Let f: R → R be the function f(x) = x². Fill in the blanks in the following proof. Claim: f is continuous at 1 E R. Proof: Fix € We have |x² - 1| = |x + 1||x-1|. If |x-1|< 1 then by the triangle inequality Set 6 = ≤|x-1| +2< {1, €/3}. Then for |x-1| < 6 we have |x + 1| < |x-1| < From this we find x² - 1] = |x+2| 1 3 2 min x+1| max 6 ≤ 3(e/3) = e. 0
Let f: R → R be the function f(x) = x². Fill in the blanks in the following proof. Claim: f is continuous at 1 E R. Proof: Fix € We have |x² - 1| = |x + 1||x-1|. If |x-1|< 1 then by the triangle inequality Set 6 = ≤|x-1| +2< {1, €/3}. Then for |x-1| < 6 we have |x + 1| < |x-1| < From this we find x² - 1] = |x+2| 1 3 2 min x+1| max 6 ≤ 3(e/3) = e. 0
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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![Let f: R → R be the function f(x) = x². Fill in the blanks in the following proof.
Claim: f is continuous at 1 E R.
Proof: Fix €
We have |x² - 1| = |x + 1||x-1|.
If |x-1|< 1 then by the triangle inequality
Set 6 =
≤|x-1| +2<
{1, €/3}. Then for |x-1| < 6 we have |x + 1| <
|x-1| <
From this we find x² - 1] =
|x+2| 1
3
2
min x+1| max
8 ≤ 3(e/3) = e.
0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6fe0f2e6-a3fe-44ed-a932-d5c5d1248f5a%2F4513c574-ecce-4a62-8722-5e71734a6f4a%2Fh4lqav4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Let f: R → R be the function f(x) = x². Fill in the blanks in the following proof.
Claim: f is continuous at 1 E R.
Proof: Fix €
We have |x² - 1| = |x + 1||x-1|.
If |x-1|< 1 then by the triangle inequality
Set 6 =
≤|x-1| +2<
{1, €/3}. Then for |x-1| < 6 we have |x + 1| <
|x-1| <
From this we find x² - 1] =
|x+2| 1
3
2
min x+1| max
8 ≤ 3(e/3) = e.
0
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