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(x−7)(x−3)* Give a step-by-step e-6 proof that lim Q(x) = Q(2). #-2

(x−7)(x−3)* Give a step-by-step e-6 proof that lim Q(x) = Q(2). #-2

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Wksp 5 Q3

PLEASE PLEASE include labeled definition(AS IT APPLIES TO THE QUESTION), scratchwork, graph/sign-chart and proof 

 

5(2x+1) Let Q(x) = (x-7)(x-3)* Give a step-by-step - proof that lim Q(x) = Q(2). Z→2 Start by stating the appropriate definition with the given values substituted (this is the definition for a rational function having real limit at a point, i.e. the - definition). Definition. Next, sketch the graph or use a sign chart to understand the behaviour near 2. Use this knowledge to do the scratch work to determine & based on . (Hint: simplify |Q(z) - Q(2)| to a rational function R(z) times |x - 2). The r - 2| can be controlled by 6, while the rational function R(x) must be controlled by introducing a temporary bound). Scratch work.
Transcribed Image Text:5(2x+1) Let Q(x) = (x-7)(x-3)* Give a step-by-step - proof that lim Q(x) = Q(2). Z→2 Start by stating the appropriate definition with the given values substituted (this is the definition for a rational function having real limit at a point, i.e. the - definition). Definition. Next, sketch the graph or use a sign chart to understand the behaviour near 2. Use this knowledge to do the scratch work to determine & based on . (Hint: simplify |Q(z) - Q(2)| to a rational function R(z) times |x - 2). The r - 2| can be controlled by 6, while the rational function R(x) must be controlled by introducing a temporary bound). Scratch work.
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can you please write this part in handwriting? the characters are in different format so im having a hard time understanding

 

Given function: Q(x)=52x+1x-7x-3

To show: limx→2Q(x)=Q(2) byε-δ method.

 

Definition: ∀ε ∃δ  |x-a|<δ  ⇒  |f(x)-f(a)|<ε

Q(x)=52x+1x-7x-3

Now, Q(2)=55-5-1=5

Then Q(x)-Q(2)=52x+1x-7x-3-5=-5x-7x-3+52x+1x-7x-3=-5x2+60x-100x-7x-3

Choose δ≤ε where -5x2+60x-100x-7x-3<δ

Therefore Q(x)-Q(2)=-5x2+60x-100x-7x-3<ε

Thus, Q(x)-Q(2)<ε whenever -5x2+60x-100x-7x-3<δ

Q(x) is continuous at x=2

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