Why Cancellation Changes the Limit
A rational function can be undefined at an input even when its values nearby approach a finite number. This happens when the numerator and denominator share a factor that vanishes there. Canceling the shared factor gives an equivalent expression at nearby inputs, and that expression can reveal the limit.
But cancellation does not always remove the singular behavior. If the denominator contains more copies of the vanishing factor than the numerator does, some copies remain after cancellation. The remaining denominator determines whether the function approaches a finite value or becomes unbounded on one or both sides.
The key is to count factors before substituting. In particular, a denominator that is zero at the input does not by itself tell us whether the limit is finite or infinite.
Factor Multiplicity Near a Singular Input
Suppose \(a\) is a real number and \(f(x)=N(x)/D(x)\) is a rational function, where \(N\) and \(D\) are polynomials and \(D(a)=0\). If both polynomials contain a factor \(x-a\), write them as
where \(m\) and \(n\) are nonnegative integers, and \(P(a)\ne 0\) and \(Q(a)\ne 0\). The integers \(m\) and \(n\) count how many copies of \(x-a\) occur in the numerator and denominator, respectively. They are the multiplicities of that factor.
For \(x\ne a\), the common powers can be canceled algebraically. The original function remains undefined at \(x=a\), because its original denominator is zero there. Cancellation gives an expression that agrees with the original function at nearby inputs, not a new value for the original function at \(a\).
Once the factors are counted, there are three possibilities:
- If \(m=n\), all copies of \(x-a\) cancel, leaving a finite nonzero limit.
- If \(m>n\), the numerator has extra copies, so the reduced expression tends to zero.
- If \(m<n\), some copies remain in the denominator, and the one-sided limits are infinite with signs determined by the remaining factors.
A Factor-Comparison Result
The factorization gives a direct test for the limit. Since \(Q(a)\ne 0\), the quotient \(P(x)/Q(x)\) is defined and continuous for all \(x\) sufficiently close to \(a\). For \(x\ne a\),
Here is why the test works. If \(m=n\), the power of \(x-a\) disappears, and continuity of \(P/Q\) gives the limit \(P(a)/Q(a)\). If \(m>n\), the remaining factor \((x-a)^{m-n}\) tends to zero while \(P/Q\) tends to the finite number \(P(a)/Q(a)\), so the product tends to zero.
If \(m<n\), set \(k=n-m\), which is a positive integer, and write \(R(x)=P(x)/Q(x)\). Then \(R(x)\to c=P(a)/Q(a)\), where \(c\ne0\). Therefore, sufficiently close to \(a\), \(R(x)\) has the same sign as \(c\), and its magnitude is bounded below by a positive number, for example \(|c|/2\). Consequently,
As \(x\to a\), the denominator on the right tends to zero, so this lower bound tends to infinity. Thus the function is unbounded in magnitude on either side. For the sign, the factor \(R(x)\) keeps the sign of \(c\) near \(a\), while \((x-a)^k\) is positive on both sides when \(k\) is even and changes sign across \(a\) when \(k\) is odd. This establishes the one-sided behavior.
Worked Examples
Worked Example: Equal Multiplicities Leave a Finite Limit
Find \(\lim_{x\to1}\dfrac{x^2-1}{x-1}\).
At \(x=1\), the numerator and denominator are both zero, so direct substitution does not determine the limit. Factor the numerator:
For \(x\ne1\), cancel the common factor \(x-1\):
The numerator and denominator each contain one copy of \(x-1\), so the multiplicities are equal. The reduced expression tends to \(1+1=2\). Therefore, the two-sided limit is \(2\).
Check: For every \(x\ne1\), the original quotient equals \(x+1\), whose value approaches \(2\) as \(x\to1\). The original function is still undefined at \(x=1\); it is the limit, not the function value, that equals \(2\).
Worked Example: Extra Numerator Factors Give a Zero Limit
Find \(\lim_{x\to2}\dfrac{(x-2)^3(x+1)}{(x-2)^2(x+4)}\).
The numerator contains three copies of \(x-2\), and the denominator contains two. For \(x\ne2\), cancel the two common copies:
The denominator remaining after cancellation approaches \(2+4=6\), which is nonzero. Substituting into the reduced expression now gives
Check: The uncanceled power is one copy of \(x-2\) in the numerator, and it tends to zero; the other factors tend to \(3\) and \(6\), with the denominator limit nonzero. Thus the product tends to zero, as calculated. The original quotient is undefined at \(x=2\), even though its limit is zero.
Worked Example: A Remaining Denominator Factor Gives Opposite One-Sided Limits
Determine the one-sided limits of \(\dfrac{(x-3)^2(x+2)}{(x-3)^3(x+1)}\) as \(x\to3\).
There are two copies of \(x-3\) in the numerator and three in the denominator. Cancel two copies for \(x\ne3\):
The remaining factor \(x-3\) is in the denominator. The other quotient tends to
Since \(5/4\) is positive, the sign near \(3\) is controlled by \(x-3\). On the left, \(x-3\) is negative, so the quotient decreases without bound. On the right, \(x-3\) is positive, so the quotient increases without bound:
Check: At \(x=29/10\), the reduced expression is \(-490/39\), which is negative. At \(x=31/10\), it is \(510/41\), which is positive. These values confirm the signs on the left and right, respectively. The factor \(x-3\) remains in the denominator, so the limit is not finite.
A Common Cancellation Mistake
In the last example, a tempting but incorrect calculation is to cancel \((x-3)^2\) and then substitute \(x=3\) into the remaining quotient as if no singular factor remained:
This calculation has used \(5/4\), the limit of \((x+2)/(x+1)\), as though it were the limit of the entire reduced expression. It has omitted the remaining factor \(1/(x-3)\). The correction is to keep every uncanceled factor visible:
The second factor tends to \(5/4\), but the first diverges with opposite signs on the two sides. The full expression therefore has left-hand limit \(-\infty\) and right-hand limit \(+\infty\), not a finite limit.
A Reliable Procedure
For a rational function near an input where its denominator is zero, use this sequence of checks:
Express numerator and denominator using powers of \(x-a\), where \(a\) is the input approached.
Record the number of copies in the numerator and denominator before canceling.
The canceled expression agrees with the original only for \(x\ne a\), which is enough for finding a limit.
If the denominator factor is fully removed, evaluate the reduced expression’s limit. If a power remains below the fraction bar, determine its sign on each side.
This procedure matters because the form \(0/0\) is not itself a limit value or a conclusion that the limit fails to exist. It signals that the original numerator and denominator both vanish, so their factors must be compared. Conversely, canceling a common factor does not guarantee a finite limit: the denominator may have had higher multiplicity and retained a vanishing factor.
Check Your Understanding
For each question, factor first and keep track of any denominator factor that remains after cancellation.
- Find \(\lim_{x\to-2}\dfrac{x^2-4}{x+2}\). Is the original function defined at \(x=-2\)?
- For \(\dfrac{(x-1)^4}{(x-1)^2(x+3)}\), determine the limit as \(x\to1\).
- For \(\dfrac{(x+1)^2}{(x+1)^3}\), determine the left- and right-hand limits as \(x\to-1\).
- In a rational function, the numerator contains two copies of \(x-a\), and the denominator contains four. How many copies remain in the denominator after cancellation?
- Why does canceling a factor that vanishes at \(a\) not establish the value of the original function at \(a\)?
