Five Calculus Proofs Using the ε-δ Definition

[Impérialiste]

This is dedicated to the Spirit of G. H. Hardy. May the Sperg Lord play much cricket in Heaven using Trigonometry to become the champion.

Challenge: Disprove any of these "proofs."

All of the five proofs presented premise a definition that lets f: D > ℝ and lets c be an accumulation point of D. A real number L is a limit of f at c if for each ε > 0, there exists a δ > 0 such that |f(x) – L| < ε whenever x ∈ D and 0 < |x – c| < δ.

Proof I

Find a δ > 0 so that |x – 3| < δ implies that |x^2 + 2x – 15| < (1/4).

Let 0 < |x – c| < δ be 0 < |x – 3| < δ.
Let |f(x) – L| < ε be |x^2 + 2x – 15| < (1/4) so ε = (1/4).

|x^2 + 2x – 15| = |(x + 5)(x – 3)| = |x + 5||x – 3| < (1/4).

Let |x – 3| be less than one.

The relationship between |x + 5| and |x – 3| is:
|x + 5| = |x – 3 + 8| ≤ |x – 3| + |8| < 1 + |8| = 1 + 8 = 9.

|x + 5||x – 3| < (1/4); 9*|x – 3|< (1/4); |x - 3| < (1/(9*4)); |x - 3| < (1/36).

If |x – 3| < δ and |x - 3| < (1/36), then δ = (1/36).

QED

[Impérialiste]

Proof II

Find a δ > 0 so that |x – 2| < δ implies that |x^2 – 7x + 10| < (1/3).

Let 0 < |x – c| < δ be 0 < |x – 2| < δ.
Let |f(x) – L| < ε be |x^2 – 7x – 15| < (1/3) so ε = (1/3).

|x^2 – 7x + 10| = |(x – 2)(x – 5)| = |x – 5||x – 2| < (1/3).

Let |x – 2| be less than one.

The relationship between |x – 5| and |x – 2| is:
|x – 5| = |x – 2 - 3| ≤ |x – 2| + |-3| < 1 + |-3| = 1 + 3 = 4.

|x – 5||x – 2| < (1/3); 4*|x – 2|< (1/3); |x - 2| < (1/(3*4)); |x - 2| < (1/12).

If |x – 2| < δ and |x – 2| < (1/12), then δ = (1/12).

QED

[Impérialiste]

Proof III

Prove that the limit as x approaches 5 for (x^2 – 3x + 1) = 11.

Let f(x) = x^2 – 3x + 1, L = 11, and c = 5.
Let 0 < |x – c| < δ be 0 < |x – 5| < δ.
Let |f(x) – L| < ε be |x^2 – 3x + 1 - 11| < ε.

|x^2 – 3x + 1 - 11| = |x^2 – 3x – 10| = |(x – 5)(x + 2)| = |x – 5||x + 2| < ε.

Let |x – 5| be less than 1.

The relationship between |x – 5| and |x + 2| is:
|x + 2| = |x – 5 + 7| ≤ |x – 5| + |7| < 1 + |7| = 1 + 7 = 8.

|x – 5||x + 2| < ε; |x – 5|*8 < ε; 8|x – 5| < ε; |x – 5| < (ε/8).

If |x – 5| < δ and |x – 5| < (ε/8), then δ = (ε/8).

8|x – 5| < ε
|x – 5| < δ

8|x – 5| < ε
(|x – 5| < δ)*-1

8|x – 5| < ε
-|x – 5| < -δ

8|x – 5| - |x – 5| < ε – δ; |x – 5|(8 – 1) < ε – δ; |x – 5|*7 < ε – δ; 7|x – 5| < ε – δ.

Equation 1 under the premise that ε = 8δ:
7|x – 5| < ε – δ; 7|x – 5| < (8δ – δ); 7|x – 5| < 7δ; |x – 5| < δ.

Equation 2 under the premise that δ = (ε/8):
7|x – 5| < ε – δ; 7|x – 5| < ε – (ε/8); 7|x – 5| < (8ε – ε)/ 8; 7|x – 5| < 7ε/8; |x – 5| < (ε/8).

|f(x) – L| = |x^2 – 3x + 1 – 11| = |x – 5||x + 2| < |x – 5|*8 = 8|x – 5| < 8δ = 8(ε/8) = ε.

QED

[Impérialiste]

Proof IV

Prove that the limit as x approaches -2 for (x^2 + 2x + 7) = 7.

Let f(x) = x^2 + 2x + 7, L = 7, and c = -2.
Let 0 < |x – c| < δ be 0 < |x – (-2)| < δ so 0 < |x + 2)| < δ.
Let |f(x) – L| < ε be |x^2 + 2x + 7 - 7| < ε.

|x^2 + 2x + 7 – 7| = |x^2 + 2x| = |x(x + 2)| = |x||x + 2| < ε.

Let |x + 2| be less than 1.

The relationship between |x| and |x + 2| is:
|x| = |x + 2 – 2| ≤ |x + 2| + |-2| < 1 + |-2| = 1 + 2 = 3.

|x||x + 2| < ε; 3*|x + 2| < ε; |x + 2| < (ε/3).

If |x + 2| < δ and |x + 2| < (ε/3), then δ = (ε/3).

3|x + 2| < ε
|x + 2| < δ

3|x + 2| < ε
(|x + 2| < δ)*-1

3|x + 2| < ε
-|x + 2| < -δ

3|x + 2| - |x + 2| < ε – δ; |x + 2|(3 – 1) < ε – δ; |x + 2|(2) < ε – δ; 2|x + 2| < ε – δ.

Equation 1 under the premise that ε = 3δ:
2|x + 2| < ε – δ; 2|x + 2| < 3δ – δ; 2|x + 2| < 2δ; |x + 2| < δ.

Equation 2 under the premise that δ = (ε/3):
2|x + 2| < ε – δ; 2|x + 2| < ε – (ε/3); 2|x + 2| < (3ε – ε)/3; 2|x + 2| < (2ε)/3; |x + 2| < (ε/3).

|f(x) – L| = |x^2 + 2x + 7 – 7| = |x||x + 2| < 3|x + 2| < 3δ = 3(ε/3) = ε.

QED

[Impérialiste]

Proof V

Prove that the limit as x approaches 2 for x^3 = 8.

Let f(x) = x^3, L = 8, and c = 2.
Let 0 < |x – c| < δ be 0 < |x + 2)| < δ.
Let |f(x) – L| < ε be |x^3 - 8| < ε.

|x^3 – 8| = |(x – 2)(x^2 + 2x + 4)| = |x – 2||x^2 + 2x + 4| < ε.

Let |x – 2| be less than 1.

The relationship between |x – 2| and |x^2 + 2x + 4| is:
|x^2 + 2x + 4| = |x(x + 2) + 4| ≤ |x(x + 2)| + |4| = |x||x + 2| + |4|…

{If |x – 2| < 1, then…
|x – 2 + 2| ≤ |x – 2| + |2| < 1 + |2| = 1 + 2 = 3, so |x| < 3.
|x – 2 + 4| ≤ |x – 2| + |4| < 1 + |4| = 1 + 4 = 5, so |x + 2| < 5}.

A revisit: |x||x + 2| + |4| < (3*5) + 4 = 15 + 4 = 19.

|x – 2||x^2 + 2x + 4| < ε; |x – 2|*19 < ε; 19|x – 2| < ε; |x – 2| < (ε/19).

If |x – 2| < δ and |x – 2| < (ε/19), then δ = ε/19.

19|x – 2| < ε
|x – 2| < δ

19|x – 2| < ε
(|x – 2| < δ)*-1

19|x – 2| < ε
-|x – 2| < -δ

19|x – 2| - |x – 2| < ε – δ; |x – 2|(19 – 1) < ε – δ; |x – 2|(18) < ε – δ; 18|x – 2| < ε – δ.

Equation 1 under the premise that ε = 19δ:

18|x – 2| < ε – δ; 18|x – 2| < (19δ) – δ; 18|x – 2| < 18δ; |x – 2| < δ.

Equation 2 under the premise that δ = (ε/19):
18|x – 2| < ε – δ; 18|x – 2| < ε – (ε/19); 18|x – 2| < (19ε – ε)/19; 18|x – 2| < 18ε/19;
|x – 2| < (ε/19).

|f(x) – L| = |x^3 – 8| = |x – 2||x^2 + 2x + 4| < |x – 2|*19 = 19|x – 2| < 19δ = 19(ε/19) = ε.

QED