Why Factoring Helps with Limits
A limit can present an obstacle when direct substitution gives \(0/0\). That result does not tell us the limit; it tells us that substitution alone has not evaluated it. When the expressions are polynomials, factoring may reveal a linear factor shared by the numerator and denominator. Simplifying then leaves an expression whose limit can be evaluated by substitution.
The important work is not merely spotting a factor to cancel. You must factor each quadratic correctly, identify a factor that occurs in both expressions, and check that the simplified denominator is nonzero at the target value.
Factoring Quadratics
A quadratic expression has the form \(ax^2+bx+c\), where \(a\ne 0\). In the examples here, the coefficients allow it to be written as a product of linear factors. For a monic quadratic \(x^2+bx+c\), look for two numbers whose product is \(c\) and whose sum is \(b\). For a quadratic with a leading coefficient other than \(1\), splitting the middle term can make grouping possible.
For instance, to factor \(x^2-5x+6\), find two numbers whose product is \(6\) and whose sum is \(-5\). Those numbers are \(-2\) and \(-3\), so the factorization is \((x-2)(x-3)\). Multiplying these factors back gives \(x^2-5x+6\), confirming the result.
For \(ax^2+bx+c\), the splitting method looks for numbers \(m\) and \(n\) such that \(m+n=b\) and \(mn=ac\). Replace \(bx\) with \(mx+nx\), then group terms. This method works when the resulting groups can be factored to reveal a common binomial.
Why Splitting the Middle Term Works
The method follows from expanding a product of two linear factors. If \(ax^2+bx+c=(px+q)(rx+s)\), then expansion gives
Thus \(a=pr\), \(b=ps+qr\), and \(c=qs\). The product \(ac\) is
The two terms \(ps\) and \(qr\) therefore have product \(ac\), while their sum is the middle coefficient \(b\). This explains why the splitting method searches for two numbers with product \(ac\) and sum \(b\). Once they are found, regrouping reverses the expansion.
Worked Example: Factoring a Monic Quadratic in a Limit
Evaluate \(\displaystyle\lim_{x\to 2}\frac{x^2-5x+6}{x-2}\).
First substitute \(x=2\) into the numerator: \(2^2-5(2)+6=4-10+6=0\). The denominator is \(2-2=0\), so direct substitution gives \(0/0\).
Factor the numerator. We need two numbers whose product is \(6\) and sum is \(-5\): \(-2\) and \(-3\). Hence
Check by multiplying: \((x-2)(x-3)=x^2-3x-2x+6=x^2-5x+6\). For \(x\ne 2\), the expression simplifies to
The simplified expression is defined at \(x=2\), so its limit there is \(2-3=-1\). Therefore the limit is \(-1\). The check is consistent: the simplified expression approaches \(-1\) as \(x\) approaches \(2\), even though the original quotient is undefined at \(x=2\).
Factoring When the Leading Coefficient Is Not One
When the coefficient of \(x^2\) is not \(1\), check the product \(ac\) as well as the middle coefficient. Splitting the middle term is useful only if the chosen terms can be grouped to produce a common binomial.
Worked Example: Splitting the Middle Term
Evaluate \(\displaystyle\lim_{x\to 3}\frac{2x^2-7x+3}{x-3}\).
At \(x=3\), the numerator is \(2(3^2)-7(3)+3=18-21+3=0\), and the denominator is \(3-3=0\). To factor the numerator, compute \(ac=2\cdot3=6\). The numbers \(-6\) and \(-1\) have product \(6\) and sum \(-7\). Split the middle term and group:
Check by multiplication: \((2x-1)(x-3)=2x^2-6x-x+3=2x^2-7x+3\). For \(x\ne3\), cancel the shared factor \(x-3\), leaving \(2x-1\). Substitution into that expression gives \(2(3)-1=6-1=5\). Thus the limit is \(5\).
The remaining denominator after cancellation is \(1\), which is nonzero at \(x=3\); equivalently, the simplified expression is a polynomial and can be evaluated there.
Factoring Both Quadratics
Sometimes both the numerator and denominator are quadratic. Factor each one separately, then compare the factors. A shared factor is useful only if it is genuinely the same linear expression in both.
Worked Example: Finding a Shared Factor in Two Quadratics
Evaluate \(\displaystyle\lim_{x\to 3}\frac{x^2-4x+3}{x^2-5x+6}\).
Substitution gives a numerator \(3^2-4(3)+3=9-12+3=0\), and a denominator \(3^2-5(3)+6=9-15+6=0\). Factor the numerator using the numbers \(-1\) and \(-3\), whose product is \(3\) and sum is \(-4\). Factor the denominator using \(-2\) and \(-3\), whose product is \(6\) and sum is \(-5\):
Check the products:
Both expressions contain \(x-3\). For \(x\ne3\), cancel that factor:
At the target, the remaining denominator is \(3-2=1\), which is nonzero. The limit is therefore \(\frac{3-1}{3-2}=\frac{2}{1}=2\). Substitution into the simplified quotient confirms the value.
Conditions and Careful Cancellation
The cancellation in these examples is performed for \(x\) values near the target, but not equal to it. At the target itself, the original denominator is zero, so the original quotient is not defined. The simplified expression is used to evaluate the limit, not to claim that the original expression has acquired a value at the missing point.
After cancellation, check the denominator that remains. If it is nonzero at the target, direct substitution evaluates the limit using the limit laws for polynomials and quotients. If the remaining denominator is still zero, this procedure has not yet produced a value by direct substitution; more analysis would be needed.
A Common Mistake: Cancelling Terms
Cancellation applies to common factors in a product, not to matching-looking pieces inside sums or differences. For example, it is incorrect to cancel \(x\) in \(x^2-5x+6\) against \(x-2\) term by term. The numerator is a sum of three terms, not a product containing a visible factor \(x\).
A concrete incorrect calculation would be
That incorrect expression would give \(2-5+6=3\) at \(x=2\), not the correct limit \(-1\). The correction is to factor the entire numerator first:
Now the limit is \(2-3=-1\). The multiplication check \((x-2)(x-3)=x^2-5x+6\) confirms that the cancelled expression was a factor of the whole numerator.
A Reliable Workflow
Check whether the numerator and denominator both become zero.
Use the sum-and-product method for a monic quadratic, or split the middle term when the leading coefficient is not \(1\).
Multiply them back and confirm that they reproduce the original quadratic.
Cancel only a factor of the entire numerator and denominator, and remember that the original expression remains undefined at a point where its denominator is zero.
If it is nonzero at the target, substitute into the simplified expression.
Factoring is useful because it exposes the structure hidden by expanded polynomials. A shared linear factor explains why a quotient that initially gives \(0/0\) can simplify to an expression with an ordinary, evaluable limit. Accurate factoring and careful checks make the method reliable.
Check Your Understanding
For each question, show the factoring steps and check any factorization by multiplying it back.
- Factor \(x^2-7x+12\). Which two numbers determine its factors?
- Evaluate \(\displaystyle\lim_{x\to 2}\frac{x^2-3x+2}{x-2}\).
- Factor \(3x^2-8x+4\) by splitting the middle term.
- Evaluate \(\displaystyle\lim_{x\to 1}\frac{x^2-1}{x^2-3x+2}\), and state whether the remaining denominator is nonzero at the target.
- Why is it invalid to cancel a term from a sum? Give an example of a common factor that can be cancelled correctly.
