Why Factor Near a Denominator Zero?
A rational function can become unbounded near a number where its denominator is zero. To determine what happens on each side of that number, it is not enough to notice that the denominator approaches zero. The numerator and the remaining denominator factors also affect the sign of the quotient.
Factoring separates the factor that approaches zero from factors whose signs stay fixed nearby. We can then check the expression from the left and from the right independently.
One-Sided Infinite Limits
For a function \(f\), the notation \(\lim_{x\to c^-}f(x)=+\infty\) means that \(f(x)\) becomes larger than any prescribed positive number when \(x\) is sufficiently close to \(c\) from the left. The notation \(\lim_{x\to c^+}f(x)=+\infty\) has the same meaning when \(x\) approaches from the right. A limit of \(-\infty\) means that the function becomes less than any prescribed negative number on the indicated side.
These are statements about unbounded behavior, not claims that the function takes the value \(+\infty\) or \(-\infty\). A rational function is undefined at a point where its original denominator is zero.
Isolating the Factor That Approaches Zero
Suppose factoring gives a rational function in the form
where \(N\) and \(Q\) are continuous near \(c\), and \(N(c)\ne 0\) and \(Q(c)\ne 0\). For \(x\ne c\), rewrite this as
Since \(Q(c)\ne0\), the function \(H\) is defined and continuous near \(c\), and \(H(c)=N(c)/Q(c)\ne0\). Thus \(H(x)\) keeps the sign of \(H(c)\) sufficiently close to \(c\). Meanwhile, \(x-c\) is negative to the left of \(c\) and positive to the right. Comparing these signs determines the direction of each one-sided limit.
Why do the nonzero factors keep their signs? If a continuous function \(Q\) satisfies \(Q(c)\ne0\), continuity ensures that sufficiently nearby values differ from \(Q(c)\) by less than \(|Q(c)|/2\). They therefore cannot cross zero. The same reasoning applies to any other continuous factor whose value at \(c\) is nonzero.
Why the Sign Test Gives an Infinite Limit
The sign test does more than suggest an answer. It explains why the quotient becomes unbounded. Suppose \(H\) is continuous at \(c\) and \(H(c)=A>0\). For \(x\) sufficiently close to \(c\), continuity gives \(H(x)>A/2\). On the right, \(x-c>0\), so \(H(x)/(x-c)>0\). Given any \(M>0\), take \(x\) close enough that \(0<x-c<A/(2M)\). Then
So the right-hand limit is \(+\infty\). On the left, \(x-c<0\), and the quotient is negative; choosing \(0<c-x<A/(2M)\) makes its magnitude exceed \(M\), so the left-hand limit is \(-\infty\). If \(A<0\), the signs reverse. This justifies the one-sided sign procedure whenever the factor approaching zero is \(x-c\) and the remaining quotient approaches a nonzero value.
Worked Examples
Worked Example: Different Infinite Limits on Either Side
Find \(\lim_{x\to 2^-}\frac{2x+1}{(x-2)(x+3)}\) and \(\lim_{x\to 2^+}\frac{2x+1}{(x-2)(x+3)}\).
The denominator is already factored. The factor \(x-2\) approaches zero, while \(2x+1\) and \(x+3\) approach nonzero values:
Thus, near \(x=2\), the numerator and the factor \(x+3\) are positive. The sign of the whole quotient is controlled by \(x-2\).
From the left, \(x-2<0\), so the quotient is negative and its magnitude grows without bound as \(x-2\) approaches zero:
From the right, \(x-2>0\), and the other factors remain positive:
As a sign check, at \(x=1.9\), the numerator is \(4.8\) and the denominator is \((-0.1)(4.9)=-0.49\), giving a negative value. At \(x=2.1\), the numerator is \(5.2\) and the denominator is \((0.1)(5.1)=0.51\), giving a positive value. These checks agree with the limits.
Worked Example: The Nonzero Factors Determine the Overall Sign
Find the one-sided limits of \(f(x)=\frac{x+2}{(x-1)(x+4)}\) as \(x\) approaches \(-4\).
The factor \(x+4\) approaches zero. The other factors approach
Both are negative near \(-4\), so their quotient is positive. More explicitly, for \(x\ne-4\), write
From the left, \(x+4<0\), while the remaining quotient is positive, so \(f(x)\) is negative and unbounded:
From the right, \(x+4>0\), so the function is positive and unbounded:
For a numerical sign check, at \(x=-4.1\) the numerator is \(-2.1\), and the denominator is \((-5.1)(-0.1)=0.51\), giving a negative value. At \(x=-3.9\), the numerator is \(-1.9\), and the denominator is \((-4.9)(0.1)=-0.49\), giving a positive value. The signs match the conclusions.
Worked Example: A Negative Nonzero Factor Reverses the Signs
Find the one-sided limits of \(g(x)=\frac{x^2+1}{x(x-3)}\) as \(x\) approaches \(0\).
The factor \(x\) approaches zero. The numerator and the other denominator factor approach
Combine the factors that do not approach zero:
From the left, \(x<0\). The numerator-like remaining quotient is negative, so the whole expression is positive. Its denominator approaches zero in magnitude, giving
From the right, \(x>0\), so the negative remaining quotient makes the expression negative and unbounded:
As a check, at \(x=-0.01\), the denominator is \((-0.01)(-3.01)=0.0301>0\), while the numerator is \(1.0001>0\). At \(x=0.01\), the denominator is \((0.01)(-2.99)=-0.0299\), while the numerator is \(1.0001>0\). The respective signs are positive and negative, as found.
When the Numerator Also Vanishes
The method above requires that the remaining quotient approach a nonzero value. If the numerator also becomes zero, it is not enough to say “the denominator approaches zero, so the limit is infinite.” First factor and simplify the expression for \(x\ne c\), then evaluate the resulting limit.
Worked Example: A Vanishing Numerator Does Not Give an Infinite Limit
Evaluate \(\lim_{x\to 1}\frac{(x-1)(x+2)}{(x-1)(x+3)}\).
At \(x=1\), both the numerator and denominator of the original expression are zero. For \(x\ne1\), cancel the common factor \(x-1\):
The simplified expression has a nonzero denominator at \(x=1\), so substitute into it:
Check: substituting \(x=1\) into the simplified expression gives \(3/4\), confirming the limit. The original function is still undefined at \(x=1\); cancellation gives an expression equal to it only for \(x\ne1\), which is sufficient for the limit.
A Common Sign Mistake
A frequent error is to see that a denominator approaches zero and conclude that the function approaches \(+\infty\), without checking signs. For the function \(\frac{x+2}{(x-1)(x+4)}\) near \(-4\), that reasoning would incorrectly claim \(+\infty\) from both sides. The factor \(x+4\) does change sign, and the other factors must also be included in the check.
The correction is to evaluate the nonzero factors at the approach point and record the sign of the factor approaching zero on each side. Here, \((x+2)/(x-1)\) approaches \(2/5>0\), while \(x+4\) is negative on the left and positive on the right. Therefore the limits are \(-\infty\) from the left and \(+\infty\) from the right.
Check Your Understanding
For each question, factor where useful and determine the one-sided behavior from signs.
- Find the left-hand and right-hand limits of \(\frac{x+1}{(x-3)(x+2)}\) as \(x\to3\).
- Find the left-hand and right-hand limits of \(\frac{2x-1}{(x+1)(x-4)}\) as \(x\to-1\).
- For \(\frac{x^2+4}{(x-2)(x+5)}\), identify the signs of the numerator and the nonzero denominator factor near \(x=2\), then determine both one-sided limits.
- Why does a factor that is nonzero at the approach point keep its sign sufficiently close to that point?
- What should you check if both the numerator and denominator are zero at the approach point?
