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Let Q(x) = 4+2). Give a step-by-step - proof that _lim Q(x) = -1. 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 e-6 definition). Definition.

Let Q(x) = 4+2). Give a step-by-step - proof that _lim Q(x) = -1. 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 e-6 definition). Definition.

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 Q2

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

. Let Q(x) = 4(+2). Give a step-by-step - proof that lim Q(x) = -1. 24-2 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 e. (Hint: factoring the denominator simplifies the expression Q(x) − (−1)| to. The numerator can be controlled by 6, while the denominator must be controlled by introducing a temporary bound). Scratch work.
Transcribed Image Text:. Let Q(x) = 4(+2). Give a step-by-step - proof that lim Q(x) = -1. 24-2 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 e. (Hint: factoring the denominator simplifies the expression Q(x) − (−1)| to. The numerator can be controlled by 6, while the denominator must be controlled by introducing a temporary bound). Scratch work.
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would you please write this part in handwriting? because im not sure i understand.

Let Qx=4x+2x2-4=4x-2

Then Q-2=4-2-2=-1

Now, Qx-Q-2=4x-2+1=x+2x-2

Choose δ≤ε where x+2x-2<δ

Qx-Q-2=x+2x-2<ε

Therefore Qx-Q-2<ε whenever x+2x-2<δ

Therefore, Q(x) is continuous at x=-2

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