Curriculum
Pure Mathematics — Second Secondary
Unit 1: Real Functions and Curve Sketching
0/8
Unit 2: Exponents and Logarithms
0/12
Unit 3: Limits
0/6
Unit 4: Trigonometry
0/4
Text lesson
Limits at Infinity
Limits at Infinity
Worked examples
Worked examples
Example 1: equal degrees
- Find limx → ∞ (3x2 + x) / (x2 − 5).
- Solution Divide the numerator and the denominator by x2: (3 + 1/x) / (1 − 5/x2). As x → ∞ the terms 1/x and 5/x2 approach zero, so the limit is 3 / 1 = 3.
Example 2: a larger denominator
- Find limx → ∞ (2x + 1) / (5x − 3).
- Solution Divide by x: (2 + 1/x) / (5 − 3/x). As x → ∞ the terms 1/x and 3/x approach zero, so the limit is 2/5.
Example 3: a larger numerator
- Find limx → ∞ x3 / (x2 + 1).
- Solution The numerator has degree 3 and the denominator has degree 2. Divide by x2: x / (1 + 1/x2). As x grows, the numerator grows without bound while the denominator approaches 1. So the limit is infinite.
Example 4: identifying a horizontal asymptote
| Comparison aspect | Main idea | Keywords |
|---|---|---|
| • | Let f(x) = (2x − 1) / (x + 3). Find limx → ∞ f(x) and identify the horizontal asymptote. | Findlimxidentifyhorizontal |
| Solution | Divide by x: (2 − 1/x) / (1 + 3/x). As x → ∞ the limit is 2 / 1 = 2. So the line y = 2 is a horizontal asymptote | Dividelimitlinehorizontal |
| Check |
| 1000value19991003 |
Example 5: a limit at minus infinity
- Find limx → −∞ (3 / x + 2).
- Solution As x decreases without bound, 3/x gets close to zero, so the expression gets close to 2. The limit is 0 + 2 = 2.
Example 6: finding an unknown coefficient
- Find the value of k for which limx → ∞ (kx2 + 3) / (2x2 − 1) = 4.
Key points
SolutionDivide the numerator and the denominator by x2: (k + 3/x2) / (2 − 1/x2). The limit is k / 2. Setting k / 2 = 4 gives k = 8.
CheckWith k = 8 the function is (8x2 + 3) / (2x2 − 1), and its limit is 8 / 2 = 4.
Limits at Infinity · 3 / 5