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Problem type

Limit of a sequence from the definition

What it solves
Prove that a sequence of the form (an + b) / (cn + d) converges to its limit, using only the definition of convergence: for every tolerance epsilon, find the position M after which every term is within epsilon of the limit.
Subject
Real Analysis
Topic
Limits from the definition
Also called
epsilon N, convergence of a sequence, sequence converges, limit of a sequence
Your problem

Prove this limit using only the definition of convergence, then try the proof out with the tolerance ε=1100\varepsilon = \frac{1}{100}.

lim⁡n→∞2n−114n−6=12\lim_{n \to \infty} \frac{2 n - 11}{4 n - 6} = \frac{1}{2}
Invent a new one:
Femtobytes
Prove from the definition that (2n - 11)/(4n - 6) converges to 1/2: epsilon-N proof, step by step
femtobytes.com/solve/epsilon-n-limit/2_-11_4_-6_1~100
The problem

Prove this limit using only the definition of convergence, then try the proof out with the tolerance ε=1100\varepsilon = \frac{1}{100}.

lim⁡n→∞2n−114n−6=12\lim_{n \to \infty} \frac{2 n - 11}{4 n - 6} = \frac{1}{2}
The result
lim⁡n→∞2n−114n−6=12\lim_{n \to \infty} \frac{2 n - 11}{4 n - 6} = \frac{1}{2}

Proved. Given any tolerance ε>0\varepsilon > 0, every natural number M>84ε+32M > \frac{8}{4 \varepsilon} + \frac{3}{2} works.

checked 22 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.
Picture 1 of 2Step 1: Guess the limit, then say exactly what must be proved
102030405060708090100110120130140150160What the definition is asking for∀ε > 0 (for EVERY tolerance, however small) ∃M (there is a position) ∀n ≥ M (every term from there on): |xₙ − L| < ε (is inside)Shrink the tolerance and M moves further out. That dependence of M on ε IS the definition.0.440.460.480.520.540.560.50.510.49n, the position in the sequence (1st term, 2nd term, and so on)the value of the term xₙLL + εL − ε
The definition drawn twice over. The wider band is one tolerance and the line marked for it is the position from which every term stays inside. The narrower band is a stricter tolerance, and its position sits further out. The definition demands that a position can be found for EVERY tolerance, however small, which is why no single picture can prove it and why the rest of this page is an argument rather than a drawing. Here L=12L = \frac{1}{2} and ε=1100\varepsilon = \frac{1}{100}. A small triangle at the edge marks a term too far away to fit on the drawing.
2468101214161820The terms of the sequence, and the level they approachL is the limit being claimed.-4-2240.5n, the position in the sequence (1st term, 2nd term, and so on)the value of the term xₙL
The terms of the sequence, and the level they seem to be heading for. Seeming is not proof: the rest of this is what turns a picture like this into one. Here L=12L = \frac{1}{2}. A small triangle at the edge marks a term too far away to fit on the drawing.
1

Guess the limit, then say exactly what must be proved

Why this step

A proof from the definition needs a target before it can start: the definition only lets us CHECK a candidate limit, it never produces one. So we first find the candidate by asking what happens to the fraction when nn is enormous.

When nn is huge, the fixed numbers −11-11 and −6-6 are tiny luggage next to the terms that grow with nn. Dividing the top and the bottom by nn makes this visible:

xn=2n−114n−6=2−11n4−6nx_n = \frac{2 n - 11}{4 n - 6} = \frac{2 - \frac{11}{n}}{4 - \frac{6}{n}}
Divide top and bottom by nn. This changes nothing, because it is multiplying by 1/n1/n=1\frac{1/n}{1/n} = 1.

As nn grows, anything of the form fixed numbern\frac{\text{fixed number}}{n} shrinks toward 00, so the fraction heads toward 24=12\frac{2}{4} = \frac{1}{2}. That is a guess, not yet a proof. The proof is the rest of this page.

Claim: lim⁡n→∞2n−114n−6=12\text{Claim: } \lim_{n \to \infty} \frac{2 n - 11}{4 n - 6} = \frac{1}{2}
This is what we will prove, with L=12L = \frac{1}{2}.
102030405060708090100110120130140150160What the definition is asking for∀ε > 0 (for EVERY tolerance, however small) ∃M (there is a position) ∀n ≥ M (every term from there on): |xₙ − L| < ε (is inside)Shrink the tolerance and M moves further out. That dependence of M on ε IS the definition.0.440.460.480.520.540.560.50.510.49n, the position in the sequence (1st term, 2nd term, and so on)the value of the term xₙLL + εL − ε
The definition drawn twice over. The wider band is one tolerance and the line marked for it is the position from which every term stays inside. The narrower band is a stricter tolerance, and its position sits further out. The definition demands that a position can be found for EVERY tolerance, however small, which is why no single picture can prove it and why the rest of this page is an argument rather than a drawing. Here L=12L = \frac{1}{2} and ε=1100\varepsilon = \frac{1}{100}. A small triangle at the edge marks a term too far away to fit on the drawing.
2468101214161820The terms of the sequence, and the level they approachL is the limit being claimed.-4-2240.5n, the position in the sequence (1st term, 2nd term, and so on)the value of the term xₙL
The terms of the sequence, and the level they seem to be heading for. Seeming is not proof: the rest of this is what turns a picture like this into one. Here L=12L = \frac{1}{2}. A small triangle at the edge marks a term too far away to fit on the drawing.
2

Measure the distance between a term and the limit

Why this step

The definition talks about one quantity only, the distance ∣xn−L∣\left| x_n - L \right|. Right now it is a difference of two fractions, and we cannot see how it depends on nn. We simplify it until nn appears in exactly one place, because then we can solve for nn.

∣2n−114n−6−12∣\displaystyle \left| \frac{2 n - 11}{4 n - 6} - \frac{1}{2} \right|==∣2(2n−11)−1(4n−6)2(4n−6)∣\displaystyle \left| \frac{2 \left( 2 n - 11 \right) - 1 \left( 4 n - 6 \right)}{2 \left( 4 n - 6 \right)} \right|
Give both fractions the same bottom, 2(4n−6)2 \left( 4 n - 6 \right): multiply the first fraction top and bottom by 22, and the second top and bottom by (4n−6)\left( 4 n - 6 \right). Then subtract top from top. checked
==∣4n−22−4n+62(4n−6)∣\displaystyle \left| \frac{4 n - 22 - 4 n + 6}{2 \left( 4 n - 6 \right)} \right|
Multiply out the brackets on top. Take care with the second bracket: the minus sign in front of it reaches every term inside. checked
==∣−22+62(4n−6)∣\displaystyle \left| \frac{-22 + 6}{2 \left( 4 n - 6 \right)} \right|
The two terms containing nn are 4n4 n and −4n-4 n. They are opposites, so together they make 00. This is no accident. We chose L=12L = \frac{1}{2} precisely so that the growing parts match; any other candidate would leave an nn on top and the distance would not shrink. checked
==∣−162(4n−6)∣\displaystyle \left| \frac{-16}{2 \left( 4 n - 6 \right)} \right|
Work out the top: −22+6=−16-22 + 6 = -16. checked
==162∣4n−6∣\displaystyle \frac{16}{2 \left| 4 n - 6 \right|}
The absolute value of a quotient is the quotient of the absolute values: ∣−16∣=16\left| -16 \right| = 16, and 22 is already positive so it comes out of the bars unchanged. checked
==8∣4n−6∣\displaystyle \frac{8}{\left| 4 n - 6 \right|}
Divide top and bottom by their common factor 22. checked
2468101214161820How far one term is from the limitL is the limit being claimed.-4-2240.5n, the position in the sequence (1st term, 2nd term, and so on)the value of the term xₙL|x3 − L| = 1.333x3 = -0.8333
The distance between one term and the limit, measured on the drawing. That measured length is what the bars mean, and it is the thing the whole proof must get under any tolerance. Here L=12L = \frac{1}{2}. A small triangle at the edge marks a term too far away to fit on the drawing.
3

Remove the absolute value bars on the bottom

4

Rough work: how large must n be?

5

Announce the rule for M

6

The proof, written forwards

7

Test the rule with a real tolerance