x2, x+2 If (a) f(x) = x² and (b) f(x)= 0, prove that lim f(x) = 4. x2 x = 2 (a) We must show that, given any e > 0, we can find 8 > 0 (depending on e in general) such that |x - 4|
x2, x+2 If (a) f(x) = x² and (b) f(x)= 0, prove that lim f(x) = 4. x2 x = 2 (a) We must show that, given any e > 0, we can find 8 > 0 (depending on e in general) such that |x - 4|
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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
3.10) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.
![Limits
(x², x+2
If (a) f(x) = x² and (b) f(x)=
(0,
prove that lim f(x) = 4.
x→2
3.10.
x = 2
(a) We must show that, given any e > 0, we can find 8 > 0 (depending on e in general) such that |x - 4|<e
when 0 < |x – 2| < 8.
Choose 8 < 1 so that 0< |x- 2| <1 or 1<x<3, x + 2. Then |x² – 4| = |(x– 2)(x + 2)| = |x– 2|
|x + 2| < 8]x+2] <58.
Take 8 as 1 or e/5, whichever is smaller. Then we have |x² – 4| < e whenever 0< |x– 2| < 8, and the
required result is proved.
It is of interest to consider some numerical values. If, for example, we wish to make |x – 4| <.05,
we can choose d = €/5 = .05/5 = .01. To see that this is actually the case, note that if 0 < |x – 2| < .01,
then 1.99 <x< 2.01 (x + 2), and so 3.9601 <x² < 4.0401, – .0399 <x² – 4< .0401, and certainly |x –- |
<.05 (x + 4). The fact that these inequalities also happen to hold at x = 2 is merely coincidental.
If we wish to make |x2 – 4| < 6, we can choose 8 = 1, and this will be satisfied.
(b) There is no difference between the proof for this case and the proof in (a), since in both cases we exclude
x= 2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad530511-9758-42d9-9994-894bc4eea3f3%2F618bcce8-d9c5-4301-87f0-27c02ee1e956%2Feqpdsbn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Limits
(x², x+2
If (a) f(x) = x² and (b) f(x)=
(0,
prove that lim f(x) = 4.
x→2
3.10.
x = 2
(a) We must show that, given any e > 0, we can find 8 > 0 (depending on e in general) such that |x - 4|<e
when 0 < |x – 2| < 8.
Choose 8 < 1 so that 0< |x- 2| <1 or 1<x<3, x + 2. Then |x² – 4| = |(x– 2)(x + 2)| = |x– 2|
|x + 2| < 8]x+2] <58.
Take 8 as 1 or e/5, whichever is smaller. Then we have |x² – 4| < e whenever 0< |x– 2| < 8, and the
required result is proved.
It is of interest to consider some numerical values. If, for example, we wish to make |x – 4| <.05,
we can choose d = €/5 = .05/5 = .01. To see that this is actually the case, note that if 0 < |x – 2| < .01,
then 1.99 <x< 2.01 (x + 2), and so 3.9601 <x² < 4.0401, – .0399 <x² – 4< .0401, and certainly |x –- |
<.05 (x + 4). The fact that these inequalities also happen to hold at x = 2 is merely coincidental.
If we wish to make |x2 – 4| < 6, we can choose 8 = 1, and this will be satisfied.
(b) There is no difference between the proof for this case and the proof in (a), since in both cases we exclude
x= 2.
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