Why the Choice of Rewrite Matters
A limit calculation can begin with direct substitution and still leave the main question unanswered. If substitution gives zero divided by zero, that result does not determine the limit. It signals that the expression must be examined more closely.
Several rewrites may be available, but the expression itself usually suggests which one will help. A difference of polynomial terms often invites factoring. A difference involving a square root often becomes simpler after multiplying by a conjugate. A trigonometric expression such as \(1-\cos u\) may be better handled with an identity that produces a sine quotient.
The goal is not to apply every technique in turn. It is to choose a rewrite that removes the obstruction while keeping the expression equal to the original one near the point of interest.
What Zero over Zero Tells You
Suppose a quotient has the form \(f(x)/g(x)\), and direct substitution at \(x=a\) gives \(f(a)=0\) and \(g(a)=0\). The resulting form \(0/0\) is called an indeterminate form. It is not a value, and it does not by itself say whether the limit exists or what the limit is.
For instance, different quotients can give different limits after substitution produces \(0/0\): \((x-2)/(x-2)\) has limit \(1\) as \(x\) approaches \(2\), while \((x-2)^2/(x-2)\) has limit \(0\). The common substitution result identifies a need for further work, not a particular answer.
Before choosing a technique, look at what is causing each zero. Is there a polynomial difference with a shared factor? Is there a difference of radicals? Does a trigonometric identity turn the expression into a familiar quotient? These questions point toward an efficient rewrite.
A Diagnostic Guide
| Expression structure | Useful first choice | What to look for |
|---|---|---|
| Polynomial numerator and denominator, both zero at \(a\) | Factor | A shared factor that vanishes at \(a\) |
| A subtraction involving a square root | Multiply by the conjugate | A difference of squares after multiplication |
| A trigonometric difference such as \(1-\cos u\) | Use a trigonometric identity | A sine expression that can use the sine quotient limit |
| More than one of these structures | Combine rewrites as needed | Simplify one obstruction at a time |
This guide is a starting point, not an automatic rule. Factoring is useful only if it actually reveals a factor that can be removed from the quotient. A conjugate helps when its product turns a radical difference into an algebraic difference. A trigonometric identity helps when it connects the expression to a limit already known, such as \(\lim_{u\to 0}\sin u/u=1\), with angles measured in radians.
Why Rewriting Away from the Point Is Valid
Factoring and cancellation often produce a new formula that does not look like the original one. The justification is that a limit concerns values arbitrarily close to the input, not necessarily the value at the input itself.
To justify this, let \(\varepsilon>0\). Since \(g(x)\) tends to \(L\), there is a number \(\eta>0\) such that \(0<|x-a|<\eta\) implies \(|g(x)-L|<\varepsilon\). Choose \(r=\min(\eta,\delta)\). Whenever \(0<|x-a|<r\), both \(f(x)=g(x)\) and \(|g(x)-L|<\varepsilon\) hold. Therefore \(|f(x)-L|=|g(x)-L|<\varepsilon\). This is precisely the definition that \(f(x)\) tends to \(L\).
This argument explains why canceling a factor that vanishes at \(a\) can still be legitimate for a limit: the cancellation is performed for nearby inputs with \(x\ne a\). It does not claim the original expression is defined at \(a\).
Worked Example: Factor a Polynomial Difference
Worked Example: Factor a Polynomial Difference
Evaluate \(\lim_{x\to 3}\frac{x^2-9}{x-3}\).
Direct substitution gives \(3^2-9=0\) in the numerator and \(3-3=0\) in the denominator. Because the numerator is a difference of squares, factoring is the natural choice:
For \(x\ne 3\), cancel the common factor \(x-3\):
The original quotient and \(x+3\) agree for all nearby inputs except \(x=3\). Thus the limit is
Check: The factorization expands back to \(x^2-9\), since \((x-3)(x+3)=x^2-9\). Also, for every \(x\ne3\), the original quotient equals \(x+3\). As \(x\) approaches \(3\), this simplified expression approaches \(6\), confirming the result.
Worked Example: Use a Conjugate with a Radical Difference
Worked Example: Use a Conjugate with a Radical Difference
Evaluate \(\lim_{x\to 2}\frac{\sqrt{x+7}-3}{x-2}\).
At \(x=2\), the numerator is \(\sqrt{9}-3=3-3=0\), and the denominator is \(2-2=0\). The numerator is a difference involving a square root. Multiply by the conjugate of that numerator, \(\sqrt{x+7}+3\), in both numerator and denominator:
The numerator simplifies to \(x-2\). For \(x\ne2\), cancel this factor:
The remaining denominator approaches \(\sqrt{9}+3=6\), which is nonzero. Therefore,
Check: The conjugate product in the numerator is \((\sqrt{x+7}-3)(\sqrt{x+7}+3)=(x+7)-9=x-2\), so the simplification is exact. For nearby \(x\ne2\), the canceled factor is nonzero, and the resulting expression tends to \(1/6\).
Worked Example: Use an Identity for a Trigonometric Difference
Worked Example: Use an Identity for a Trigonometric Difference
Evaluate \(\lim_{x\to 0}\frac{1-\cos(4x)}{x^2}\), where angles are measured in radians.
Substitution gives \(1-\cos(0)=0\) and \(0^2=0\). Factoring polynomials does not address the trigonometric numerator. Instead, use the identity \(1-\cos u=2\sin^2(u/2)\), with \(u=4x\):
Rewrite the quotient to use the sine quotient limit:
As \(x\to0\), \(2x\to0\). The standard sine quotient limit gives \(\sin(2x)/(2x)\to1\). Hence,
Check: Since \(2\sin^2(2x)=1-\cos(4x)\) and \(8(\sin(2x)/(2x))^2=2\sin^2(2x)/x^2\) for \(x\ne0\), the rewritten expression is exactly the original quotient near zero. The limit calculation uses the sine quotient limit with the angle in radians, giving \(8\).
When the Best Choice Is a Combination
Expressions do not always fit into just one category. A radical difference may produce a polynomial factor after conjugation, so the calculation uses both conjugation and factoring. In such cases, apply the rewrite that addresses the most immediate obstruction, then inspect what remains.
Worked Example: Combine a Conjugate and Factoring
Evaluate \(\lim_{x\to2}\frac{\sqrt{x^2+5}-3}{x-2}\).
Substitution gives \(\sqrt{2^2+5}-3=\sqrt{9}-3=0\) and \(2-2=0\). First use the conjugate of the numerator:
Now factor the polynomial difference \(x^2-4=(x-2)(x+2)\). For \(x\ne2\), cancel \(x-2\):
Substitute into the simplified expression:
Check: The conjugate product gives \(x^2+5-9=x^2-4\), and \((x-2)(x+2)=x^2-4\). Thus both transformations are exact; the canceled factor is nonzero for nearby \(x\ne2\). The simplified quotient tends to \(2/3\), verifying the limit.
A Common Mistake: Canceling Before Rewriting
In the radical example, it would be incorrect to claim that \(\sqrt{x+7}-3\) cancels with \(x-2\). They are not the same factor, so the following is not a valid algebraic step:
The correction is to multiply by the conjugate and show the product explicitly. Only then does the numerator become \(x-2\), a factor that matches the denominator. Likewise, obtaining \(0/0\) by substitution never justifies assigning a value to the quotient; it is a signal to seek a valid rewrite.
A Practical Selection Process
When a limit produces zero over zero, use a short diagnostic sequence rather than guessing:
Check whether numerator and denominator both approach zero. If they do, do not treat \(0/0\) as the limit.
Look for polynomial factors, a difference involving radicals, or a trigonometric form that matches an identity.
Factor, multiply by a conjugate, or use an identity. Cancel only a factor that is common and nonzero for nearby inputs away from the limit point.
Use limit laws or a standard limit only after the expression has been simplified enough for those laws to apply.
The method matters because a well-chosen rewrite reveals the relationship hidden by the original form. Factoring exposes polynomial structure, conjugates convert radical differences into algebraic ones, and trigonometric identities connect expressions to known trigonometric limits. A good calculation also makes clear why each new expression represents the original one near the point.
Check Your Understanding
For each question, identify the useful first step and explain why it applies.
- For \(\lim_{x\to4}\frac{x^2-16}{x-4}\), what feature suggests a particular rewrite?
- For \(\lim_{x\to1}\frac{\sqrt{x+8}-3}{x-1}\), which expression should be used as the conjugate?
- Why does a result of zero over zero after substitution not determine a limit?
- For \(\lim_{x\to0}\frac{1-\cos(6x)}{x^2}\), which trigonometric identity can connect the expression to the sine quotient limit?
- When a factor is canceled even though it is zero at the limit point, what condition makes that cancellation valid for evaluating the limit?
