Words to know
sequence
An ordered list of numbers t₁, t₂, t₃, …; a function whose domain is the natural numbers.
limit of a sequence
The number L that the terms tₙ get closer and closer to, and stay close to, as n gets larger and larger. Written lim (n → ∞) tₙ = L.
converge
A sequence converges if it has a limit.
diverge
A sequence diverges if it has no limit, for example because its terms grow without bound or keep jumping between values.
golden ratio (φ)
φ = (1 + √5)/2 ≈ 1.618 034, the limit of the ratios of consecutive Fibonacci numbers.
e
The number e ≈ 2.718 28, the limit of (1 + 1/n)ⁿ as n → ∞.
The idea of a limit
All of calculus rests on one idea: the limit. A limit describes the value something approaches, even if it never gets there.
The easiest place to meet limits is in a sequence. Look at 1/n:
1, 0.5, 0.333…, 0.25, 0.2, …, 0.01 (n = 100), …, 0.000 001 (n = 1 000 000), …
No term is ever 0. But the terms get as close to 0 as you like, and stay that close, if you go far enough along. We say the limit of 1/n as n approaches infinity is 0, and write
lim (n → ∞) 1/n = 0
The symbol ∞ (infinity) is not a number. "n → ∞" just means "as n gets larger without bound".
Remember from Grade 11: in the MCR3U unit Sequences and Series you met the Fibonacci sequence, saw that ratios of consecutive terms approach the golden ratio, and saw that adding more and more terms of some geometric series changes the sum less and less. In this lesson you will make those observations precise.
Three behaviours
As n → ∞, a sequence can:
- converge to a limit L: the terms settle down near one number. Example: (n + 1)/n = 2, 1.5, 1.333, 1.25, … → 1.
- diverge by growing without bound: n² = 1, 4, 9, 16, … gets bigger than any number you name. We sometimes write lim n² = ∞, but this means "no limit, because it grows without bound".
- diverge by oscillating: (−1)ⁿ = −1, 1, −1, 1, … keeps jumping and never settles, so it has no limit.
A sequence that alternates can still converge if the jumps shrink. For example, (−1)ⁿ/n = −1, 0.5, −0.333, 0.25, … → 0.
Dividing by the highest power
For a fraction of polynomials in n, divide the top and bottom by the highest power of n in the denominator. Then every term of the form (number)/nᵏ goes to 0.
- Same degree on top and bottom: the limit is the ratio of the leading coefficients.
- Higher degree on the bottom: the limit is 0.
- Higher degree on top: the terms grow without bound, so there is no limit.
It matches how you found horizontal asymptotes by comparing degrees in the MHF4U unit Rational Functions and Equations (for example, y = (ax + b)/(cx + d) has horizontal asymptote y = a/c, the ratio of the leading coefficients). The lesson Limits Involving Infinity and Asymptotes uses it again.
Worked exampleExample 2: Fibonacci ratios (curriculum example)
The Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, …, with tₙ = tₙ₋₁ + tₙ₋₂. Investigate rₙ = tₙ₊₁ ÷ tₙ and find its limit exactly.
- Numerically: r₁ = 1, r₂ = 2, r₃ = 1.5, r₄ ≈ 1.6667, r₅ = 1.6, r₆ = 1.625, r₇ ≈ 1.6154, r₈ ≈ 1.6190, r₉ ≈ 1.6176, r₁₀ ≈ 1.6182, …, r₁₄ = 610 ÷ 377 ≈ 1.618 04.
- The ratios zigzag above and below about 1.618, with smaller and smaller jumps. They seem to converge.
- Exactly: divide the rule tₙ₊₁ = tₙ + tₙ₋₁ by tₙ: tₙ₊₁/tₙ = 1 + tₙ₋₁/tₙ, which says rₙ = 1 + 1/rₙ₋₁.
- If the ratios approach a limit r, then for large n both rₙ and rₙ₋₁ are close to r, so r = 1 + 1/r.
- Multiply by r: r² = r + 1, so r² − r − 1 = 0. By the quadratic formula, r = (1 ± √5)/2. The ratios are positive, so r = (1 + √5)/2.
lim (n → ∞) tₙ₊₁/tₙ = (1 + √5)/2 = φ ≈ 1.618 034, the golden ratio.
Limits of sums
An infinite sum such as 1/2 + 1/4 + 1/8 + … is defined by a limit. Add one term at a time to get the partial sums S₁ = 0.5, S₂ = 0.75, S₃ = 0.875, S₄ = 0.9375, … The partial sums form a sequence, and the infinite sum is its limit. Here the limit is 1.
The same idea explains why 0.999… = 1. The sequence 0.9, 0.99, 0.999, … has limit 1, and "0.999…" means exactly that limit.
Hands-on activityFibonacci ratios with a spreadsheet or calculator
You need: a calculator, or a free spreadsheet (Google Sheets or LibreOffice), and paper.
- In column A, type 1 in A1 and 1 in A2. In A3 type
=A1+A2 and fill down to A30. - In B2 type
=A2/A1 and fill down. (On a calculator, divide each term by the one before.) - Watch the ratios. After how many terms do the first 6 decimal places stop changing?
- Change the starting values to any two positive numbers you like (for example, 7 and 2). What do the ratios approach now?
- In another column, compute (1 + 1/n)ⁿ for n = 1, 10, 100, 1000, 10 000.
What to look for: the Fibonacci ratios lock onto 1.618 034 (to six decimal places) from about the 16th ratio, that is, within about 17 terms. Changing the starting values does not change the limit: any positive starting pair gives φ, because the proof in Example 2 only uses the rule tₙ₊₁ = tₙ + tₙ₋₁. The (1 + 1/n)ⁿ column creeps toward 2.718 28.
Check your understandingQuiz
1. What is lim (n → ∞) (5n + 2)/(n − 1)?
3. The ratios of consecutive Fibonacci numbers approach:
4. What is lim (n → ∞) (1 + 1/n)ⁿ?
5. What is lim (n → ∞) (3n² + 1)/(n³ − 4)?
6. Which statement about 0.999… is true?