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Lesson 3: Limits at Infinity and Asymptotes

Estimated time: 30-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Limits at Infinity

A limit at infinity describes the long-run behavior of a function as x grows without bound (positive or negative).

Limit at Infinity

We write limx→∞ f(x) = L if f(x) gets arbitrarily close to L as x increases without bound. Similarly for x → −∞.

Example 1: A Basic Limit at Infinity

Evaluate limx→∞ 1/x.

Solution: As x grows larger (10, 100, 1000, ...), 1/x gets smaller (0.1, 0.01, 0.001, ...). Therefore limx→∞ 1/x = 0.

Similarly, limx→−∞ 1/x = 0.

Key Fact

For any positive integer n: limx→∞ 1/xn = 0 and limx→−∞ 1/xn = 0.

Horizontal Asymptotes

If limx→∞ f(x) = L or limx→−∞ f(x) = L, then the line y = L is a horizontal asymptote of f. The graph of f approaches but never quite reaches this line.

Horizontal Asymptote

The line y = L is a horizontal asymptote of y = f(x) if limx→∞ f(x) = L or limx→−∞ f(x) = L (or both).

A function can have at most two horizontal asymptotes (one as x → ∞ and one as x → −∞), or fewer.

Limits of Rational Functions at Infinity

For a rational function p(x)/q(x), the limit at infinity depends on the degrees of the numerator and denominator. The technique is to divide every term by the highest power of x in the denominator.

Rational Function Rules at Infinity

Let f(x) = (anxn + ...)/( bmxm + ...) where an, bm ≠ 0.

  • If n < m (degree of numerator < degree of denominator): limit = 0
  • If n = m (degrees equal): limit = an/bm (ratio of leading coefficients)
  • If n > m (degree of numerator > degree of denominator): limit = ±∞ (no horizontal asymptote)

Example 2: Degrees Equal

Evaluate limx→∞ (3x² + 5x − 1)/(7x² − 2).

Solution: Both numerator and denominator have degree 2. Divide every term by x²:

= (3 + 5/x − 1/x²)/(7 − 2/x²)

As x → ∞, the terms 5/x, 1/x², and 2/x² all → 0:

= (3 + 0 − 0)/(7 − 0) = 3/7

Horizontal asymptote: y = 3/7.

Example 3: Degree of Numerator Less Than Denominator

Evaluate limx→∞ (2x + 1)/(x³ − 4).

Solution: Degree 1 < degree 3, so the limit is 0.

Horizontal asymptote: y = 0.

Example 4: Degree of Numerator Greater Than Denominator

Evaluate limx→∞ (x³ + 2x)/(x − 1).

Solution: Degree 3 > degree 1. Divide: x³/x = x², which grows without bound.

limx→∞ = . No horizontal asymptote.

Vertical Asymptotes

A vertical asymptote occurs at x = a when the function grows without bound near that point. This happens when the denominator approaches zero while the numerator does not.

Vertical Asymptote

The line x = a is a vertical asymptote of f if at least one of the following is true:

  • limx→a+ f(x) = ±∞
  • limx→a f(x) = ±∞

Example 5: Finding Vertical Asymptotes

Find the vertical asymptotes of f(x) = 1/(x² − 4).

Step 1: Factor the denominator: x² − 4 = (x − 2)(x + 2).

Step 2: The denominator is zero at x = 2 and x = −2.

Step 3: The numerator (1) is not zero at these points, so both x = 2 and x = −2 are vertical asymptotes.

Step 4: Check signs near x = 2:

  • As x → 2+: (x − 2) is small positive, (x + 2) ≈ 4, so f(x) → +∞
  • As x → 2: (x − 2) is small negative, (x + 2) ≈ 4, so f(x) → −∞

Limits at Infinity for Non-Rational Functions

Example 6: Exponential Function

Evaluate limx→∞ e−x.

Solution: As x → ∞, −x → −∞, so e−x → 0. Therefore the limit is 0.

Horizontal asymptote: y = 0 (to the right).

Example 7: Square Root at Infinity

Evaluate limx→∞ (x − √(x² + 3x)).

Solution: This is an ∞ − ∞ indeterminate form. Rationalize:

Multiply by (x + √(x² + 3x))/(x + √(x² + 3x)):

= [x² − (x² + 3x)] / [x + √(x² + 3x)] = −3x / [x + √(x² + 3x)]

Divide top and bottom by x: = −3 / [1 + √(1 + 3/x)]

As x → ∞: = −3 / [1 + √1] = −3/2 = −3/2

End Behavior Summary

Knowing limits at infinity tells us the "end behavior" of a function — what happens as you move far to the left or far to the right on the graph.

Key Takeaways

Check Your Understanding

1. Find limx→∞ (5x³ − 2x)/(4x³ + x − 7).

Answer: Degrees are equal (both 3). The limit equals the ratio of leading coefficients: 5/4. Horizontal asymptote: y = 5/4.

2. Find the horizontal asymptote(s) of f(x) = (x + 1)/(x² + 1).

Answer: Degree of numerator (1) < degree of denominator (2), so limx→±∞ f(x) = 0. Horizontal asymptote: y = 0.

3. Find all vertical asymptotes of g(x) = (x + 3)/[(x − 1)(x + 5)].

Answer: The denominator is zero at x = 1 and x = −5. The numerator is not zero at either point. So the vertical asymptotes are x = 1 and x = −5.

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