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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→∞10n−20−8n+12=−54\lim_{n \to \infty} \frac{10 n - 20}{-8 n + 12} = -\frac{5}{4}
Invent a new one:
Femtobytes
Prove from the definition that (10n - 20)/(-8n + 12) converges to -5/4: epsilon-N proof, step by step
femtobytes.com/solve/epsilon-n-limit/10_-20_-8_12_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→∞10n−20−8n+12=−54\lim_{n \to \infty} \frac{10 n - 20}{-8 n + 12} = -\frac{5}{4}
The result
lim⁡n→∞10n−20−8n+12=−54\lim_{n \to \infty} \frac{10 n - 20}{-8 n + 12} = -\frac{5}{4}

Proved. Given any tolerance ε>0\varepsilon > 0, every natural number M>58ε+32M > \frac{5}{8 \varepsilon} + \frac{3}{2} works.

checked 24 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.
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 −20-20 and 1212 are tiny luggage next to the terms that grow with nn. Dividing the top and the bottom by nn makes this visible:

xn=10n−20−8n+12=10−20n−8+12nx_n = \frac{10 n - 20}{-8 n + 12} = \frac{10 - \frac{20}{n}}{-8 + \frac{12}{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 10−8=−54\frac{10}{-8} = -\frac{5}{4}. That is a guess, not yet a proof. The proof is the rest of this page.

M for ε102030405060708090100110120130140150160What 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.-1.3-1.28-1.22-1.2-1.25-1.24-1.26n, the position in the sequence (1st term, 2nd term, and so on)the value of the term xₙLL + εL − εx65 = -1.24, the first inside
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=−54L = -\frac{5}{4} and ε=1100\varepsilon = \frac{1}{100}. A small triangle at the edge marks a term too far away to fit on the drawing. Position M = 65 lies beyond the right edge of this drawing.
Claim: lim⁡n→∞10n−20−8n+12=−54\text{Claim: } \lim_{n \to \infty} \frac{10 n - 20}{-8 n + 12} = -\frac{5}{4}
This is what we will prove, with L=−54L = -\frac{5}{4}.
2468101214161820The terms of the sequence, and the level they approachL is the limit being claimed.-20-1.25n, 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=−54L = -\frac{5}{4}. A small triangle at the edge marks a term too far away to fit on the drawing.
2

Make the bottom grow upward instead of downward

Why this step

The bottom −8n+12-8 n + 12 becomes more and more NEGATIVE as nn grows. Later we will multiply and divide inequalities by the bottom, and doing that with a negative quantity flips the inequality sign every time. One move now removes that trap for the rest of the proof.

10n−20−8n+12\displaystyle \frac{10 n - 20}{-8 n + 12}==−1(10n−20)−1(−8n+12)\displaystyle \frac{-1 \left( 10 n - 20 \right)}{-1 \left( -8 n + 12 \right)}
Multiply top and bottom by −1-1. The value is unchanged because −1−1=1\frac{-1}{-1} = 1. checked
==−10n+208n−12\displaystyle \frac{-10 n + 20}{8 n - 12}
Multiply out: every sign on top and on the bottom switches. checked

From here on we use this form of xnx_n. It is the same sequence, term for term.

3

Measure the distance between a term and the limit

4

Remove the absolute value bars on the bottom

5

Rough work: how large must n be?

6

Announce the rule for M

7

The proof, written forwards

8

Test the rule with a real tolerance