Factoring Out a Shared Linear Factor

Why a Shared Factor Helps

When direct substitution into a quotient gives \(0/0\), the result does not determine the limit. It signals that substitution alone is inconclusive. Sometimes the numerator and denominator both contain the same linear factor, \(x-a\). Factoring can reveal that shared structure and reduce the limit to one that can be evaluated by substitution.

The key is to factor the whole numerator and the whole denominator. A factor is something multiplied by another expression; it is not an individual term inside a sum. Once the shared factor is exposed, the quotient can be simplified for values of \(x\) near \(a\), with \(x\ne a\).

Factoring by Grouping

One way to expose \(x-a\) is to group terms so that each group contains that factor. For instance, in \(x(x-2)+4(x-2)\), both terms contain \(x-2\). Factoring it out gives \((x-2)(x+4)\). The same idea works in the denominator.

Before factoring, check the direct-substitution result. If substituting \(x=a\) makes both numerator and denominator zero, a shared factor \(x-a\) may be present. This is a useful clue, not a guarantee that this particular method will finish the problem: after factoring and simplifying, the remaining denominator must be nonzero at \(a\) for direct substitution to apply.

Worked Example: Grouping to Reveal a Shared Factor

Why the Simplification Works

The cancellation is an algebraic identity for nearby inputs, not a claim that the original quotient is defined at \(x=a\). If \(x\ne a\), then \(x-a\ne 0\), so the common factor may be divided out. Provided the remaining denominator does not vanish at \(a\), the simplified quotient has a value that can be approached by substitution.

Here is the reasoning in steps. The factorization makes the shared factor explicit. For all nearby \(x\ne a\), the original quotient equals the simplified one. The simplified quotient is a rational function whose denominator is nonzero at \(a\), so its limit follows from direct substitution and the quotient limit law.

Worked Example: A Limit at a Negative Number

Worked Example: Checking the Remaining Denominator

A Common Mistake: Cancelling Terms Instead of Factors

Cancellation applies to factors multiplying an entire expression, not to matching-looking terms in sums. For example, it is not valid to cross out the \(x\) in \(x^2+2x-8\) against the \(x\) in \(x^2+x-6\). Those \(x\)'s are terms within sums, not common factors of the complete numerator and denominator.

A concrete wrong calculation would be to cancel the visible \(x\)'s term by term and claim that the limit is \(\frac{2-8}{1-6}=\frac{6}{5}\) at \(x=2\). The expression on that line is not obtained by a valid algebraic cancellation. In this particular example, the correct method is to factor the full expressions first; it gives \(\frac{(x-2)(x+4)}{(x-2)(x+3)}\), and then the limit is \(\frac{6}{5}\). A coincidentally correct number would not make an invalid cancellation valid.

When This Method Is Not Enough

After cancelling \(x-a\), always inspect the remaining denominator. If it is still zero at \(a\), the quotient limit law cannot yet be applied by direct substitution. The factorization may have exposed only part of the structure, and more algebra may be needed. If no shared factor is exposed, this method alone does not determine the limit.

Also distinguish a limit from the function's value at the target. The original quotient may be undefined at \(x=a\) because its denominator is zero there. Cancelling a factor helps evaluate what the quotient approaches nearby; it does not assign a value to the original expression at \(a\).