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Inleveropdracht uitwerking week 13 fundamenten van de wiskunde $5.93   Add to cart

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Inleveropdracht uitwerking week 13 fundamenten van de wiskunde

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Inleveropdracht uitwerking week 13 fundamenten van de wiskunde.

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  • October 21, 2023
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  • 2021/2022
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Fundamenten uitwerkingen inleveropgave week 13

18 december 2021


Inleveropgave:

Gegeven en te doen opschrijven: 1pt

1
Gegeven: f : (1, ∞) → R gegeven door x → x + x2 .
Te doen: toon aan dat limx→2 f (x) = 92 .

(we merken op dat f welgedefinieerd is voor x ̸= 0 en 0 zit niet in het domein).

Idee van wat we moeten aantonen: 2pt

We willen laten zien dat limx→2 f (x) = 92 .
Volgens de definitie moeten we dan aantonen dat er voor iedere ϵ > 0 er een δ > 0 bestaat
zodat voor iedere x ∈ (1, ∞) − {2} geldt dat |x − 2| < δ impliceert dat |f (x) − 29 | < ϵ.

Opzetje voor ϵ > 0 (willekeurig zodat het voor iedere ϵ > 0 geldt): 0.5pt

Laat ϵ > 0 willekeurig gegeven zijn.

Een correcte delta geven: 1.5pt

Dan nemen we δ = min{1, 5ϵ }.

Uitwerking: 4pt

We beschouwen nu x ∈ (1, ∞) − {2} met |x − 2| < δ.
We bekijken nu

9 1 9 2x3 − 9x + 2 (2 − x)(2x2 + 4x − 1) 2x2 + 4x − 1
|f (x) − | = | + x2 − | = | |=| | = |2 − x| · | |
2 x 2 2x 2x 2x
1
= |x − 2| · |x + 2 − |.
2x
We weten dat x ∈ (2 − δ, 2 + δ) en met δ ≤ 1 dat x ∈ (1, 3) daarmee dat x + 2 ∈ (3, 5)
1
en ook dat 2x ∈ (2, 6), met dit laatste geldt dat 2x ∈ (1/6, 1/2). Hierme vinden we dat
1 1 5 1 1
x + 2 − 2x ∈ (2 2 , 4 6 ), wat weer geeft dat 2 2 < |x + 2 − 2x | < 4 56 .


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