Learning Objectives
Apply sum, product, and quotient limit laws when their conditions hold.
Resolve 0/0 forms by factoring or rationalizing.
Describe one-sided unbounded behavior without treating infinity as a real number.
Prerequisites: Understanding Limits, polynomial factoring, and conjugates. Keep the original domain visible while simplifying an expression.
1. Start with Direct Substitution
Polynomials are continuous at every real input, so their limits can be found by substitution. For a rational function, substitution also works wherever its denominator is nonzero. For example, lim(x → 2) (x² + 3)/(x + 1) = 7/3.
If f(x) approaches L and g(x) approaches M, both finite, then their sum approaches L + M and their product approaches LM. Their quotient approaches L/M provided M ≠ 0. That final condition is essential: division by a limiting value of zero needs separate analysis.
2. A 0/0 Result Is a Signal to Simplify
Substitution that produces 0/0 has not evaluated the limit. It is an indeterminate form, meaning that different expressions with that form can have different limiting behavior. Look for an equivalent expression valid near the target input.
Worked Example: Factoring
Evaluate lim(x → 3) (x² − 9)/(x − 3). Direct substitution gives 0/0.
Factor x² − 9 = (x − 3)(x + 3). For x ≠ 3, the quotient equals x + 3.
The limit is therefore 3 + 3 = 6. The original quotient remains undefined at x = 3, but equality near 3 is enough for the limit.
Worked Example: A Conjugate
Evaluate lim(x → 4) (√x − 2)/(x − 4). Multiply numerator and denominator by √x + 2.
The numerator becomes x − 4. For x ≠ 4, cancel to obtain 1/(√x + 2).
Substitute into the simplified expression: the limit is 1/(2 + 2) = 1/4. Nearby inputs are positive, so the square root is defined.
3. Nonzero over Zero Requires a Sign Check
For f(x) = 1/(x − 2), the numerator stays positive. As x approaches 2 from below, the denominator is negative and tends to zero, so the outputs decrease without bound. From above, the denominator is positive and tends to zero, so the outputs increase without bound.
Approach | Denominator sign | Behavior |
|---|---|---|
x → 2⁻ | Negative | f(x) → −∞ |
x → 2⁺ | Positive | f(x) → +∞ |
There is no finite two-sided limit in this example, and the two infinite directions disagree. The notation +∞ describes growth without bound; it is not a real number that the function reaches. The line x = 2 is a vertical asymptote.
Common Mistakes
Cancel factors, not terms. You cannot cancel the x in (x + 1)/x.
Do not conclude that every zero denominator gives a vertical asymptote. A shared factor can produce a hole instead.
Do not use the quotient law when the denominator's limit is zero.
4. Practice
1. Evaluate lim(x → −1) (3x² − x + 2).
2. Evaluate lim(x → 2) (x² − x − 2)/(x − 2). State the excluded input.
3. Evaluate lim(x → 0) (√(x + 9) − 3)/x.
4. Describe the left- and right-hand behavior of 1/(x + 1)² at x = −1.
Worked Solutions
1. Substitution gives 3(1) − (−1) + 2 = 6.
2. Factor the numerator as (x − 2)(x + 1). For x ≠ 2, the quotient is x + 1, whose limit is 3. The excluded input is still 2.
3. Multiply by the conjugate to obtain 1/(√(x + 9) + 3) for x ≠ 0. The limit is 1/6.
4. The squared denominator is positive on both sides and approaches zero. Both one-sided limits are +∞; there is no finite limit.
5. Summary
- Try substitution first and check whether the expression is defined.
- For 0/0, find a valid algebraic simplification near the target.
- When outputs are unbounded, use signs to distinguish the two sides.