x+2}{(x-1)(x+3)}\) at \(a=1\).
The denominator is zero at \(x=1\), while the numerator there is \(1+2=3\ne0\). The factor \(x+3\) is positive near \(1\), since its value at \(1\) is \(4\). The factor \(x+2\) is also positive near \(1\), since its value there is \(3\). Only \(x-1\) changes sign across \(1\).
| Factor or expression | Left of \(1\) | Right of \(1\) |
|---|---|---|
| \(x+2\) | Positive | Positive |
| \(x-1\) | Negative | Positive |
| \(x+3\) | Positive | Positive |
| Denominator \((x-1)(x+3)\) | Negative | Positive |
| Quotient \(f(x)\) | Negative | Positive |
The denominator tends to zero from below on the left and from above on the right, while the numerator tends to \(3>0\). Therefore,
Check the chart with exact nearby inputs. At \(x=9/10\), the numerator is \(29/10\) and the denominator is \((-1/10)(39/10)=-39/100\), so \(f(9/10)=-290/39<0\). At \(x=11/10\), the numerator is \(31/10\) and the denominator is \((1/10)(41/10)=41/100\), so \(f(11/10)=310/41>0\). These values verify the chart’s signs; the theorem and the behavior as \(x\to1\) justify the limits.
Worked Example: A Positive Left-Hand Limit
Find the one-sided limits of \(g(x)=\dfrac{2-x}{(x+1)(x-2)}\) at \(a=-1\).
At \(x=-1\), the numerator is \(2-(-1)=3>0\), and the factor \(x-2\) is \(-3<0\). Both keep those signs sufficiently close to \(-1\). The factor \(x+1\) is negative to the left and positive to the right.
| Factor or expression | Left of \(-1\) | Right of \(-1\) |
|---|---|---|
| \(2-x\) | Positive | Positive |
| \(x+1\) | Negative | Positive |
| \(x-2\) | Negative | Negative |
| Denominator \((x+1)(x-2)\) | Positive | Negative |
| Quotient \(g(x)\) | Positive | Negative |
Thus the quotient tends to positive infinity on the left and negative infinity on the right:
For an exact check, at \(x=-11/10\), the numerator is \(31/10\), and the denominator is \((-1/10)(-31/10)=31/100\). The quotient is \(10>0\). At \(x=-9/10\), the numerator is \(29/10\), and the denominator is \((1/10)(-29/10)=-29/100\). The quotient is \(-10<0\). The values confirm the left and right signs.
Worked Example: A Negative Right-Hand Limit
Find the one-sided limits of \(h(x)=\dfrac{x-5}{(x-2)(x+4)}\) at \(a=2\).
The numerator at \(2\) is \(-3<0\). The factor \(x+4\) is positive there, with value \(6\), and remains positive close to \(2\). The factor \(x-2\) is negative to the left and positive to the right.
| Factor or expression | Left of \(2\) | Right of \(2\) |
|---|---|---|
| \(x-5\) | Negative | Negative |
| \(x-2\) | Negative | Positive |
| \(x+4\) | Positive | Positive |
| Denominator \((x-2)(x+4)\) | Negative | Positive |
| Quotient \(h(x)\) | Positive | Negative |
The signs and the theorem give
Check using \(x=19/10\): the numerator is \(-31/10\), and the denominator is \((-1/10)(59/10)=-59/100\), so \(h(19/10)=310/59>0\). Using \(x=21/10\), the numerator is \(-29/10\), and the denominator is \((1/10)(61/10)=61/100\), so \(h(21/10)=-290/61<0\). Again, the exact checks agree with the chart.
A Common Sign-Chart Mistake
A frequent error is to assign the same sign to a factor on both sides of its zero. For example, someone might reason that \(x-1\) is “the denominator factor near \(1\)” and label it positive on both sides. That would give the wrong left-hand sign in the first example.
The correction is to compare \(x\) with \(1\) on each side separately. If \(x<1\), then \(x-1<0\); if \(x>1\), then \(x-1>0\). The factor’s value at the singular input is zero, but that value does not supply its sign nearby.
A related caution is that testing one nearby input is only a check, not a proof of a limit. The sign chart establishes the signs throughout the relevant intervals, and the limit argument establishes that the magnitude exceeds every positive bound as the input approaches the singular point.
Why the Method Is Useful
A sign chart separates two questions that are easy to confuse: whether the quotient becomes unbounded, and which direction it takes. The numerator and denominator signs answer the direction question; the fact that the numerator stays away from zero while the denominator tends to zero accounts for unbounded magnitude.
The method also prevents left-hand and right-hand behavior from being collapsed into one conclusion. A rational expression may tend to opposite infinities on the two sides, or its signs may agree on both sides. In either case, the one-sided charts show exactly how the signs combine.
For a new rational expression, proceed in order: identify the singular input, factor the numerator and denominator, record each factor’s sign on each side, combine signs, and then state the one-sided limits. Before making a signed infinite-limit conclusion, check that the numerator is nonzero at the input and that the denominator has no zeros in the punctured one-sided interval under consideration.
Check Your Understanding
For each question, make a separate sign chart for the left and right sides where appropriate.
- For \(f(x)=\dfrac{x+4}{(x-3)(x+1)}\), what are the signs of \(f(x)\) immediately to the left and right of \(x=3\)?
- For \(g(x)=\dfrac{1-x}{(x+2)(x-1)}\), determine the two one-sided limits at \(x=-2\).
- Why is checking the value of a rational expression at one nearby input not enough to establish its one-sided limit?
- In the theorem, why must the denominator be nonzero sufficiently close to \(a\) on the chosen side?
- What additional fact, beyond unbounded magnitude, is needed to conclude a limit of \(+\infty\) rather than \(-\infty\)?
