Why Combine Terms Before Substitution?
A limit expression can contain several rational terms whose values approach one another as \(x\) approaches a particular number. Substituting into the separate terms may show that their difference approaches zero, but if that difference is then divided by another quantity approaching zero, direct substitution does not determine the limit. The useful information is often hidden in the numerator of the combined fraction.
Combining the terms over a common denominator reveals that numerator. If it contains the same factor that makes the whole expression have a zero denominator, the factor may cancel. The cancellation can turn an indeterminate \(0/0\) form into an expression whose limit can be evaluated by substitution.
Common Denominators and Domain Restrictions
For rational terms, the common-denominator identity is
This equality holds only when \(q(x)\ne 0\) and \(s(x)\ne 0\). Those restrictions come from the original terms: the combined expression does not restore values at which either original denominator is zero. If the entire difference is also divided by \(x-a\), then \(x\ne a\) is an additional restriction.
In a limit as \(x\) approaches \(a\), the expression need not be defined at \(a\). It must, however, be defined at the nearby \(x\)-values used to approach \(a\). If \(q(a)\ne0\) and \(s(a)\ne0\), then both denominators remain nonzero sufficiently close to \(a\), by continuity. This lets us use the combined expression on a punctured interval around \(a\), meaning an interval around \(a\) with \(a\) itself removed.
A General Pattern
Consider a difference of two rational terms divided by \(x-a\). If the two rational terms have the same value at \(a\), their combined numerator is zero at \(a\). When that numerator is a polynomial, the factor theorem says it has a factor of \(x-a\). That factor can then cancel the \(x-a\) in the denominator for \(x\ne a\).
If \(q(a)\ne0\), \(s(a)\ne0\), and \(\dfrac{p(a)}{q(a)}=\dfrac{r(a)}{s(a)}\), then \(p(a)s(a)-r(a)q(a)=0\). Thus the polynomial \(p(x)s(x)-r(x)q(x)\) has a factor of \(x-a\).
Worked Examples
Worked Example: Combining a Reciprocal and a Constant
Evaluate \(\displaystyle \lim_{x\to1}\frac{\frac{1}{x+1}-\frac12}{x-1}\).
Substitution into the two terms in the numerator gives \(\frac12-\frac12=0\), while the outer denominator approaches \(0\). Combine the numerator before deciding what the limit is:
The first equality uses the common denominator \(2(x+1)\). Since \(1-x=-(x-1)\), the factor cancels for \(x\ne1\). The original expression requires \(x\ne-1\) and \(x\ne1\). Therefore, the reduced expression gives the limit by substitution:
Check: At the allowed input \(x=0\), the original expression is \(\frac{1-\frac12}{-1}=-\frac12\). The reduced expression gives \(-\frac{1}{2(0+1)}=-\frac12\), confirming the algebra at that input. Hence the limit is \(-\frac14\).
Worked Example: A Rational Term Minus a Constant
Evaluate \(\displaystyle \lim_{x\to2}\frac{\frac{4}{x+2}-1}{x-2}\).
The numerator approaches \(\frac44-1=0\), and the denominator approaches \(0\). Use the common denominator \(x+2\) in the numerator:
Here \(2-x=-(x-2)\), so the factor \(x-2\) cancels on the original domain. That domain requires \(x\ne-2\) and \(x\ne2\). Evaluating the reduced expression gives
Check: At \(x=0\), the original expression is \(\frac{2-1}{-2}=-\frac12\). The reduced expression is \(-\frac{1}{0+2}=-\frac12\). The values agree, and the limit is \(-\frac14\).
Worked Example: Combining Two Reciprocal Terms and a Constant
Evaluate \(\displaystyle \lim_{x\to1}\frac{\frac{1}{x+1}-\frac{1}{x+2}-\frac16}{x-1}\).
At \(x=1\), the three terms in the numerator combine to \(\frac12-\frac13-\frac16=0\). First combine the two reciprocal terms, then include the constant:
After dividing by \(x-1\), cancel that factor for \(x\ne1\):
The original expression requires \(x\ne-1\), \(x\ne-2\), and \(x\ne1\). Substitution into the reduced expression is valid for the limit:
Check: At \(x=0\), the original expression is \(\frac{1-\frac12-\frac16}{-1}=-\frac13\). The reduced expression is \(-\frac{0+4}{6(0+1)(0+2)}=-\frac{4}{12}=-\frac13\). This verifies the simplification at an allowed input, and the limit is \(-\frac{5}{36}\).
Why the Cancellation Works
The common-denominator step is an algebraic identity wherever the original denominators are nonzero. Suppose the combined numerator is a polynomial \(H(x)\) and \(H(a)=0\). The factor theorem gives a polynomial \(K(x)\) such that \(H(x)=(x-a)K(x)\). If the original expression also has a factor \(x-a\) in its denominator, then the two expressions are equal whenever \(x\ne a\) and the original denominators are nonzero.
Exclude values that make any original denominator zero, including an outer denominator such as \(x-a\).
Use a common denominator and simplify the numerator carefully.
If it vanishes at \(a\), factor out \(x-a\) when the numerator is a polynomial.
The cancellation gives an equal expression for nearby \(x\ne a\), not a new value for the original expression at \(a\).
When the reduced expression has a denominator nonzero at \(a\), substitute \(a\) into that expression.
To justify the final limit step, the original and reduced expressions agree for every sufficiently nearby \(x\ne a\). Their values therefore approach the same number as \(x\) approaches \(a\). The reduced expression is continuous at \(a\) when its denominator is nonzero there, so its limit is its value at \(a\). The cancellation does not claim the original expression is defined at \(a\); it establishes the limit from nearby inputs.
A Common Mistake: Treating \(0/0\) as an Answer
For the first example, substituting \(x=1\) into the original form gives
A common incorrect conclusion is to say that the limit equals \(0\) because the numerator equals \(0\). But division by zero is undefined, and \(0/0\) is an indeterminate form: it does not by itself give the limit. In this example, combining the terms shows that the limit is \(-\frac14\), not \(0\).
Another error is to cancel \(x-1\) before showing where it comes from. There is no such factor to cancel in the separate terms \(\frac{1}{x+1}\) and \(\frac12\). It appears only after combining them and rewriting \(1-x=-(x-1)\). The order of the algebra matters: combine first, factor second, and cancel only on the domain where the original expression is defined.
When This Method Is Useful
Combining rational terms is useful when a limit contains a sum or difference inside a numerator and direct substitution produces \(0/0\). It makes the numerator visible as a single expression, so you can determine whether it has a factor that cancels with the vanishing denominator. This is especially important when several fractions or a constant are involved: evaluating the pieces separately can hide the relationship that makes the cancellation possible.
Before substituting, ask three questions: What values are excluded by the original denominators? What is the correct common denominator? Does the combined numerator vanish at the limit point, and can its factorization expose a matching factor? If the factorization does not produce a valid cancellation, do not force one; the combined form may indicate that this particular method has not resolved the limit.
Check Your Understanding
For each question, keep the original denominator restrictions in mind as you simplify.
- Combine \(\frac{1}{x+3}-\frac14\) into one fraction. Which values must be excluded from its original domain?
- Evaluate \(\displaystyle \lim_{x\to3}\frac{\frac{6}{x+3}-1}{x-3}\). Show the common-denominator step and the cancellation.
- For \(\displaystyle \frac{\frac{1}{x+1}-\frac13}{x-2}\), list the values excluded by the original expression. Is substitution into its numerator at \(x=2\) equal to zero?
- Why does canceling a factor \(x-a\) not imply that the original expression is defined at \(x=a\)?
- A simplification gives \(0/0\) after direct substitution. What does that form tell you, and what does it not tell you?
