MAT 137 LEC 0601 Instructor: Alessandro Malus TA: Muhammad Mohid - PowerPoint PPT Presentation

MAT 137 — LEC 0601 Instructor: Alessandro Malusà TA: Muhammad Mohid October 1st, 2020 Warm-up question : Suppose you need to prove that x → a f ( x ) = L . lim 1 Write down the formal definition of the statement; 2 Write down what the structure of the formal proof should be. This is for your reference: please wait before sharing your answer.

Preparation: choosing deltas 1 Find one value of δ > 0 such that | x − 3 | < δ = ⇒ | 5 x − 15 | < 1 . 2 Find all values of δ > 0 such that | x − 3 | < δ = ⇒ | 5 x − 15 | < 1 . 3 Find all values of δ > 0 such that | x − 3 | < δ = ⇒ | 5 x − 15 | < 0 . 1 . 4 Let us fix ε > 0. Find all values of δ > 0 such that | x − 3 | < δ = ⇒ | 5 x − 15 | < ε.

What is wrong with this “proof"? Prove that x → 3 (5 x + 1) = 16 lim “Proof:" Let ε > 0. WTS ∀ ε > 0, ∃ δ > 0 s.t. 0 < | x − 3 | < δ = ⇒ | (5 x + 1) − (16) | < ε | (5 x + 1) − (16) | < ε ⇐ ⇒ | 5 x + 15 | < ε ⇒ δ = ε ⇐ ⇒ 5 | x + 3 | < ε = 3 �

Your first ε − δ proof Goal We want to prove that x → 3 (5 x + 1) = 16 lim directly from the definition.

Your first ε − δ proof Goal We want to prove that x → 3 (5 x + 1) = 16 lim directly from the definition. 1 Write down the formal definition of the statement ( ?? ).

Your first ε − δ proof Goal We want to prove that x → 3 (5 x + 1) = 16 lim directly from the definition. 1 Write down the formal definition of the statement ( ?? ). 2 Write down what the structure of the formal proof should be, without filling the details.

Your first ε − δ proof Goal We want to prove that x → 3 (5 x + 1) = 16 lim directly from the definition. 1 Write down the formal definition of the statement ( ?? ). 2 Write down what the structure of the formal proof should be, without filling the details. 3 Write down a complete formal proof.

A harder proof Goal We want to prove that x 3 + x 2 � � lim = 0 x → 0 directly from the definition.

A harder proof Goal We want to prove that x 3 + x 2 � � lim = 0 x → 0 directly from the definition. 1 Write down the formal definition of the statement ( ?? ).

A harder proof Goal We want to prove that x 3 + x 2 � � lim = 0 x → 0 directly from the definition. 1 Write down the formal definition of the statement ( ?? ). 2 Write down what the structure of the formal proof should be, without filling the details.

A harder proof Goal We want to prove that x 3 + x 2 � � lim = 0 x → 0 directly from the definition. 1 Write down the formal definition of the statement ( ?? ). 2 Write down what the structure of the formal proof should be, without filling the details. 3 Rough work: What is δ ?

A harder proof Goal We want to prove that x 3 + x 2 � � lim = 0 x → 0 directly from the definition. 1 Write down the formal definition of the statement ( ?? ). 2 Write down what the structure of the formal proof should be, without filling the details. 3 Rough work: What is δ ? 4 Write down a complete formal proof.

Is this proof correct? Claim: | x 3 + x 2 | < ε. ∀ ε > 0 , ∃ δ > 0 s.t. 0 < | x | < δ = ⇒ Proof: • Let ε > 0. � ε • Take δ = | x + 1 | . • Let x ∈ R . Assume 0 < | x | < δ . Then ε | x 3 + x 2 | = x 2 | x + 1 | < δ 2 | x + 1 | = | x + 1 || x + 1 | = ε. • I have proven that | x 3 + x 2 | < ε .

Choosing deltas again Let us fix numbers A , ε > 0. Find: ⇒ | Ax 2 | < ε 1 a value of δ > 0 s.t. | x | < δ =

Choosing deltas again Let us fix numbers A , ε > 0. Find: ⇒ | Ax 2 | < ε 1 a value of δ > 0 s.t. | x | < δ = ⇒ | Ax 2 | < ε 2 all values of δ > 0 s.t. | x | < δ =

Choosing deltas again Let us fix numbers A , ε > 0. Find: ⇒ | Ax 2 | < ε 1 a value of δ > 0 s.t. | x | < δ = ⇒ | Ax 2 | < ε 2 all values of δ > 0 s.t. | x | < δ = 3 a value of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10

Choosing deltas again Let us fix numbers A , ε > 0. Find: ⇒ | Ax 2 | < ε 1 a value of δ > 0 s.t. | x | < δ = ⇒ | Ax 2 | < ε 2 all values of δ > 0 s.t. | x | < δ = 3 a value of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10 4 all values of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10

Choosing deltas again Let us fix numbers A , ε > 0. Find: ⇒ | Ax 2 | < ε 1 a value of δ > 0 s.t. | x | < δ = ⇒ | Ax 2 | < ε 2 all values of δ > 0 s.t. | x | < δ = 3 a value of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10 4 all values of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10 � | Ax 2 | < ε 5 a value of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10

Choosing deltas again Let us fix numbers A , ε > 0. Find: ⇒ | Ax 2 | < ε 1 a value of δ > 0 s.t. | x | < δ = ⇒ | Ax 2 | < ε 2 all values of δ > 0 s.t. | x | < δ = 3 a value of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10 4 all values of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10 � | Ax 2 | < ε 5 a value of δ > 0 s.t. | x | < δ = ⇒ | x + 1 | < 10 ⇒ | ( x + 1) x 2 | < ε 6 a value of δ > 0 s.t. | x | < δ =

Before next class... • Watch videos 2.5, 2.6. • Download the next class’s slides (no need to look at them!)

Before next class... • Watch videos 2.5, 2.6. • Download the next class’s slides (no need to look at them!)