Why a Known Root Helps with a Limit
A cubic polynomial can be difficult to factor by inspection. Knowing one of its roots changes the task: the corresponding linear factor is known, and polynomial division reduces the cubic to a quadratic. In a limit problem, that linear factor may match a factor in the denominator, revealing a simpler expression to evaluate.
The key is to check the result of cancellation carefully. A shared factor may not be the only factor that vanishes at the limit point. After simplifying, direct substitution is appropriate only when the remaining denominator is nonzero there.
From a Root to a Factor
The factor theorem follows from polynomial division. Dividing a polynomial \(P(x)\) by \(x-r\) gives a quotient \(Q(x)\) and a constant remainder \(R\), so \(P(x)=(x-r)Q(x)+R\). Setting \(x=r\) makes the first term zero, giving \(P(r)=R\). Thus, when \(r\) is a root, \(P(r)=0\), so the remainder is zero and \(P(x)=(x-r)Q(x)\). Conversely, if \(P(x)=(x-r)Q(x)\), then substituting \(x=r\) gives \(P(r)=0\).
For a cubic \(P(x)=ax^3+bx^2+cx+d\), where \(a\ne 0\), a known root \(r\) lets us divide by \(x-r\). Synthetic division organizes the calculation. The final number in the synthetic-division row is the remainder; it must equal \(P(r)\).
When \(r\) is a root, the remainder is zero. The quotient is then \(ax^2+(b+ar)x+(c+br+ar^2)\). Multiplying \(x-r\) by this quotient is a useful check: the product should reproduce the original cubic.
Worked Example: A Cubic Factor in a Limit
Worked Example: A Cubic Factor in a Limit
Evaluate \(\displaystyle \lim_{x\to 2}\frac{x^3+x^2-5x-2}{x^2-4}\).
First check the numerator at the limit point:
Therefore, \(2\) is a root of the cubic, and \(x-2\) is a factor. Synthetic division gives:
The quotient is \(x^2+3x+1\). Check by multiplication:
Factor the denominator as well:
For \(x\ne 2\), and also \(x\ne -2\) so the original denominator is nonzero, the expression simplifies to:
The remaining denominator at \(x=2\) is \(2+2=4\), which is nonzero. Direct substitution into the simplified expression is valid:
Check the arithmetic in the numerator: \(4+6+1=11\), and the denominator is \(4\). The limit is \(\boxed{\frac{11}{4}}\).
Using the Factor Theorem Reliably
A proposed root should be checked by substitution before it is used. If the value does not make the polynomial zero, then the corresponding linear expression is not a factor, and synthetic division will leave a nonzero remainder. Once the root is confirmed, division gives a quadratic quotient. That quotient can be left in polynomial form or factored further if its factors help match the denominator.
In a limit problem, factor the numerator and denominator separately. Look for the factor associated with the limit point, then simplify the quotient for nearby \(x\)-values where the original denominator is nonzero. Before substituting into the simplified expression, evaluate its remaining denominator at the limit point.
Substitute the limit point \(r\) into the cubic. A zero value confirms that \(x-r\) is a factor.
Use synthetic division or polynomial division to find the quadratic quotient, and check the factorization by multiplication.
Identify whether it contains the same linear factor and whether it contains any additional factors that vanish at \(r\).
Cancel shared factors only for values where the original denominator is nonzero. Then check the remaining denominator at the limit point before using direct substitution.
Worked Example: A Negative Known Root
Worked Example: A Negative Known Root
Evaluate \(\displaystyle \lim_{x\to -1}\frac{x^3-x^2+x+3}{x^2-1}\).
Check whether \(-1\) is a root of the numerator:
So \(x+1\) is a factor. Synthetic division by \(x-(-1)=x+1\) gives:
The quotient is \(x^2-2x+3\). Verify the factorization:
Factor the denominator:
For \(x\ne -1\) and \(x\ne 1\), cancel the shared factor \(x+1\). The remaining denominator at \(-1\) is \(-1-1=-2\), so it is nonzero:
Multiplying the proposed value by the denominator checks the final quotient: \((-3)(-2)=6\), the numerator value. Thus the limit is \(\boxed{-3}\).
When One Cancellation Is Not Enough
Canceling a shared factor does not guarantee that the simplified denominator is nonzero at the limit point. If another factor also vanishes there, direct substitution still cannot evaluate the limit. The expression needs further analysis: perhaps another common factor can be canceled, or perhaps the limit does not exist as a finite number.
Worked Example: Checking for Another Vanishing Factor
Consider \(\displaystyle \lim_{x\to 1}\frac{x^3-3x+2}{(x-1)^2}\). The numerator vanishes at \(x=1\):
Divide by \(x-1\): synthetic division gives coefficients \(1,1,-2\) and remainder \(0\), so the quotient is \(x^2+x-2\). Factor the quadratic:
Hence \(x^3-3x+2=(x-1)^2(x+2)\). Check by expansion:
The original denominator is \((x-1)^2\), so after simplifying for \(x\ne 1\), the expression is \(x+2\). There is no remaining denominator, and substitution gives:
Check by substituting \(x=1\) into the simplified expression: it gives \(3\). The limit is \(\boxed{3}\).
A Common Pitfall: Substituting Too Soon After Cancellation
Consider the expression \(\displaystyle \frac{(x-1)(x^2+1)}{(x-1)^2}\) as \(x\) approaches \(1\). Canceling one factor gives \(\displaystyle \frac{x^2+1}{x-1}\), but the remaining denominator is still zero at \(x=1\). The incorrect calculation is:
Division by zero does not give a finite value, so this is not a valid substitution. In fact, when \(x\ne 1\), the simplified expression has numerator \(x^2+1>0\). As \(x\) approaches \(1\) from the right, \(x-1\) is positive and approaches zero, so the quotient grows without bound. From the left, \(x-1\) is negative and approaches zero, so the quotient decreases without bound. The two-sided limit is not a finite number.
Why This Method Matters
Factoring a cubic with a known root turns a higher-degree expression into a product that can be compared directly with the denominator. The factor theorem explains why the root supplies a linear factor, while division supplies the rest of the factorization. In a limit, this algebra can uncover a simplified expression whose value is accessible through direct substitution.
The method has clear limits. A root must actually be known or found; a guessed value must be verified. Also, cancellation only simplifies the expression where the original denominator is nonzero. The value at the limit point must be assessed from the simplified expression, and a remaining zero denominator requires additional work rather than automatic substitution.
Check Your Understanding
For each question, show enough work to verify the factorization and the limit conclusion.
- Verify that \(x=1\) is a root of \(x^3-2x^2-5x+6\), then use synthetic division to find the quadratic quotient.
- Evaluate \(\displaystyle \lim_{x\to 2}\frac{x^3-4x^2+x+6}{x^2-4}\), or explain why direct substitution after cancellation needs further work.
- What remainder does synthetic division produce when dividing \(P(x)\) by \(x-r\), and what does that remainder equal?
- After canceling a shared factor in a limit expression, what must be checked before using direct substitution?
- Explain why finding one shared factor does not guarantee that the simplified denominator is nonzero at the limit point.
