Why These Limits Need an Algebraic Rewrite
When direct substitution into a quotient gives zero in both the numerator and denominator, the quotient is undefined at that input. But that fact alone does not tell us what happens to the function as the input approaches it. A useful next step is to rewrite the expression so that its behavior near the input becomes visible.
In this workshop, the main tools are factoring and multiplication by a conjugate. Factoring is especially useful when a polynomial vanishes at the input. Conjugates are useful when a difference involving square roots creates a zero numerator. In either case, the aim is to obtain an expression equal to the original one at nearby inputs, then evaluate its limit.
A Reliable Sequence of Steps
For a limit of a quotient at \(x=a\), begin by substituting \(a\) into the numerator and denominator separately. If both are zero, look for a common factor or a difference of square roots that can be simplified using a conjugate. After simplifying, check that the new expression agrees with the old one for inputs near \(a\), except possibly at \(a\) itself. Finally, substitute \(a\) into the simplified expression if it is defined and continuous there.
Evaluate the numerator and denominator at the approach value. Confirm whether both are zero.
Factor polynomial expressions, or multiply by a conjugate when a square-root difference is present.
Note where the original and simplified expressions are defined and where any canceled factor is nonzero.
Evaluate the simplified expression at the approach value, then verify the algebra that produced it.
Why Canceling a Factor Preserves a Limit
Suppose the numerator and denominator both contain a factor \(x-a\). The original quotient may not be defined at \(x=a\), but for every \(x\ne a\) where the remaining denominator is nonzero, the factor can be canceled. The two expressions are then equal at all such nearby inputs.
Here is the essential justification. If \(g(x)=(x-a)h(x)\) and \(f(x)=(x-a)k(x)\), then whenever \(x\ne a\) and \(h(x)\ne0\),
Thus cancellation establishes equality on the nearby inputs where the quotient is defined; it does not claim that the original quotient has a value at \(a\). If \(k(x)/h(x)\) is continuous at \(a\), its limit as \(x\to a\) is its value at \(a\). This is why cancellation can reveal the limit.
Worked Examples
Worked Example: Factoring a Difference of Squares
Evaluate \(\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}\).
Direct substitution gives \(3^2-9=0\) in the numerator and \(3-3=0\) in the denominator, so the quotient has the zero-over-zero form. Factor the numerator as a difference of squares:
For \(x\ne3\), cancel the common factor \(x-3\):
The cancellation is valid for nearby inputs other than \(3\). The simplified expression is defined at \(3\), so substitute:
Check: Multiplying the factors back gives \((x-3)(x+3)=x^2-9\), the original numerator. Therefore the rewrite is correct wherever \(x\ne3\), and the limit is \(6\).
Worked Example: Factoring a Quadratic Trinomial
Evaluate \(\displaystyle\lim_{x\to2}\frac{x^2+x-6}{x-2}\).
Substitution gives \(2^2+2-6=0\) and \(2-2=0\). To factor the trinomial, find two numbers whose product is \(-6\) and whose sum is \(1\). Those numbers are \(3\) and \(-2\), so
For \(x\ne2\), cancel the factor \(x-2\):
Now substitute into the simplified expression:
Check: Expanding \((x+3)(x-2)\) gives \(x^2-2x+3x-6=x^2+x-6\). The factorization is correct, and the canceled expression equals the original quotient for \(x\ne2\). The limit is \(5\).
Worked Example: Conjugating a Square-Root Difference
Evaluate \(\displaystyle\lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}\).
At \(x=4\), the numerator is \(\sqrt{9}-3=0\), and the denominator is \(4-4=0\). The conjugate of \(\sqrt{x+5}-3\) is \(\sqrt{x+5}+3\). Multiply numerator and denominator by this conjugate, which is equivalent to multiplying by \(1\) wherever it is defined and nonzero:
Use \((u-v)(u+v)=u^2-v^2\) in the numerator:
For \(x\ne4\), cancel \(x-4\). The resulting expression is defined near \(4\), including at \(4\):
Substitute \(x=4\):
Check: The difference of conjugates is \((\sqrt{x+5})^2-3^2=x+5-9=x-4\), exactly the factor needed to cancel the denominator. The limit is \(\frac16\).
Worked Example: Conjugating with a Linear Inner Expression
Evaluate \(\displaystyle\lim_{x\to1}\frac{\sqrt{2x+7}-3}{x-1}\).
Direct substitution gives \(\sqrt{2(1)+7}-3=\sqrt9-3=0\), while \(1-1=0\). Multiply by the conjugate of the numerator over itself:
Simplify the numerator carefully: \((2x+7)-9=2x-2=2(x-1)\). Therefore, for \(x\ne1\),
Substitute \(x=1\) into this simplified expression:
Check: The conjugate product in the numerator is \((2x+7)-9=2x-2\), which factors as \(2(x-1)\). This confirms the cancellation and the limit \(\frac13\).
Choosing the Useful Rewrite
Factoring and conjugate multiplication have the same broad purpose: expose a factor that vanishes at the approach value and then remove it on the punctured neighborhood. The visible form of the expression suggests which tool to try.
| Expression feature | Useful first move | What to check |
|---|---|---|
| A polynomial numerator that is zero at the approach value | Factor the polynomial | Confirm that the factorization expands back to the original polynomial. |
| A difference involving a square root | Multiply by the conjugate | Use the difference-of-squares identity and simplify every resulting term. |
| A common factor in numerator and denominator | Cancel it for nearby inputs where it is nonzero | Do not claim cancellation gives a value at the excluded input. |
These are algebraic methods, not automatic recipes. After a rewrite, check whether the remaining denominator is nonzero at the approach value. If it is still zero, the expression has not yet been simplified enough for direct substitution; another valid algebraic step may be needed. If the remaining denominator is nonzero and the expression is built from continuous operations near the input, substitution evaluates its limit.
A Common Cancellation Mistake
A frequent error is to cancel terms that are added or subtracted, rather than common factors. For example, this calculation is wrong:
The numerator \(x^2-9\) is a difference, not a product with a visible factor \(x-3\). Cancellation applies to common factors of an entire numerator and denominator, not to selected terms. The correct calculation factors the whole numerator first:
Substituting \(x=3\) into the correct simplified expression gives \(6\), not \(3\). The factorization step is what makes the cancellation legitimate.
Takeaway
When direct substitution gives zero in both parts of a quotient, do not treat that form as an answer. Factor polynomial expressions or use a conjugate to find a valid simplification. Keep track of where cancellation is allowed, verify the algebra, and only then substitute into an expression defined at the approach value.
Check Your Understanding
For each question, identify the algebraic step that makes direct substitution possible and keep track of any input excluded during cancellation.
- Evaluate \(\displaystyle\lim_{x\to5}\frac{x^2-25}{x-5}\). What factor is canceled?
- Evaluate \(\displaystyle\lim_{x\to-1}\frac{x^2+4x+3}{x+1}\). Show the factorization before canceling.
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}\). Which conjugate do you use?
- In one sentence, explain why canceling a factor \(x-a\) can help evaluate a limit even though the original quotient is undefined at \(x=a\).
- What is wrong with canceling terms in \(\frac{x^2-9}{x-3}\) before factoring the numerator?
