Why Factor a Quotient?
For many quotient limits, direct substitution gives an immediate answer: if the denominator approaches a nonzero number, the quotient law applies. A different situation arises when both numerator and denominator approach zero. The quotient law does not determine the limit in that case, because its nonzero-denominator condition is not met.
When the numerator and denominator are quadratic polynomials, factoring can reveal a shared linear factor. The quotient can then be rewritten for inputs near the approach point, and the limit may be evaluated from the remaining factors.
Recognizing the Starting Point
Suppose we want to evaluate \(\lim_{x\to a}\frac{P(x)}{Q(x)}\), where \(P\) and \(Q\) are quadratic polynomials. First substitute \(x=a\) into each polynomial. If \(Q(a)\ne 0\), polynomial limits and the quotient law give
The interesting case here is \(P(a)=0\) and \(Q(a)=0\). The notation \(0/0\) does not mean the limit is zero, or that the limit fails to exist. It means direct substitution has not determined the limit. Factoring may provide more information.
A quadratic can sometimes be written as a product of linear factors. For example, \(x^2-5x+6=(x-2)(x-3)\). When substitution at \(a\) gives zero for a quadratic, \(x-a\) is one of its factors. Factoring both quadratics can therefore expose the factors that vanish at the same input.
Why the Simplification Works
The key justification is that a limit depends on the values of a function for inputs close to \(a\), but not equal to \(a\). Suppose the two polynomials factor as
For \(x\ne a\), the common factor \(x-a\) is nonzero, so ordinary algebra gives
whenever the original quotient is defined. If \(Q_1(a)\ne 0\), then \(Q_1(x)\) remains nonzero for inputs sufficiently close to \(a\), and the quotient law and polynomial limits give the finite limit \(P_1(a)/Q_1(a)\).
This argument explains both the method and its condition. The factorization alone does not guarantee a finite limit: after removing the shared factor, we must check that the remaining denominator is nonzero at the approach point.
A Reliable Procedure
Evaluate the numerator and denominator separately. If the denominator is nonzero, use the quotient law. If both are zero, continue by factoring.
Write each polynomial as a product of linear factors, keeping its coefficients and signs intact.
Identify shared factors and simplify only where the original quotient is defined.
If the remaining denominator is nonzero at the approach value, substitute into the simplified expression. If it is still zero, the finite-limit argument above does not apply.
Worked Examples
Worked Example: Two Quadratics with a Shared Factor
Evaluate \(\lim_{x\to 2}\frac{x^2-5x+6}{x^2-4}\).
Direct substitution gives \(2^2-5(2)+6=4-10+6=0\) in the numerator, and \(2^2-4=4-4=0\) in the denominator. Factor both quadratics:
For \(x\ne 2\), the quotient is therefore
The remaining denominator at \(x=2\) is \(2+2=4\), which is nonzero. Thus the quotient law applies to the simplified expression:
Check: The factored numerator and denominator both vanish at \(x=2\), as found by direct substitution. For nearby \(x\ne2\), the original quotient equals \((x-3)/(x+2)\), whose value approaches \(-1/4\). Hence the limit is \(-1/4\).
Worked Example: Factoring When the Leading Coefficient Is Not One
Evaluate \(\lim_{x\to -1}\frac{2x^2+5x+3}{x^2+3x+2}\).
Substitution gives \(2(-1)^2+5(-1)+3=2-5+3=0\), and \((-1)^2+3(-1)+2=1-3+2=0\). Factor the numerator by splitting its middle term. Since \(2\cdot 3=6\) and \(2+3=5\),
Factor the denominator by finding two numbers whose product is \(2\) and sum is \(3\):
For \(x\ne -1\), cancel the shared factor \(x+1\), giving \(\frac{2x+3}{x+2}\). The remaining denominator at \(-1\) is \(-1+2=1\), so substitution is valid:
Check: Substituting \(-1\) into the original numerator gives \(0\), and into the original denominator gives \(0\). For nearby \(x\ne-1\), the original quotient equals \((2x+3)/(x+2)\), which approaches \(1/1=1\). The limit is \(1\).
Worked Example: A Remaining Zero in the Denominator
Determine whether \(\lim_{x\to 1}\frac{x^2+x-2}{x^2-2x+1}\) exists as a finite two-sided limit.
Substitution gives \(1+1-2=0\) in the numerator and \(1-2+1=0\) in the denominator. Factor:
For \(x\ne1\), the quotient becomes
The remaining denominator is still zero at \(x=1\), so direct substitution into this expression cannot give a finite limit. To determine the behavior, note that when \(x\) is sufficiently close to \(1\), \(x+2>0\). If \(x<1\), then \(x-1<0\), so the quotient is negative; as \(x\) approaches \(1\) from the left, its magnitude grows without bound. If \(x>1\), then \(x-1>0\), so the quotient is positive and grows without bound as \(x\) approaches \(1\) from the right.
Because the one-sided behaviors disagree, the finite two-sided limit does not exist.
Check: The original numerator and denominator both equal zero at \(1\). For \(x\ne1\), their quotient is exactly \((x+2)/(x-1)\). The opposite signs on the two sides confirm that there is no common finite limit.
A Common Mistake: Treating Zero Over Zero as an Answer
A frequent error is to substitute first and treat the resulting \(0/0\) as if it were a value. For instance, in the first example, one might write
This is incorrect: \(0/0\) is undefined, and substitution has not evaluated the limit. The correction is to factor and compare the expressions for \(x\ne2\):
The resulting limit is \(-1/4\), not \(0\). A related point is that the original quotient need not be defined at the approach value for its limit to exist. The simplification describes nearby inputs, which is what determines the limit.
When the Method Does and Does Not Apply
| After factoring and simplifying | What to conclude |
|---|---|
| The remaining denominator is nonzero at \(a\). | Use polynomial limits and the quotient law to substitute \(a\). |
| The remaining denominator is zero at \(a\). | The finite-limit result does not apply; inspect the nearby behavior before concluding. |
| The original denominator approaches a nonzero value. | Factoring is unnecessary; direct substitution and the quotient law suffice. |
Factoring is useful because it changes an unresolved substitution into a question about the remaining factors. The crucial check is not merely whether a common factor appeared, but whether the denominator left after simplification is nonzero at the approach value.
Check Your Understanding
For each question, factor carefully and keep track of the condition that makes substitution valid.
- Evaluate \(\lim_{x\to3}\frac{x^2-7x+12}{x^2-9}\).
- Evaluate \(\lim_{x\to -2}\frac{x^2+5x+6}{x^2+4x+4}\), or explain why the finite-limit rule does not apply.
- Why does simplifying a shared factor not assert that the original quotient is defined at the approach value?
- After factoring, what should you check before substituting into the simplified quotient?
- If the remaining denominator is zero at the approach value, what additional question must you answer before deciding whether a two-sided limit exists?
