Cubic Factorization in Limit Evaluations

Why Cubic Factorization Helps

A polynomial quotient can be undefined at the point where its limit is being evaluated. If direct substitution gives zero in both the numerator and denominator, the quotient law does not apply there: its denominator would have limit zero. But that substitution alone does not tell us whether a limit exists or what its value is.

Factoring can reveal that the numerator and denominator share a factor that is zero at the point. The original quotient may then agree, at nearby points, with a simpler expression whose limit can be found by substitution. For cubic numerators, the sum-of-cubes and difference-of-cubes identities often provide the needed factor.

Key idea: If a cubic numerator and a denominator share a factor that vanishes at the limit point, cancel that factor only for nearby points where it is nonzero. Then evaluate the limit of the simplified expression, provided its denominator has a nonzero limit.

The Two Cubic Identities

For any real expressions \(a\) and \(b\), the difference of cubes and the sum of cubes factor as follows:

$$ a^3-b^3=(a-b)(a^2+ab+b^2) $$

$$ a^3+b^3=(a+b)(a^2-ab+b^2) $$

The factors have different signs in the middle term of the quadratic factor. In the difference-of-cubes identity, the middle term is \(+ab\); in the sum-of-cubes identity, it is \(-ab\). Keeping track of this distinction is essential.

These identities can be checked by multiplying out. For the difference, distribute each term of the first factor:

\begin{aligned} (a-b)(a^2+ab+b^2) &=a^3+a^2b+ab^2-a^2b-ab^2-b^3\\ &=a^3-b^3. \end{aligned}

The middle terms cancel. For the sum, multiplication gives:

\begin{aligned} (a+b)(a^2-ab+b^2) &=a^3-a^2b+ab^2+a^2b+ab^2+b^3\\ &=a^3+2ab^2+b^3. \end{aligned}

This line does not simplify to \(a^3+b^3\), so the expansion as written signals an error in the multiplication: the term \(b(-ab)\) is \(-ab^2\), not \(+ab^2\). With the correct distribution,

\begin{aligned} (a+b)(a^2-ab+b^2) &=a^3-a^2b+ab^2+a^2b-ab^2+b^3\\ &=a^3+b^3. \end{aligned}

Thus both identities are verified by expansion. Substituting a variable and a constant gives particularly useful forms. For a real number \(c\),

$$ x^3-c^3=(x-c)(x^2+cx+c^2) $$

and

$$ x^3+c^3=(x+c)(x^2-cx+c^2). $$

These forms expose the linear factor that becomes zero at \(x=c\) or \(x=-c\), respectively.

How Cancellation Gives a Limit

Suppose a quotient has the form \(\frac{(x-c)P(x)}{(x-c)Q(x)}\), where \(P\) and \(Q\) are polynomials. The original quotient is defined only at points where its denominator is nonzero. In particular, the factor \(x-c\) makes the original expression undefined at \(x=c\). For \(x\ne c\), however, that factor is nonzero and may be canceled:

\frac{(x-c)P(x)}{(x-c)Q(x)} = \frac{P(x)}{Q(x)}.

For the limit at \(c\), equality at \(x=c\) is not required. What matters is that the expressions agree at all sufficiently close points in the original domain, other than possibly \(c\). This is why cancellation can simplify a limit without changing its value.

Condition for evaluating after cancellation: The simplified expression must be defined near \(c\) and its denominator must have a nonzero limit at \(c\). Then the quotient law and polynomial limit laws allow direct substitution into the simplified expression.

This is an application of limit laws established earlier in the material on limits. The key additional step here is the algebraic factorization that makes a common factor visible.

Worked Examples

Worked Example: A Difference of Cubes

Evaluate \(\displaystyle \lim_{x\to 2}\frac{x^3-8}{x-2}\).

Substituting \(x=2\) into the numerator gives \(2^3-8=8-8=0\), and the denominator gives \(2-2=0\). Direct substitution therefore produces the indeterminate form \(0/0\), not the value of the limit.

Since \(8=2^3\), apply the difference-of-cubes identity:

\begin{aligned} x^3-8 &=x^3-2^3\\ &=(x-2)(x^2+2x+4). \end{aligned}

For \(x\ne 2\), cancel the common factor \(x-2\):

\frac{x^3-8}{x-2} = \frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4.

The simplified expression is a polynomial, so its limit is found by substitution:

\begin{aligned} \lim_{x\to 2}(x^2+2x+4) &=2^2+2(2)+4\\ &=4+4+4\\ &=12. \end{aligned}

Therefore, the original limit is \(12\). As a check, multiplying the factors back gives \((x-2)(x^2+2x+4)=x^3-8\), and substituting \(2\) into the simplified expression gives \(12\). The original quotient is still undefined at \(x=2\); the limit describes its behavior as \(x\) approaches \(2\), not its value there.

Worked Example: A Sum of Cubes

Evaluate \(\displaystyle \lim_{x\to -1}\frac{x^3+1}{x+1}\).

At \(x=-1\), the numerator is \((-1)^3+1=-1+1=0\), and the denominator is \(-1+1=0\). Write \(1\) as \(1^3\) and use the sum-of-cubes identity:

\begin{aligned} x^3+1 &=x^3+1^3\\ &=(x+1)(x^2-x+1). \end{aligned}

For \(x\ne -1\), cancel \(x+1\):

\frac{x^3+1}{x+1} = \frac{(x+1)(x^2-x+1)}{x+1} = x^2-x+1.

Now evaluate the limit of the polynomial:

\begin{aligned} \lim_{x\to -1}(x^2-x+1) &=(-1)^2-(-1)+1\\ &=1+1+1\\ &=3. \end{aligned}

Thus the limit is \(3\). To check the factorization, multiplying gives \(x^3-x^2+x+x^2-x+1=x^3+1\). The simplified expression has value \(3\) at \(-1\), so it is defined there even though the original quotient is not.

Worked Example: A Cubic Factor Shared with a Quadratic Denominator

Evaluate \(\displaystyle \lim_{x\to 3}\frac{x^3-27}{x^2-9}\).

At \(x=3\), the numerator is \(3^3-27=27-27=0\), and the denominator is \(3^2-9=9-9=0\). Factor the numerator as a difference of cubes and the denominator as a difference of squares:

\begin{aligned} x^3-27&=(x-3)(x^2+3x+9),\\ x^2-9&=(x-3)(x+3). \end{aligned}

For the original quotient, both \(x=3\) and \(x=-3\) are excluded because they make the denominator zero. At points near \(3\) with \(x\ne 3\), cancel \(x-3\):

\frac{x^3-27}{x^2-9} = \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)} = \frac{x^2+3x+9}{x+3}.

The denominator of the simplified quotient tends to \(3+3=6\), which is nonzero. The quotient law therefore applies to the simplified expression:

\begin{aligned} \lim_{x\to 3}\frac{x^2+3x+9}{x+3} &=\frac{3^2+3(3)+9}{3+3}\\ &=\frac{9+9+9}{6}\\ &=\frac{27}{6}\\ &=\frac{9}{2}. \end{aligned}

The limit is \(\frac{9}{2}\). To check the cancellation, multiplying \((x-3)(x^2+3x+9)\) gives \(x^3-27\), while \((x-3)(x+3)=x^2-9\). The simplified denominator is nonzero at \(3\), as required for the final substitution.

A Reliable Procedure

When a polynomial quotient gives \(0/0\) at the point of interest, use the following sequence. The \(0/0\) result is a signal to investigate the algebra, not a conclusion about the limit.

Identify the zero-producing value.
Substitute the limit point into the numerator and denominator. If both are zero, record that direct substitution does not evaluate the limit.

Match the cubic pattern.
Use \(x^3-c^3=(x-c)(x^2+cx+c^2)\) or \(x^3+c^3=(x+c)(x^2-cx+c^2)\), as appropriate.

Factor the entire quotient.
Factor the denominator as well, and identify factors common to numerator and denominator.

Cancel only on the punctured domain.
Cancel a common factor only at points where it is nonzero. The resulting equality need not hold at the limit point itself.

Check the remaining denominator and evaluate.
If the simplified denominator has a nonzero limit at the point, apply the quotient and polynomial limit laws.

Common Mistake: Reversing the Signs

A frequent error is to use the difference-of-cubes quadratic factor for a sum of cubes. For instance, the incorrect calculation

$$ x^3+1=(x+1)(x^2+x+1) $$

does not pass an expansion check:

$$ (x+1)(x^2+x+1)=x^3+2x^2+2x+1, $$

which is not \(x^3+1\). The correct sum-of-cubes factor has a minus sign in its middle term:

$$ x^3+1=(x+1)(x^2-x+1). $$

Multiplying the corrected factors gives \(x^3-x^2+x+x^2-x+1=x^3+1\). Checking by expansion is a quick way to catch a sign error before it leads to an incorrect cancellation or limit.

When This Method Applies—and When It Does Not

Cubic factorization is useful when a numerator contains a sum or difference of cubes and the resulting linear factor is shared with the denominator. It does not guarantee that every quotient limit can be evaluated by this method. If the denominator has no matching factor, or if the simplified denominator still tends to zero, cancellation has not yet produced an expression to which direct substitution can be applied.

Also, a common factor may be canceled in a quotient only where that factor is nonzero. Cancellation establishes equality near the limit point, not necessarily at the point itself. If no common factor is present, the sum- or difference-of-cubes identity alone does not authorize cancellation.

The method matters because it separates two questions: what value the original formula takes at a point, and how its values behave near that point. A removable hole can prevent the original quotient from being defined at one point while leaving a well-defined limit. Factoring makes this behavior visible.

Takeaway: Factor a cubic sum or difference, verify the factorization, cancel only a common factor that is nonzero away from the limit point, and substitute only after the remaining denominator is nonzero at that point.

Check Your Understanding

For each question, show the factorization and indicate why cancellation is valid near the limit point.

Evaluate \(\displaystyle \lim_{x\to 1}\frac{x^3-1}{x-1}\).
Factor \(x^3+8\) using the sum-of-cubes identity. What is the quadratic factor?
Evaluate \(\displaystyle \lim_{x\to -2}\frac{x^3+8}{x+2}\).
Why does canceling \(x-c\) in a quotient not establish the value of the original function at \(x=c\)?
What must be true of the denominator after cancellation before direct substitution can evaluate the limit?

Cubic Factorization in Limit Evaluations

What You'll Learn

Why Cubic Factorization Helps

The Two Cubic Identities

How Cancellation Gives a Limit