Epsilon-delta proof that the limit of x^2 - 3x + 5 as x approaches 2 is 3, step by step

1

Find the candidate limit, then say exactly what must be proved

Why this step

The definition can only CHECK a number we propose; it never hands us one. For a polynomial the natural candidate is the value at the point itself, because a polynomial is built from nothing but adding and multiplying, and neither of those can make a sudden jump. That is a belief. The proof turns it into a fact.

Put $x = 2$ into the function:

\displaystyle 2^{2} - 3 \cdot 2 + 5

=

\displaystyle 4 - 3 \cdot 2 + 5

Work out each power of

2

first. Powers come before multiplying, and multiplying before adding. checked

=

\displaystyle 4 - 6 + 5

Do each multiplication. checked

=

\displaystyle 3

Add the numbers up. checked

\text{Claim: } \lim_{x \to 2} \left( x^{2} - 3 x + 5 \right) = 3

This is what we will prove, with

c = 2

and

L = 3

.

The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here

c = 2

,

L = 3

,

\varepsilon = \frac{1}{10}

and

\delta = \frac{1}{20}

. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.

2

Rewrite everything in terms of the distance from c

Why this step

We control one thing only: how far $x$ is from $2$ . But $f$ is written in powers of $x$ , which hides that distance. So we give the distance a name, $h = x - 2$ , and rewrite the function in powers of $h$ . After this the quantity we control is the only letter on the page.

From $h = x - 2$ we get $x = 2 + h$ . The condition $0 < \left| x - c \right| < \delta$ becomes simply $0 < \left| h \right| < \delta$ .

\displaystyle x^{2} - 3 x + 5 - 3

=

\displaystyle \left( 2 + h \right)^{2} - 3 \left( 2 + h \right) + 5 - 3

Replace every

x

by

2 + h

. checked

=

\displaystyle \left( 4 + 4 h + h^{2} \right) - 3 \left( 2 + h \right) + 5 - 3

Multiply out the square:

(u + h)^2 = u^2 + 2uh + h^2

with

u = 2

. checked

=

\displaystyle h^{2} + 4 h + 4 - 3 h - 6 + 5 - 3

Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked

=

\displaystyle h^{2} + h

Collect the terms with the same power of

h

. The plain numbers add up to exactly

0

, and that is no accident: they are

f(2) - L

, and we CHOSE

L = f(2)

. A wrong candidate for

L

would leave a number here that no

\delta

could ever shrink. checked

=

\displaystyle h \left( h + 1 \right)

Every remaining term contains

h

, so take

h

out as a common factor. checked

\left| x^{2} - 3 x + 5 - 3 \right| = \left| h \right| \cdot \left| h + 1 \right|

Take absolute values. The absolute value of a product is the product of the absolute values. checked

The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here

c = 2

,

L = 3

,

\varepsilon = \frac{1}{10}

and

\delta = \frac{1}{20}

. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.

3

Cage the second factor

Why this step

$\left| h + 1 \right|$ depends on $h$ , so it has no single value. But we do not need its value, only a ceiling it can never pass. A ceiling exists only if $h$ is kept inside some fixed region, so we make a promise now: whatever $\delta$ we finally choose, it will be at most $1$ . Then $\left| h \right| < 1$ .

\displaystyle \left| h + 1 \right|

\le

\displaystyle \left| h \right| + \left| 1 \right|

The triangle inequality. checked

=

\displaystyle \left| h \right| + 1

In each term, the absolute value of a product is the product of the absolute values. checked

\le

\displaystyle 1 + 1

Because

\left| h \right| < 1

: a number below

1

only gets smaller when raised to a power, so every

\left| h \right|

and

\left| h \right|^2

is at most

1

. checked

=

\displaystyle 2

Add the numbers up. checked

The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here

c = 2

,

L = 3

,

\varepsilon = \frac{1}{10}

and

\delta = \frac{1}{20}

. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.

4

Rough work: choose delta

Why this step

Inside the cage the output distance is at most $2 \left| h \right|$ . That is now a straight-line problem: a fixed number times the input distance. We ask how small $\left| h \right|$ must be to push it under $\varepsilon$ .

we want

2 \left| h \right| < \varepsilon

What would be enough.

which is true exactly when

\left| h \right| < \frac{\varepsilon}{2}

Divide both sides by

2

, which is positive, so the sign stays. checked

\text{Choose } \delta = \min\left\{ 1, \; \frac{\varepsilon}{2} \right\}

\min

means the smaller of the two numbers. It is positive because both

1

and

\frac{\varepsilon}{2}

are.

The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here

c = 2

,

L = 3

,

\varepsilon = \frac{1}{10}

and

\delta = \frac{1}{20}

. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.

5

The proof, written forwards

Why this step

Rough work convinces us; a proof must convince the sceptic, in the order the definition demands: $\varepsilon$ arrives first, we name $\delta$ , then ANY permitted $x$ must obey. Every relation below is justified.

Let $\varepsilon > 0$ be given. Choose $\delta = \min\left\{ 1, \; \frac{\varepsilon}{2} \right\}$ . Suppose $0 < \left| x - 2 \right| < \delta$ , and write $h = x - 2$ .

Since $\delta \le 1$ we have $\left| h \right| < 1$ , so the cage holds. Since $\delta \le \frac{\varepsilon}{2}$ we have $\left| h \right| < \frac{\varepsilon}{2}$ . Now the distance itself:

\displaystyle \left| x^{2} - 3 x + 5 - 3 \right|

=

\displaystyle \left| h \right| \cdot \left| h + 1 \right|

The rewritten distance. checked

\le

\displaystyle \left| h \right| \cdot 2

The cage: the second factor is at most

2

. Multiplying the non-negative number

\left| h \right|

by something larger cannot make the product smaller. checked

<

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

Because

\left| h \right| < \frac{\varepsilon}{2}

, and multiplying both sides by the positive number

2

keeps the sign. checked

=

\displaystyle \varepsilon

The

2

on top cancels the

2

underneath. checked

The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here

c = 2

,

L = 3

,

\varepsilon = \frac{1}{10}

and

\delta = \frac{1}{20}

. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.

6

Test the rule with a real tolerance

Why this step

A rule for $\delta$ that has never been run on a number is only a claim. We play the sceptic ourselves, pick $\varepsilon = \frac{1}{10}$ , and check inputs spread across the whole permitted window.

With $\varepsilon = \frac{1}{10}$ the rule $\delta = \min\left\{ 1, \; \frac{\varepsilon}{2} \right\}$ gives $\delta = \frac{1}{20}$ , which is $0.05$ . Every $x$ below is inside the window $0 < \left| x - 2 \right| < \delta$ , and the outputs are computed exactly.

$x$	$f(x)$	$\left\| f(x) - L \right\|$	under $\varepsilon$ ?
$1.9505$	$2.95295025$	$0.04704975$	yes
$1.975$	$2.975625$	$0.024375$	yes
$2.005$	$3.005025$	$0.005025$	yes
$2.025$	$3.025625$	$0.025625$	yes
$2.0495$	$3.05195025$	$0.05195025$	yes

Inputs across the window, tolerance 1/10. Decimals are rounded for display; the comparison uses exact fractions.

The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here

c = 2

,

L = 3

,

\varepsilon = \frac{1}{10}

and

\delta = \frac{1}{20}

. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.

Limit of a function from the epsilon-delta definition