Learning Objectives
Distinguish a limit at infinity from an infinite limit near a finite input.
Evaluate rational-function limits by comparing leading terms.
Handle square roots and signs when x approaches negative infinity.
Prerequisites: rational expressions, polynomial degree, and absolute value. Infinity is not a number to substitute into a formula. A limit at infinity describes what outputs do as the input becomes arbitrarily large in one direction.
1. Inputs Can Escape in Two Directions
The notation lim(x → +∞) f(x) = L means outputs approach L as x increases without bound. The notation x → −∞ follows inputs decreasing without bound. If either limit is finite, y = L is a horizontal asymptote in that direction.
For f(x) = 1/x, both limits at infinity are 0. This differs from the behavior near x = 0, where the one-sided outputs are unbounded. In a limit at infinity the input is unbounded; in an infinite limit at a finite point the output is unbounded.
2. Divide by the Highest Denominator Power
Worked Example: Equal Degrees
Evaluate lim(x → +∞) (3x² − 2x + 1)/(5x² + 4). Divide numerator and denominator by x².
The expression becomes (3 − 2/x + 1/x²)/(5 + 4/x²). Each reciprocal-power term tends to zero.
The limit is 3/5. The same calculation works as x → −∞, so y = 3/5 is a horizontal asymptote in both directions.
Numerator degree versus denominator degree | Rational-function end behavior |
|---|---|
Smaller | Limit 0 in both directions |
Equal | Ratio of leading coefficients |
Larger | No horizontal asymptote; inspect the leading power and sign |
The degree comparison assumes nonzero polynomials and their actual leading terms. When the numerator degree is larger, the quotient behaves like a constant times a positive power of x. Odd and even powers behave differently as x tends to negative infinity.
Example: Different Signs at the Two Ends
For f(x) = (2x³ + 1)/(x² + 1), the leading-term ratio is 2x.
Therefore f(x) tends to +∞ as x → +∞ and to −∞ as x → −∞. There is no finite horizontal asymptote.
3. Remember That √(x²) = |x|
Worked Example: A Square Root
Find both limits at infinity of √(9x² + 1)/x.
Factor x² inside the square root: √(9x² + 1) = |x|√(9 + 1/x²). The quotient is (|x|/x)√(9 + 1/x²).
For positive x, |x|/x = 1, so the limit at +∞ is 3. For negative x, |x|/x = −1, so the limit at −∞ is −3.
Replacing √(x²) with x without checking the sign would incorrectly give the same answer at both ends. The square root itself is nonnegative; division by a negative x makes the whole quotient negative.
4. Subtracting Large Terms May Hide a Small Difference
Worked Example: Rationalize a Difference
Evaluate lim(x → +∞) [√(x² + 4) − x]. Multiplying by the conjugate gives 4/[√(x² + 4) + x].
The denominator grows without bound through positive values, so the limit is 0.
The phrase infinity minus infinity does not evaluate this limit. Algebra reveals how the two large terms cancel.
Common Mistakes
Do not substitute infinity as if it were an ordinary real number.
A horizontal asymptote describes end behavior; a graph may cross it at finite inputs.
For square roots at negative infinity, preserve |x| until you determine its sign.
5. Practice
1. Find lim(x → −∞) (4x − 1)/(x² + 2).
2. Find lim(x → +∞) (−2x³ + x)/(5x³ − 7).
3. Find lim(x → −∞) √(4x² + 9)/x.
4. Find lim(x → +∞) [√(x² + 6x) − x].
Worked Solutions
1. The numerator degree is smaller, so the limit is 0. Dividing by x² gives the same result.
2. Equal degrees give the leading-coefficient ratio −2/5.
3. The expression equals (|x|/x)√(4 + 9/x²). At negative infinity the first factor is −1, so the limit is −2.
4. Rationalize to obtain 6x/[√(x² + 6x) + x]. For positive x, divide by x to get 6/[√(1 + 6/x) + 1], whose limit is 3.
6. Summary
- Analyze positive and negative infinity separately when signs matter.
- For rational functions, leading terms control end behavior.
- Preserve √(x²) = |x| when simplifying radicals.
- Use a conjugate when large terms are subtracted.