When Direct Substitution Works
A rational function is a quotient of polynomials. When finding its limit as \(x\) approaches a number \(a\), direct substitution is often the shortest method: evaluate the numerator at \(a\), evaluate the denominator at \(a\), and divide.
The essential check is that the denominator’s value at \(a\) is not zero. This condition allows the quotient law for limits to be used. If the denominator is zero, direct substitution does not provide an answer by this method, even if the numerator and denominator are both polynomials.
This rule combines two familiar facts. Polynomial limits can be evaluated by direct substitution, and the quotient law allows the limits of a numerator and denominator to be divided when the denominator limit is nonzero. The examples below show how to check and use the condition carefully.
The Direct-Substitution Method
Write the rational function as \(P(x)/Q(x)\), with \(P\) and \(Q\) polynomials. For a limit as \(x\) approaches \(a\), evaluate \(P(a)\) and \(Q(a)\). If \(Q(a)\ne 0\), the polynomial limit rule and quotient law give the answer \(P(a)/Q(a)\).
Separate the numerator \(P(x)\) from the denominator \(Q(x)\).
Calculate \(P(a)\) and \(Q(a)\), showing the arithmetic.
Confirm that \(Q(a)\ne 0\). If so, the quotient law applies.
Divide \(P(a)\) by \(Q(a)\), then verify the result by multiplying it by \(Q(a)\).
The order matters. In particular, do not divide before checking that the denominator value is nonzero. Direct substitution into the polynomials is justified by the polynomial limit rule; division is justified by the quotient law only after its condition has been checked.
Worked Example: A Quadratic Rational Function
Evaluate \(\displaystyle \lim_{x\to 2}\frac{3x^2-1}{x^2+x+1}\).
The numerator and denominator are polynomials. Substituting \(x=2\) into the numerator gives
Substituting into the denominator gives
Since the denominator value is \(7\ne 0\), the quotient law applies. Therefore,
Check: multiplying the proposed quotient by the denominator value gives \(\frac{11}{7}\cdot 7=11\), the numerator value. Thus the quotient is consistent with the two polynomial limits.
Why the Nonzero Condition Is Enough
The quotient law requires that the denominator’s limit exist and be nonzero. Here, because \(Q\) is a polynomial, its limit at \(a\) is \(Q(a)\). Thus \(Q(a)\ne 0\) supplies exactly the denominator condition needed for the law. It also has a useful consequence: the denominator stays nonzero for all \(x\) sufficiently close to \(a\).
To see this consequence, let \(Q(a)\ne 0\). Polynomial functions are continuous, so \(Q(x)\) approaches \(Q(a)\) as \(x\) approaches \(a\). In particular, for \(x\) sufficiently close to \(a\), the difference between \(Q(x)\) and \(Q(a)\) is less than half the size of \(Q(a)\). The triangle inequality then gives a positive lower bound for \(|Q(x)|\).
Therefore \(Q(x)\ne 0\) in some neighborhood of \(a\). The quotient is defined there, and the denominator limit is the nonzero number \(Q(a)\). Applying the quotient law to the polynomial limits now yields
This explains why the test is local and why it concerns the value at the approach point. We do not need \(Q(x)\) to be nonzero for every real number. We need \(Q(a)\ne 0\), which guarantees that it is nonzero near \(a\), where the limit is being evaluated.
More Worked Examples
Worked Example: A Limit at a Negative Number
Evaluate \(\displaystyle \lim_{x\to -1}\frac{x^3+2x+4}{2x^2+3x+5}\).
Evaluate the numerator at \(-1\):
Evaluate the denominator at \(-1\):
The denominator value is \(4\ne 0\), so the quotient law applies. Hence,
Check: \(\frac{1}{4}\cdot 4=1\), which recovers the numerator value. The denominator condition is satisfied, so this quotient is valid.
Worked Example: A Cubic Denominator
Evaluate \(\displaystyle \lim_{x\to 3}\frac{x^2-4x+7}{x^3+1}\).
First, substitute \(x=3\) into the numerator:
Then substitute into the denominator:
Since \(28\ne 0\), the quotient law permits division of the polynomial limits:
Check: \(\frac{1}{7}\cdot 28=4\), exactly the numerator value. The denominator limit is nonzero, as required.
Worked Example: A Nonzero Denominator That Is Negative
Evaluate \(\displaystyle \lim_{x\to 0}\frac{2x^3-x+5}{x^2-3x-4}\).
Substitute \(x=0\) into the numerator:
Substitute \(x=0\) into the denominator:
The denominator value is \(-4\ne 0\). A negative denominator is allowed; the quotient law requires a nonzero value, not a positive one. Therefore,
Check: \(\left(-\frac{5}{4}\right)(-4)=5\), which recovers the numerator value. Thus the limit is \(-\frac{5}{4}\).
A Common Mistake: Dividing by Zero
A frequent error is to substitute into a rational function and use the quotient law even when the denominator value is zero. For example, consider the attempted calculation
The last two steps are invalid: division by zero is undefined, and the quotient law cannot be used because the denominator limit is zero. In this example the numerator value is \(3\), while the denominator value is \(0\), so the stated direct-substitution rule does not apply. The correct conclusion at this stage is that this method cannot evaluate the limit; it is not a value of zero.
The important habit is to write down and check the denominator value before forming the quotient. If it is nonzero, continue with direct substitution. If it is zero, stop: a different analysis is needed, and the quotient law in this form gives no answer.
Using the Rule Reliably
For a rational-function limit with a nonzero denominator value, the computation is short, but the reasoning has distinct parts. Polynomial limits provide the numerator and denominator values. The nonzero check verifies the hypothesis of the quotient law. The quotient law then gives the limit. Keeping those parts explicit makes the method reliable, especially when expressions become more complicated.
- Substitute the approach value into the entire numerator and denominator separately.
- Show enough arithmetic to make each polynomial value clear.
- Check that the denominator value is not zero before dividing.
- Simplify the resulting fraction exactly.
- Do not treat a zero denominator value as an ordinary numerical quotient.
When the denominator value is nonzero, no factoring or other algebraic rewriting is needed to justify direct substitution. The polynomial limit rule and quotient law already establish the result. When that condition fails, this particular method does not settle the limit; the expression must be considered using an appropriate method for that situation.
Check Your Understanding
For each question, explain which polynomial values you need and whether the quotient law applies.
- Evaluate \(\displaystyle \lim_{x\to 1}\frac{2x^2+3}{x^2+2}\). State the denominator value that justifies the quotient law.
- For \(\displaystyle \lim_{x\to -2}\frac{x^2+x+1}{3x+8}\), calculate the numerator and denominator values, then evaluate the limit.
- What condition on \(Q(a)\) is required to evaluate \(\displaystyle \lim_{x\to a}\frac{P(x)}{Q(x)}\) by direct substitution using the quotient law?
- Why does \(Q(a)\ne 0\) ensure that the quotient is defined for \(x\) sufficiently close to \(a\)?
- Explain why the expression \(P(a)/Q(a)\) cannot be used as a quotient-law evaluation when \(Q(a)=0\).
