Why Check Equivalent Expressions?
A limit calculation often becomes easier after rewriting an expression. Factoring may reveal a common factor to cancel; multiplying by a conjugate may remove a radical difference. The key question is not whether the original and rewritten expressions have the same value at the target. They may not even be defined there. The important question is whether they agree at all sufficiently nearby inputs other than the target.
That distinction matters because a limit describes behavior arbitrarily close to a point, not necessarily the value at the point itself. A valid rewrite must preserve the expression on a whole punctured neighborhood of the target. In particular, canceling a factor is justified only where that factor is nonzero.
Equality Near a Target
A punctured neighborhood of \(a\) is an interval around \(a\) with the point \(a\) removed. Its radius \(\delta>0\) specifies the inputs satisfying \(0<|x-a|<\delta\). The strict inequality excludes the target itself while including points arbitrarily close to it on either side.
This definition captures exactly the type of equality needed for limit calculations. It is stronger than checking a few nearby values: the expressions must agree at every input in some entire punctured neighborhood.
Why Equivalent Expressions Have the Same Limit
The reason follows directly from the definition of a limit. A function \(f\) has limit \(L\) as \(x\) approaches \(a\) if, for every \(\varepsilon>0\), there is an \(\eta>0\) such that \(0<|x-a|<\eta\) implies \(|f(x)-L|<\varepsilon\).
Suppose \(f\) and \(g\) are equivalent near \(a\), with equality holding whenever \(0<|x-a|<\delta\), and suppose \(\lim_{x\to a}f(x)=L\). Given any \(\varepsilon>0\), the limit definition supplies an \(\eta>0\) such that \(|f(x)-L|<\varepsilon\) whenever \(0<|x-a|<\eta\). Choose the smaller radius \(\min(\delta,\eta)\). For every \(x\) in that punctured neighborhood, \(g(x)=f(x)\), so
Therefore \(\lim_{x\to a}g(x)=L\). The same argument with \(f\) and \(g\) interchanged proves the converse. The proof uses only equality near the target and the definition of a limit; it does not require the functions to agree at \(a\).
Worked Example: Cancel a Factor and Check Its Domain
Worked Example: A Removable Denominator Factor
Evaluate \(\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}\), and justify the cancellation.
At \(x=3\), the original denominator is \(3-3=0\), so direct substitution into the original quotient is not available. Factor the numerator:
For \(x\ne 3\), the denominator factor \(x-3\) is nonzero, and cancellation gives
This equality holds for every \(x\) with \(0<|x-3|<1\): in that punctured neighborhood \(x\ne3\), so the canceled factor is nonzero. The simplified expression has limit
Thus the original quotient also has limit \(6\). Check the rewrite by reversing the cancellation: for \(x\ne3\), \((x-3)(x+3)=x^2-9\), so dividing by the nonzero \(x-3\) recovers the original quotient. The simplified expression approaches \(6\), as claimed, although the original quotient is undefined at \(x=3\).
What Must Be Checked Before Canceling?
If an expression has the form \(\frac{A(x)B(x)}{C(x)B(x)}\), canceling \(B(x)\) gives \(\frac{A(x)}{C(x)}\) only at inputs where \(B(x)\ne0\) and the original quotient is defined. To use the new expression for a limit at \(a\), you must establish a radius \(\delta>0\) such that these conditions hold whenever \(0<|x-a|<\delta\).
The canceled factor may equal zero at the target itself; that is often precisely why the original expression is undefined there. But it must not vanish at other inputs arbitrarily close to the target. If it does, the proposed cancellation does not establish equivalence on a punctured neighborhood. Also check the remaining denominator: the simplified quotient must be defined on the neighborhood where it is used.
In straightforward examples, the check can be explicit. For instance, if the canceled factor is \(x-a\), it is nonzero whenever \(0<|x-a|<\delta\), for any \(\delta>0\). If other factors are involved, identify their zeros or establish a suitable neighborhood in which they do not vanish.
Worked Example: Check All Denominator Factors
Worked Example: A Rational Expression with a Remaining Denominator
Evaluate \(\displaystyle\lim_{x\to1}\frac{x^2-1}{(x-1)(x+2)}\), verifying the canceled factor and the remaining denominator.
Factor the numerator using \(x^2-1=(x-1)(x+1)\). Whenever \(x\ne1\) and \(x\ne-2\), the expression becomes
Choose the neighborhood radius \(\delta=1\). If \(0<|x-1|<1\), then \(0<x<2\). In particular, \(x-1\ne0\), so cancellation is valid, and \(x+2>2\), so the remaining denominator is also nonzero. The two expressions are therefore equal and defined throughout this punctured neighborhood.
Now evaluate the limit of the simplified quotient. Its denominator tends to \(1+2=3\), which is nonzero, so the quotient limit law applies:
Check the algebra by multiplying the simplified quotient by the canceled factor in the original denominator: \((x-1)(x+1)=x^2-1\), and the remaining factor is \(x+2\). Hence the original fraction is recovered wherever the stated denominators are nonzero. The original limit is \(\frac{2}{3}\).
Rewrites That Do Not Use Cancellation
Equivalent expressions can also come from multiplying by a quantity rather than canceling a common factor. For radical differences, multiplying numerator and denominator by a conjugate can produce a simpler quotient. This step still requires a domain check: the multiplier used as a denominator must be nonzero in the neighborhood under consideration, and the original and rewritten expressions must be defined there.
Worked Example: Rationalize and Verify the Rewrite
Evaluate \(\displaystyle\lim_{x\to1}\frac{\sqrt{x+3}-2}{x-1}\), checking the rewritten expression near \(x=1\).
The original expression is defined for \(x\ge -3\), except at \(x=1\), where its denominator is zero. Multiply numerator and denominator by the conjugate \(\sqrt{x+3}+2\):
For \(x\ne1\), cancel \(x-1\). The resulting expression is
Take \(\delta=1\). If \(0<|x-1|<1\), then \(0<x<2\), so \(x+3>0\) and the square root is defined. Also \(\sqrt{x+3}+2>2\), so the conjugate factor in the denominator is nonzero. Since \(x\ne1\), the factor canceled in the last step is nonzero as well. Thus each rewrite is valid throughout this punctured neighborhood.
The simplified expression is continuous at \(x=1\), with nonzero denominator there, so substitution gives
For a final algebra check, multiplying the rationalized numerator and denominator back gives \((x-1)(\sqrt{x+3}+2)\) in the denominator and \((\sqrt{x+3}-2)(\sqrt{x+3}+2)=x-1\) in the numerator. For \(x\ne1\), canceling that common \(x-1\) recovers the original quotient. The limit is \(\frac14\).
A Common Mistake: Treating Cancellation as Equality Everywhere
A common incorrect argument is to write
and then claim that the original function has value \(6\) at \(x=3\). This is false: at \(x=3\), the left side is undefined because its denominator is zero. Cancellation does not define the original function at that point.
The correction is to state the restriction precisely:
This is enough for the limit because it gives equality at every point in a punctured neighborhood of \(3\). The limit is \(6\), but that does not assert that the original function has value \(6\) at \(3\). A limit and a function value are different claims.
A Reliable Limit Check
Record the original and simplified expressions, including any canceled factors.
List the factors that must be nonzero and any restrictions required for the expressions to be defined.
Choose a positive radius for which all the required nonzero and domain conditions hold whenever \(0<|x-a|<\delta\).
Use the established limit laws or direct substitution when the simplified expression is defined and continuous at the target.
Confirm that reversing the algebra recovers the original expression on the punctured neighborhood, then report the limit without assigning an unsupported value at the target.
Why the Check Matters
Algebraic simplification is useful because it exposes behavior that may be hidden by an undefined expression at the target. But a correct-looking simplified formula is not, by itself, a justification. The punctured-neighborhood check links the algebra to the limit: it shows that the simpler expression describes the same nearby values that determine the original limit.
Keep the distinction clear. To cancel a factor, show it is nonzero wherever cancellation is used. To compute the limit from the simplified expression, make sure that expression is defined sufficiently near the target and that the limit evaluation is valid. The value at the target is a separate matter.
Check Your Understanding
For each question, identify the neighborhood or nonzero-factor check needed to justify the conclusion.
- In what sense can two expressions be equivalent near a target even if one is undefined at the target?
- Why does equality on a punctured neighborhood imply that two expressions have the same limit?
- For \(\frac{x^2-16}{x-4}\) as \(x\to4\), which factor is canceled, and where is that cancellation valid?
- For \(\frac{x^2-1}{(x-1)(x+2)}\) as \(x\to1\), why must you check both \(x-1\) and \(x+2\)?
- Does a limit of \(6\) imply that the original function is defined with value \(6\) at the target? Explain.
