Why a Difference of Squares Helps
A quotient can be undefined at the point we approach even when its values nearby follow a simple pattern. A common source of this difficulty is a difference of squares: an expression such as \(x^2-25\) becomes zero at \(x=5\). If the denominator also becomes zero there, direct substitution produces \(0/0\), which is not a value for the limit.
Factoring can reveal a factor shared by the numerator and denominator. We can cancel that factor for inputs near the point, as long as we do not claim the cancellation is valid at the point itself. The remaining expression may then have a well-defined limit that can be evaluated using limit laws.
Cancellation Near the Point
Suppose a quotient contains a factor \(x-a\) in both numerator and denominator, and we are taking a limit as \(x\to a\). At \(x=a\), the factor is zero, so the original quotient may be undefined. But for every \(x\ne a\), the factor is nonzero and can be canceled. Limits concern values arbitrarily close to \(a\), not necessarily the value at \(a\).
More precisely, if two expressions are equal for every \(x\) in some punctured interval \(0<|x-a|<\delta\), then they have the same limit as \(x\to a\), provided the limit exists. The point \(a\) is excluded from that interval. This is why canceling a common factor can simplify a limit without changing it, even when the original expression has a hole at \(a\).
After cancellation, evaluate the remaining expression at \(a\) only if it is defined there and the relevant limit laws apply. In particular, the quotient law requires the denominator of the simplified quotient to have a nonzero limit.
Worked Examples
Worked Example: A Basic Difference of Squares
Evaluate \(\displaystyle\lim_{x\to 5}\frac{x^2-25}{x-5}\).
Direct substitution gives numerator \(5^2-25=25-25=0\) and denominator \(5-5=0\), so it gives the indeterminate form \(0/0\). Factor the numerator using \(u^2-v^2=(u-v)(u+v)\), with \(u=x\) and \(v=5\):
For \(x\ne 5\), the denominator factor \(x-5\) is nonzero, so cancellation is valid:
Thus, for all \(x\) sufficiently close to \(5\) with \(x\ne5\), the original quotient equals \(x+5\). The simplified expression is a polynomial, so its limit follows by direct substitution:
Check: At \(x=4\), the original quotient is \((16-25)/(4-5)=(-9)/(-1)=9\). The simplified expression gives \(4+5=9\), confirming the equality at this nearby input. The original quotient is still undefined at \(x=5\); the limit is \(10\), not a value assigned to the original quotient there.
Worked Example: Factoring a Scaled Difference of Squares
Evaluate \(\displaystyle\lim_{x\to \frac{2}{3}}\frac{9x^2-4}{3x-2}\).
At \(x=\frac23\), the numerator is \(9\left(\frac23\right)^2-4=9\cdot\frac49-4=4-4=0\), and the denominator is \(3\cdot\frac23-2=2-2=0\). Rewrite the numerator as a difference of squares:
For \(x\ne\frac23\), we have \(3x-2\ne0\), so we can cancel that factor:
The simplified expression is defined and continuous at \(x=\frac23\), so:
Check: At \(x=1\), the original quotient is \((9-4)/(3-2)=5/1=5\). The simplified expression is \(3(1)+2=5\). The cancellation agrees with the original quotient wherever the canceled denominator factor is nonzero.
Worked Example: Factoring a Difference of Squares in the Denominator
Evaluate \(\displaystyle\lim_{x\to 3}\frac{x-3}{x^2-9}\).
At \(x=3\), the numerator is \(3-3=0\), and the denominator is \(3^2-9=9-9=0\). Factor the denominator:
For \(x\ne3\), cancel the common factor \(x-3\):
At \(x=3\), the remaining denominator has limit \(3+3=6\), which is nonzero. Therefore the quotient law applies to the simplified expression:
Check: At \(x=4\), the original quotient is \((4-3)/(16-9)=1/7\). The simplified expression gives \(1/(4+3)=1/7\). This confirms the simplification at an input near \(3\); the original quotient remains undefined at \(x=3\).
Why the Method Works
The argument behind each example has two parts. First, the difference-of-squares identity is an algebraic equality, so it rewrites the numerator or denominator without changing its value. Second, cancellation preserves the quotient only where the canceled factor is nonzero. In a limit as \(x\to a\), we may use that equality on a sufficiently small punctured interval around \(a\).
For instance, if \(x^2-a^2\) occurs in the numerator and \(x-a\) in the denominator, then:
Because the original quotient equals \(x+a\) at every nearby \(x\ne a\), its limit is the limit of \(x+a\), which is \(2a\). This conclusion does not require the original quotient to be defined at \(a\). It does require that the canceled expression really is nonzero at the nearby inputs used for cancellation.
This principle explains why the value at the approach point is not the deciding issue. A function can be undefined at that point, or have a different value there, while still having a limit determined by its nearby values. The cancellation does not fill in the hole by itself; it gives a simpler expression with the same nearby values.
Conditions to Check
- Factor accurately. Use \(u^2-v^2=(u-v)(u+v)\). The middle signs are opposite: one factor is a difference and the other is a sum.
- Cancel only a common factor. Terms that are added or subtracted are not individual factors and cannot be canceled separately.
- Keep track of the excluded point. If the canceled factor vanishes at \(a\), cancellation is valid for nearby \(x\ne a\), not at \(x=a\).
- Check the remaining denominator. If a denominator remains, its limit must be nonzero to use the quotient law.
- Evaluate only after simplifying. The original expression may give \(0/0\), while the simplified one has a defined limit.
A Common Mistake
Consider the first example again. A tempting but incorrect calculation is:
This is wrong because division by zero is undefined, and \(0/0\) is not the number \(0\). It is an indeterminate form: direct substitution has not determined the limit. The correction is to factor before evaluating:
and then take the limit of the simplified expression, obtaining \(5+5=10\). The condition \(x\ne5\) matters: it states exactly where the cancellation is justified.
Why This Simplification Matters
Difference-of-squares factoring is useful because it can turn a quotient that is undefined at the approach point into a simpler expression whose limit is accessible through the limit laws. It also helps separate two questions that should not be confused: whether the original function has a value at the point, and whether its values approach a limit there.
The method does not guarantee that every quotient with a difference of squares has a finite limit. The factors must actually share a vanishing factor, and after cancellation the remaining expression must have a finite limit for a finite answer to follow. Factoring and checking the remaining denominator make these conditions visible rather than hiding them behind an invalid substitution.
Check Your Understanding
For each question, identify the factorization and state where cancellation is valid before evaluating the limit.
- Evaluate \(\displaystyle\lim_{x\to4}\frac{x^2-16}{x-4}\).
- Evaluate \(\displaystyle\lim_{x\to -2}\frac{4x^2-16}{2x+4}\).
- Evaluate \(\displaystyle\lim_{x\to3}\frac{x^2-9}{x-3}\). How does the result compare with the first example?
- Why is it not valid to replace \(0/0\) by \(0\) when direct substitution gives that form?
- In \(\displaystyle\lim_{x\to3}\frac{x-3}{x^2-9}\), what condition allows the quotient law to be used after cancellation?
