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Algebraic Limit Techniques · Tutorial 25 of 999

Rational Function Limits at Defined Denominators

Check the denominator first, then evaluate the numerator and denominator at the target and use their quotient as the limit.

Beginner 9 min read

What You'll Learn

  • Identify the numerator and denominator of a rational function.
  • Check whether the denominator is nonzero at the target before substituting.
  • Use polynomial limits and the quotient law to evaluate eligible limits.
  • Distinguish the value of a rational function at a point from its limit there.
  • Recognize when direct substitution does not justify an answer.

Why Check the Denominator First?

A rational function is a quotient of polynomial expressions. When a polynomial is evaluated near a target value, its limit is found by evaluating the polynomial at that target. For a quotient, however, there is an important extra condition: the limit of the denominator must not be zero.

This means that many rational-function limits are straightforward. Check the denominator at the target; if it is nonzero, evaluate both polynomials there and divide. The check is not a formality. It is what allows the quotient limit law to apply.

Key idea: For a rational function, direct substitution gives its limit at a target \(a\) when the denominator evaluated at \(a\) is nonzero.

Rational Functions and the Substitution Rule

A rational function is a function that can be written as a quotient of polynomials. Write it as \(f(x)=\frac{P(x)}{Q(x)}\), where \(P\) and \(Q\) are polynomials. The expression is defined at any input \(x\) for which \(Q(x)\ne 0\).

Suppose the target is \(x=a\). The polynomial limit law gives \(\lim_{x\to a}P(x)=P(a)\) and \(\lim_{x\to a}Q(x)=Q(a)\). The quotient limit law applies when the denominator limit is nonzero. Therefore, if \(Q(a)\ne 0\),

$$ \lim_{x\to a}\frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)}. $$

Since \(Q(a)\ne 0\), this substitution also confirms that the rational function is defined at \(a\). The limit and the function value are therefore equal in this case.

Condition to check: Before dividing substituted values, calculate \(Q(a)\). The direct-substitution rule applies only if \(Q(a)\ne 0\).

Why the Rule Works

The argument uses two standard results from the earlier material on polynomial limits and the quotient limit law. A polynomial has limit equal to its value at the target, so the numerator approaches \(P(a)\) and the denominator approaches \(Q(a)\). When \(Q(a)\ne 0\), the quotient limit law says that the quotient approaches the quotient of those two limits.

1
Evaluate the polynomial limits.
Because \(P\) and \(Q\) are polynomials, \(\lim_{x\to a}P(x)=P(a)\) and \(\lim_{x\to a}Q(x)=Q(a)\).
2
Check the denominator limit.
The quotient law requires \(\lim_{x\to a}Q(x)\ne 0\), which here means \(Q(a)\ne 0\).
3
Apply the quotient law.
With that condition satisfied, the limit is \(P(a)/Q(a)\).

There is also a useful way to see why the denominator condition matters for the expression near \(a\). If \(Q(a)\ne 0\), then \(Q(x)\) stays away from zero for all \(x\) sufficiently close to \(a\). In fact, because \(\lim_{x\to a}Q(x)=Q(a)\), there is a neighborhood of \(a\) in which \(|Q(x)-Q(a)|<|Q(a)|/2\). The triangle inequality then gives

$$ |Q(x)| \ge |Q(a)|-|Q(x)-Q(a)| > \frac{|Q(a)|}{2} >0. $$

Thus the denominator is nonzero not only at the target but also sufficiently close to it. This is consistent with the quotient law: the quotient can be evaluated through its numerator and denominator limits because the denominator does not approach zero.

Worked Examples

Worked Example: A Quadratic Numerator and Linear Denominator

Evaluate \(\lim_{x\to 2}\frac{3x^2-x+4}{x+5}\).

First identify the denominator \(Q(x)=x+5\) and evaluate it at the target:

$$ Q(2)=2+5=7\ne 0. $$

Direct substitution is valid. Evaluate the numerator:

$$ P(2)=3(2)^2-2+4 =3(4)-2+4 =12-2+4 =14. $$

Now divide the substituted numerator by the substituted denominator:

$$ \lim_{x\to 2}\frac{3x^2-x+4}{x+5} = \frac{14}{7} =2. $$

Check: At \(x=2\), the original expression is \(14/7=2\). The function is defined there, and its value agrees with the limit.

Worked Example: A Negative Target

Evaluate \(\lim_{x\to -1}\frac{x^3+2x^2-5}{2x^2+3x+4}\).

The denominator at the target is

$$ Q(-1)=2(-1)^2+3(-1)+4 =2(1)-3+4 =3. $$

Because \(3\ne 0\), the quotient law applies. Evaluate the numerator separately:

$$ P(-1)=(-1)^3+2(-1)^2-5 =-1+2(1)-5 =-1+2-5 =-4. $$

Therefore,

$$ \lim_{x\to -1}\frac{x^3+2x^2-5}{2x^2+3x+4} = \frac{-4}{3} =-\frac{4}{3}. $$

Check: Substituting \(x=-1\) into the original expression gives numerator \(-4\) and denominator \(3\), so the function value is \(-4/3\), as found.

Worked Example: A Negative Numerator Value

Evaluate \(\lim_{x\to 3}\frac{x^2-4x+1}{x^2-2x+7}\).

Check the denominator first:

$$ Q(3)=3^2-2(3)+7 =9-6+7 =10\ne 0. $$

Now evaluate the numerator:

$$ P(3)=3^2-4(3)+1 =9-12+1 =-2. $$

Since the denominator is nonzero, divide these values:

$$ \lim_{x\to 3}\frac{x^2-4x+1}{x^2-2x+7} = \frac{-2}{10} =-\frac{1}{5}. $$

Check: The substituted denominator is \(10\), so the expression is defined at \(3\). Its value there is \(-2/10=-1/5\), matching the limit.

A Reliable Order of Work

For each rational-function limit in this situation, use the same sequence. Keeping the denominator check first helps prevent an invalid use of the quotient law.

1
Locate the target.
Read the value \(a\) in \(\lim_{x\to a}\).
2
Evaluate the denominator at the target.
Calculate \(Q(a)\) carefully, including signs and powers.
3
Confirm it is nonzero.
If \(Q(a)\ne 0\), the quotient law permits direct substitution.
4
Evaluate and divide.
Find \(P(a)\), then calculate \(P(a)/Q(a)\).
5
Check the arithmetic.
Verify the substituted numerator and denominator in the original expression and simplify the quotient.

Common Mistake: Dividing When the Denominator Is Zero

A common error is to substitute into the numerator and denominator and then divide without checking whether the denominator value is nonzero. For example, consider \(\lim_{x\to 1}\frac{x^2-1}{x-1}\). A mistaken calculation might say

$$ \frac{1^2-1}{1-1} = \frac{0}{0} =0. $$

The last equality is not valid: division by zero is undefined. More importantly for limits, the quotient law cannot be applied here because the denominator limit is zero. The substitution calculation therefore does not establish that the limit is \(0\), nor does it by itself determine the limit.

Correction: If \(Q(a)=0\), stop before dividing. The nonzero-denominator condition has failed, so another analysis is required; direct substitution under the quotient law does not answer the question.

This correction is deliberately limited: the immediate task is to identify whether the quotient law is available. A zero denominator is a signal not to claim a limit from the substitution alone.

Value at the Target and Limit Near the Target

A limit describes the behavior of a function as \(x\) approaches a target. The function value describes what happens at the target itself. These ideas are related but not identical in general. In the case treated here, \(Q(a)\ne 0\), so the rational function is defined at \(a\), and the quotient limit law shows that its limit equals its value there.

It is useful to state both conclusions accurately:

  • If \(Q(a)\ne 0\), then the rational function is defined at \(a\).
  • Under the same condition, \(\lim_{x\to a}P(x)/Q(x)=P(a)/Q(a)\).
  • If \(Q(a)=0\), the rule in this tutorial does not apply; the limit cannot be concluded by simply dividing substituted values.

The numerator may be positive, negative, or zero. None of those possibilities prevents direct substitution as long as the denominator value is nonzero. For instance, a zero numerator over a nonzero denominator gives a limit of zero. The essential check is on the denominator.

Why This Method Matters

Direct substitution is efficient because it reduces a limit calculation to polynomial evaluation and one division. But its reliability comes from respecting the condition on the quotient law. Checking first helps distinguish a routine evaluation from a case where substitution alone cannot settle the question.

For AP Calculus problems, make the condition visible in your written work. Instead of jumping straight to a quotient, write \(Q(a)=\dots\ne 0\), then evaluate \(P(a)\) and form the quotient. This shows why your method is valid, makes arithmetic errors easier to catch, and gives a clear reason if the method cannot be used.

Check Your Understanding

For each question, decide whether direct substitution using the quotient law is justified before computing a limit.

  1. For \(\lim_{x\to 2}\frac{x^2+1}{3x-1}\), what is the denominator at the target, and what is the limit?
  2. For \(\lim_{x\to -2}\frac{2x^2+x}{x^2+1}\), verify the denominator condition and evaluate the limit.
  3. For \(\lim_{x\to 4}\frac{x-4}{x^2-5}\), is the denominator nonzero at the target? If so, find the limit.
  4. What condition on \(Q(a)\) is needed to use \(\lim_{x\to a}P(x)/Q(x)=P(a)/Q(a)\)?
  5. If substitution gives a denominator of zero, what can you conclude about using the quotient limit law?

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