Prove from the definition that (3n + 1)/(2n - 5) converges to 3/2: epsilon-N proof, step by step

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The problem

Prove this limit using only the definition of convergence, then try the proof out with the tolerance $\varepsilon = \frac{1}{10}$ .

\lim_{n \to \infty} \frac{3 n + 1}{2 n - 5} = \frac{3}{2}

The result

\lim_{n \to \infty} \frac{3 n + 1}{2 n - 5} = \frac{3}{2}

Proved. Given any tolerance $\varepsilon > 0$ , every natural number $M > \frac{17}{4 \varepsilon} + \frac{5}{2}$ 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 7Step 1: Guess the limit, then say exactly what must be proved

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 = \frac{3}{2}

and

\varepsilon = \frac{1}{10}

. A small triangle at the edge marks a term too far away to fit on the drawing. Position M = 46 lies beyond the right edge of this drawing.

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 = \frac{3}{2}

. A small triangle at the edge marks a term too far away to fit on the drawing.

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 $n$ is enormous.

When $n$ is huge, the fixed numbers $1$ and $-5$ are tiny luggage next to the terms that grow with $n$ . Dividing the top and the bottom by $n$ makes this visible:

x_n = \frac{3 n + 1}{2 n - 5} = \frac{3 + \frac{1}{n}}{2 - \frac{5}{n}}

Divide top and bottom by

n

. This changes nothing, because it is multiplying by

\frac{1/n}{1/n} = 1

As $n$ grows, anything of the form $\frac{\text{fixed number}}{n}$ shrinks toward $0$ , so the fraction heads toward $\frac{3}{2}$ . That is a guess, not yet a proof. The proof is the rest of this page.

\text{Claim: } \lim_{n \to \infty} \frac{3 n + 1}{2 n - 5} = \frac{3}{2}

This is what we will prove, with

L = \frac{3}{2}

L = \frac{3}{2}

and

\varepsilon = \frac{1}{10}

. A small triangle at the edge marks a term too far away to fit on the drawing. Position M = 46 lies beyond the right edge of this drawing.

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 = \frac{3}{2}

. A small triangle at the edge marks a term too far away to fit on the drawing.

Measure the distance between a term and the limit

Why this step

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

\displaystyle \left| \frac{3 n + 1}{2 n - 5} - \frac{3}{2} \right|

=

\displaystyle \left| \frac{2 \left( 3 n + 1 \right) - 3 \left( 2 n - 5 \right)}{2 \left( 2 n - 5 \right)} \right|

Give both fractions the same bottom,

2 \left( 2 n - 5 \right)

: multiply the first fraction top and bottom by

2

, and the second top and bottom by

\left( 2 n - 5 \right)

. Then subtract top from top. checked

=

\displaystyle \left| \frac{6 n + 2 - 6 n + 15}{2 \left( 2 n - 5 \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

=

\displaystyle \left| \frac{2 + 15}{2 \left( 2 n - 5 \right)} \right|

The two terms containing

n

are

6 n

and

-6 n

. They are opposites, so together they make

0

. This is no accident. We chose

L = \frac{3}{2}

precisely so that the growing parts match; any other candidate would leave an

n

on top and the distance would not shrink. checked

=

\displaystyle \left| \frac{17}{2 \left( 2 n - 5 \right)} \right|

Work out the top:

2 + 15 = 17

. checked

=

\displaystyle \frac{17}{2 \left| 2 n - 5 \right|}

The absolute value of a quotient is the quotient of the absolute values:

\left| 17 \right| = 17

, and

2

is already positive so it comes out of the bars unchanged. checked

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 = \frac{3}{2}

. A small triangle at the edge marks a term too far away to fit on the drawing.

Remove the absolute value bars on the bottom

Why this step

Bars block algebra: we cannot multiply an inequality through by $\left| 2 n - 5 \right|$ and solve for $n$ while $n$ is locked inside them. A quantity that is positive equals its own absolute value, so we find where the bottom is positive and work there.

we want

2 n - 5 > 0

This is what we want.

which is true exactly when

2 n > 5

Add

5

to both sides. Adding the same number to both sides never changes which side is bigger. checked

which is true exactly when

n > \frac{5}{2}

Divide both sides by

2

. Dividing by a POSITIVE number keeps the inequality sign as it is. checked

So the bottom is positive exactly when $n > \frac{5}{2}$ , which for natural numbers means $n \ge 3$ . For those $n$ , $\left| 2 n - 5 \right| = 2 n - 5$ . We are free to ignore the first few terms: the definition only cares about what happens from some position $M$ onward, and we will make sure $M \ge 3$ .

\left| \frac{3 n + 1}{2 n - 5} - \frac{3}{2} \right| = \frac{17}{2 \left( 2 n - 5 \right)}

The result of the previous step, with the bars removed, valid for

n \ge 3

. checked

The same measurement further along. The bars ask for a LENGTH, so a term below the limit is no different from one above it; that is why the signs can be cleared without changing what is being measured. Here

L = \frac{3}{2}

. A small triangle at the edge marks a term too far away to fit on the drawing.

Rough work: how large must n be?

Why this step

A proof is READ forwards but FOUND backwards. We start from the inequality we want, $\left| x_n - L \right| < \varepsilon$ , and rewrite it, losing nothing at each move, until it says plainly how large $n$ must be. That tells us which $M$ to announce. This is rough work, not yet the proof.

we want

\frac{17}{2 \left( 2 n - 5 \right)} < \varepsilon

What we want, using the simplified distance.

which is true exactly when

17 < 2 \varepsilon \left( 2 n - 5 \right)

Multiply both sides by

2 \left( 2 n - 5 \right)

. It is positive, so the inequality sign stays as it is. checked

which is true exactly when

\frac{17}{2 \varepsilon} < 2 n - 5

Divide both sides by

2 \varepsilon

, which is positive because

\varepsilon > 0

. checked

which is true exactly when

\frac{17}{2 \varepsilon} + 5 < 2 n

Add

5

to both sides. The same number on both sides never changes which side is bigger. checked

which is true exactly when

n > \frac{17}{4 \varepsilon} + \frac{5}{2}

Divide both sides by

2

, which is positive, and read the inequality from right to left. Now

n

stands alone. checked

The tolerance, drawn: a band of half-width epsilon about the limit. The question this step answers is how far along the sequence one must go before every term is inside it. Here

L = \frac{3}{2}

and

\varepsilon = \frac{1}{10}

. A small triangle at the edge marks a term too far away to fit on the drawing.

Announce the rule for M

Why this step

The sceptic hands us $\varepsilon$ ; the rough work says the terms behave once $n$ is past $\frac{17}{4 \varepsilon} + \frac{5}{2}$ . So we answer with any natural number beyond that value. We must also be sure such a natural number EXISTS, and that is not free: it is a property of the real numbers.

\text{Choose a natural number } M \text{ with } \quad M > \frac{17}{4 \varepsilon} + \frac{5}{2}

Such an

M

exists by the Archimedean property: given any real number, there is a natural number larger than it. The natural numbers never run out below a ceiling.

Notice how the rule behaves: a smaller $\varepsilon$ makes $\frac{17}{4 \varepsilon}$ larger, so $M$ moves further out. A stricter sceptic forces us deeper into the sequence, but never beyond reach. That dependence of $M$ on $\varepsilon$ is the heart of convergence.

This choice also guarantees $M \ge 3$ , as we promised: $M > \frac{17}{4 \varepsilon} + \frac{5}{2} > \frac{5}{2}$ because $\frac{17}{4 \varepsilon}$ is positive.

The answer to the sceptic, drawn: the position M, and the whole tail of the sequence from it onward. The rule must produce a line like this for ANY band the sceptic chooses, however narrow. Here

L = \frac{3}{2}

and

\varepsilon = \frac{1}{10}

. A small triangle at the edge marks a term too far away to fit on the drawing.

The proof, written forwards

Why this step

Rough work convinces US; a proof must convince the sceptic, and it has to run in the order the definition demands: first $\varepsilon$ arrives, then we name $M$ , then we take ANY $n \ge M$ and show the distance is under $\varepsilon$ . Here it is, with every relation justified.

Let $\varepsilon > 0$ be given. Choose $M \in \mathbb{N}$ with $M > \frac{17}{4 \varepsilon} + \frac{5}{2}$ . Let $n \ge M$ . (Here $\mathbb{N}$ is the set of natural numbers $1, 2, 3, \ldots$ and $\in$ means "is a member of".)

First, a lower bound for the bottom:

\displaystyle 2 n - 5

\ge

\displaystyle 2 M - 5

Because

n \ge M

and the number in front,

2

, is positive: a larger

n

gives a larger bottom. checked

>

\displaystyle 2 \left( \frac{17}{4 \varepsilon} + \frac{5}{2} \right) - 5

Because

M > \frac{17}{4 \varepsilon} + \frac{5}{2}

: replace

M

by this smaller value and the expression gets smaller. checked

=

\displaystyle \frac{17}{2 \varepsilon} + 5 - 5

Multiply out the bracket:

2 \cdot \frac{17}{4 \varepsilon} = \frac{17}{2 \varepsilon}

, and

2 \cdot \frac{5}{2} = 5

. checked

=

\displaystyle \frac{17}{2 \varepsilon}

The

+5

and the

-5

are opposites and make

0

. checked

>

\displaystyle 0

A positive number divided by a positive number is positive. checked

So the bottom is positive, which lets us use the bar-free distance, and it is LARGER than $\frac{17}{2 \varepsilon}$ . Now the distance itself:

\displaystyle \left| \frac{3 n + 1}{2 n - 5} - \frac{3}{2} \right|

=

\displaystyle \frac{17}{2 \left( 2 n - 5 \right)}

The simplified distance from before; the bottom is positive, so the bars are gone. checked

<

\displaystyle \frac{17}{2 \cdot \frac{17}{2 \varepsilon}}

We replaced

2 n - 5

by the smaller positive number

\frac{17}{2 \varepsilon}

. Dividing by something smaller gives something BIGGER, so the fraction grew. checked

=

\displaystyle \frac{17}{\frac{17}{\varepsilon}}

On the bottom,

2 \cdot \frac{17}{2 \varepsilon} = \frac{17}{\varepsilon}

: the

2

on top cancels the

2

underneath. checked

=

\displaystyle \varepsilon

Dividing by a fraction is multiplying by its flip:

17 \cdot \frac{\varepsilon}{17} = \varepsilon

. checked

What the proof has established, for this tolerance: every term from M onward is inside the band. The proof is the promise that the same picture can be drawn for every epsilon, not just this one. Here

L = \frac{3}{2}

and

\varepsilon = \frac{1}{10}

. Green points are inside the band; red points are outside it. A small triangle at the edge marks a term too far away to fit on the drawing.

Test the rule with a real tolerance

Why this step

A rule for $M$ that has never been run on a number is only a claim. We play the sceptic ourselves, pick $\varepsilon = \frac{1}{10}$ , and check the terms on both sides of the position our rule gives.

\frac{17}{4 \varepsilon} + \frac{5}{2} = 45

Put

\varepsilon = \frac{1}{10}

into the rule. checked

The smallest natural number greater than that is $M = 46$ . The table computes the terms around that position exactly.

$n$	$x_n$	$\left\| x_n - L \right\|$	under $\varepsilon$ ?
$45$	$\frac{8}{5}$	$\frac{1}{10} = 0.1$	no
$46$	$\frac{139}{87}$	$\frac{17}{174} \approx 0.097701$	yes
$47$	$\frac{142}{89}$	$\frac{17}{178} \approx 0.095506$	yes
$48$	$\frac{145}{91}$	$\frac{17}{182} \approx 0.093407$	yes

Terms around position M = 46, tolerance 1/10

The terms of the sequence, the limit, and the band of half-width epsilon around it. From position M onward every term lies inside the band. Here

L = \frac{3}{2}

and

\varepsilon = \frac{1}{10}

. Green points are inside the band; red points are outside it. A small triangle at the edge marks a term too far away to fit on the drawing.

Limit of a sequence from the definition