Grade by Grade
A. Rate of Change · Limits and Rates of Change

Limits Involving Infinity and Asymptotes

A1.4About 70 minutes
Learning goals
  • I can describe the behaviour of a function near a vertical asymptote using one-sided infinite limits.
  • I can evaluate limits as x → ∞ and x → −∞ for rational and exponential functions.
  • I can connect limits at infinity to horizontal asymptotes.
  • I can interpret a limit at infinity in a real situation as a long-term value.

Words to know

vertical asymptote
A vertical line x = a near which f(x) grows without bound (toward ∞ or −∞) as x approaches a from at least one side.
horizontal asymptote
A horizontal line y = L that the graph approaches as x → ∞ or x → −∞; that is, lim f(x) = L.
infinite limit
A statement such as lim (x → a⁺) f(x) = ∞, meaning f(x) grows without bound. The limit does not exist as a number; ∞ describes how it fails.
limit at infinity
lim (x → ∞) f(x) = L: f(x) approaches L as x grows without bound.
end behaviour
What a function does as x → ∞ and as x → −∞.
long-term value
In a model, the value a quantity approaches as time goes on: a limit as t → ∞.

Two new questions

The curriculum gives "the value of a function approaching an asymptote" as a key example of a limit. Asymptotes lead to two new kinds of limit questions:

  1. What happens near a vertical asymptote? As x → a, f(x) may grow without bound. We write lim (x → a⁺) f(x) = ∞ or −∞.
  2. What happens far out to the right or left? As x → ∞ or x → −∞, f(x) may approach a number L. That is a horizontal asymptote y = L.

Remember from MHF4U: in Rational Functions and Equations you found vertical asymptotes where the denominator is 0 (and holes where a common factor cancels), found horizontal asymptotes such as y = a/c for y = (ax + b)/(cx + d) by comparing degrees and looking at very large x, and described the behaviour near each asymptote in words. Now you will write it with limits and prove it algebraically.

Infinite limits

Look at f(x) = 1/(x − 2) near x = 2.

  • x = 2.1 gives 10; x = 2.01 gives 100; x = 2.001 gives 1000. From the right, f(x) grows without bound: lim (x → 2⁺) f(x) = ∞.
  • x = 1.9 gives −10; x = 1.99 gives −100. From the left, f(x) decreases without bound: lim (x → 2⁻) f(x) = −∞.

Writing "= ∞" does not mean the limit exists. ∞ is not a number. It is a precise way of saying how the limit fails to exist.

Quick sign test: if the numerator approaches a non-zero number and the denominator approaches 0, the quotient blows up. Its sign comes from the signs of the numerator and the denominator on that side.

Worked example

Example 1: One-sided infinite limits

For f(x) = (x + 1)/(x − 2), find lim (x → 2⁺) f(x) and lim (x → 2⁻) f(x).

  1. Substituting x = 2 gives 3/0: a non-zero number over 0. So there is a vertical asymptote at x = 2.
  2. From the right (x slightly more than 2): numerator ≈ 3 (positive) and denominator is a tiny positive number. Positive ÷ tiny positive → +∞. Check: f(2.01) = 3.01/0.01 = 301.
  3. From the left (x slightly less than 2): numerator ≈ 3 and denominator is tiny negative. Positive ÷ tiny negative → −∞. Check: f(1.99) = 2.99/(−0.01) = −299.

lim (x → 2⁺) f(x) = ∞ and lim (x → 2⁻) f(x) = −∞. The two-sided limit does not exist.

Limits at infinity

For a rational function, divide every term in the numerator and denominator by the highest power of x in the denominator, then use the fact that

lim (x → ±∞) c/xᵏ = 0 for any constant c and k > 0.

For exponential functions, remember: if 0 < b < 1, then bˣ → 0 as x → ∞; if b > 1, then bˣ → 0 as x → −∞.

Worked example

Example 2: A horizontal asymptote

Find lim (x → ∞) (3x² − 5)/(2x² + x) and lim (x → −∞) of the same function.

  1. Divide top and bottom by x²: (3 − 5/x²)/(2 + 1/x).
  2. As x → ∞, 5/x² → 0 and 1/x → 0. The expression approaches (3 − 0)/(2 + 0) = 3/2.
  3. As x → −∞, the same terms still approach 0, so the limit is again 3/2.

Both limits are 3/2, so y = 3/2 is a horizontal asymptote.

−4−22468−6−4−22468xyx = 2y = 1y = (x + 1)/(x − 2)Vertical asymptote x = 2:as x → 2⁺, y → ∞as x → 2⁻, y → −∞Horizontal asymptote y = 1:as x → ∞, y → 1as x → −∞, y → 1
y = (x + 1)/(x − 2): the graph shoots up just right of x = 2 and down just left of it, and flattens toward y = 1 far to the left and right.
Worked example

Example 3: Comparing degrees

Find (a) lim (x → ∞) (5x + 2)/(x² + 1) and (b) lim (x → ∞) (x² + 1)/(x − 3).

  1. (a) Divide by x²: (5/x + 2/x²)/(1 + 1/x²) → 0/1 = 0. The bottom grows faster.
  2. (b) Divide by x (the highest power in the denominator): (x + 1/x)/(1 − 3/x). The top behaves like x, which grows without bound, and the bottom approaches 1.
  3. So the values grow without bound: lim (x → ∞) (x² + 1)/(x − 3) = ∞ (no horizontal asymptote).

(a) 0; (b) ∞ (no finite limit). Same rule as for sequences: bottom degree higher → 0; equal degrees → ratio of leading coefficients; top degree higher → no finite limit.

Worked example

Example 4: A long-term value

A 200 L tank of fresh water has brine (30 g of salt per litre) pumped in at 10 L/min. The well-mixed solution does not drain. The salt concentration after t minutes is C(t) = 300t/(200 + 10t) = 30t/(20 + t) g/L. Find C(20), C(60) and lim (t → ∞) C(t), and interpret.

  1. C(20) = 600/40 = 15 g/L. C(60) = 1800/80 = 22.5 g/L.
  2. Divide top and bottom by t: C(t) = 30/(20/t + 1). As t → ∞, 20/t → 0, so C(t) → 30/1 = 30.
  3. Interpretation: the tank's mixture gets closer and closer to the strength of the incoming brine, but never stronger.

lim (t → ∞) C(t) = 30 g/L. The horizontal asymptote C = 30 is the long-term concentration.

0408012016020005101520253035t (min)C (g/L)(20, 15)(60, 22.5)(180, 27)C = 30Salt keeps flowing in,but the concentrationcan never pass 30 g/L,the concentration ofthe incoming brine.lim C(t) = 30t→∞
The salt concentration rises quickly at first, then levels off toward its limit of 30 g/L.

Infinity is not a number

Do not substitute ∞ into a formula and do arithmetic with it. "∞/∞ = 1" and "∞ − ∞ = 0" are both wrong. For example, x²/x → ∞ but x/x² → 0, even though both look like ∞/∞. Always rewrite the expression first (divide by the highest power), then take the limit.

Just for fun

A horizontal asymptote is the ultimate patient friend. The graph keeps promising, "I'm almost there!" for all of eternity, and the asymptote just says, "Take your time."

Hands-on activity

Cooling to room temperature

You need: a mug of hot tap water (not boiling; a grown-up should handle anything hotter), a kitchen or cooking thermometer, a timer, grid paper.

  1. Note the room temperature on the thermometer first.
  2. Put the thermometer in the mug and record the temperature every 2 minutes for 30 minutes.
  3. Plot temperature against time.
  4. What value does the temperature seem to approach? Draw that horizontal asymptote as a dashed line.
  5. Compare your data with a model T(t) = R + (T₀ − R)bᵗ, where R is room temperature and T₀ is the starting temperature. Choose b (try values like 0.9 to 0.97) so the model fits your points.

What to look for: the temperature drops fast at first, then more and more slowly, approaching room temperature. In your model, bᵗ → 0 as t → ∞, so lim T(t) = R: the long-term value is the room temperature. Real data wobble a little; that is normal.

Your turn

1

Find lim (x → 3⁺) 1/(x − 3).

2

Find lim (x → 3⁻) 1/(x − 3).

3

Find lim (x → 0) 1/x². Does the limit exist as a number?

4

Find lim (x → ∞) (4x − 1)/(2x + 5).

5

Find lim (x → ∞) 7/(x + 1).

6

Find lim (x → −∞) 3ˣ.

7

For C(t) = 30t/(20 + t), find C(180) and explain what lim (t → ∞) C(t) = 30 means for the tank.

More practice: warm-up

1

Find the vertical asymptote of y = 1/(x + 4).

2

Find lim (x → ∞) 1/x.

3

Find lim (x → ∞) (6x + 1)/(3x).

4

Find the horizontal asymptote of y = 5x/(x + 1).

5

Find lim (t → ∞) (0.5)ᵗ.

More practice: on your own

1

Find lim (x → ∞) (2x³ − x)/(5x³ + 4).

2

Find lim (x → −1⁺) x/(x + 1) and lim (x → −1⁻) x/(x + 1).

3

A mug cools by T(t) = 20 + 70(0.9)ᵗ °C. Find lim (t → ∞) T(t). When is it first within 1 °C of that value?

4

For f(x) = (x² − 4)/(x² − x − 2), find lim (x → 2) f(x), the behaviour near x = −1, and lim (x → ∞) f(x).

5

A fish population is modelled by P(t) = 5000/(1 + 4(0.8)ᵗ), t in years. Find P(0) and lim (t → ∞) P(t). Interpret.

Challenge

1

Find lim (x → −∞) 2x/√(x² + 1). Be careful with the sign.

2

Write a rational function with a vertical asymptote x = 3, a horizontal asymptote y = −2 and an x-intercept at 1. Check it with limits.

3

Use the substitution u = 1/x to find lim (x → ∞) x sin(1/x).

Check your understanding

Quiz

1.

What is lim (x → 4⁻) 1/(x − 4)?

2.

What is lim (x → ∞) (6x² + 1)/(3x² − 2x)?

3.

y = 4 is a horizontal asymptote of f. Which statement must be true?

4.

What is lim (x → ∞) (2x + 7)/(x² − 1)?

5.

A model is N(t) = 800 − 600(0.7)ᵗ. What is the long-term value of N?

6.

Why is "∞/∞ = 1" not a valid step?

For parents and teachers

Big idea (A1.4): the curriculum's example "the value of a function approaching an asymptote" is a limit. Near a vertical asymptote, one-sided limits are ∞ or −∞; far out, lim (x → ±∞) f(x) = L gives a horizontal asymptote and, in models, a long-term value. This extends the MHF4U unit Rational Functions and Equations.

How to help: Ask "What happens in the long run?" about real things: a cooling drink, a tank filling with brine, a population with limited food. Then ask your teen to point to the part of the formula that goes to 0.

Common difficulties:

  • Doing arithmetic with ∞ (∞/∞, ∞ − ∞).
  • Getting the sign wrong next to a vertical asymptote: check the sign of the numerator and the denominator on each side.
  • Confusing a hole (common factor cancels) with a vertical asymptote.
  • Forgetting that √(x²) = −x when x < 0.