2. Is the proof correct? Either state that it is, or circle the first error and explain what isincorrect about it. If the proof is not correct, can it be fixed to prove the claim true?Claim:If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, andf(1) = g(1), then the function h : (0, 2) → R, defined by h(æ) = { F(x) for x € (0, 1]| 9(x) for x e [1, 2)is uniformly continuous on (0, 2).Proof:Let e > 0.Since f is uniformly continuous on (0, 1], there exists d1 > 0 such that if x, y E (0, 1] and|æ – y| < d1, then |f (x) – f(y)| < €/2.Since g is uniformly continuous on [1, 2), there exists d2 > 0 such that if æ, y E [1, 2) and|r – y| < d2, then |9(x) – 9(y)| < €/2.Let 8 = min{d1, &2}.Now suppose r, Y E (0, 2) with x < y and |x – y| < 8.If x, y E (0, 1], then |x – y| < 8 < 81 and so |h(x) – h(y)| = |f (x) – f(y)|< e/2 < e.If x, y € (1,2), then |æ – y| < 8 < d2 and so |h(x) – h(y)| = |9(x) – g(y)| < e/2 < e.If x € (0,1) and y E (1,2), then |x – 1| < |x – y| < 8 < dị and |1 – y| < |æ – y| < 8 < 82so that|h(x) – h(y)| = |f (x) – f(1) + g(1) – g(y)| < |f(x) – f (1)| + |g(1) – 9(y)| < €/2+ €/2 = e.So in all cases, |h(x) – h(y)| < e, as needed.

The given function is uniformly continuous in the intervals 0,1 and 1,0. The interval 0,2 is just a…

If ƒ : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and S (x) for æ € (0, 1] | 9(x) for æ € [1,2) ’ f(1) = g(1), then the function h : (0, 2) → R, defined by h(x): is uniformly continuous on (0, 2).

If ƒ : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and S (x) for æ € (0, 1] | 9(x) for æ € [1,2) ’ f(1) = g(1), then the function h : (0, 2) → R, defined by h(x): is uniformly continuous on (0, 2).

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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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

2. Is the proof correct? Either state that it is, or circle the first error and explain what is
incorrect about it. If the proof is not correct, can it be fixed to prove the claim true?
Claim:
If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and
f(1) = g(1), then the function h : (0, 2) → R, defined by h(æ) = { F(x) for x € (0, 1]
| 9(x) for x e [1, 2)
is uniformly continuous on (0, 2).
Proof:
Let e > 0.
Since f is uniformly continuous on (0, 1], there exists d1 > 0 such that if x, y E (0, 1] and
|æ – y| < d1, then |f (x) – f(y)| < €/2.
Since g is uniformly continuous on [1, 2), there exists d2 > 0 such that if æ, y E [1, 2) and
|r – y| < d2, then |9(x) – 9(y)| < €/2.
Let 8 = min{d1, &2}.
Now suppose r, Y E (0, 2) with x < y and |x – y| < 8.
If x, y E (0, 1], then |x – y| < 8 < 81 and so |h(x) – h(y)| = |f (x) – f(y)|< e/2 < e.
If x, y € (1,2), then |æ – y| < 8 < d2 and so |h(x) – h(y)| = |9(x) – g(y)| < e/2 < e.
If x € (0,1) and y E (1,2), then |x – 1| < |x – y| < 8 < dị and |1 – y| < |æ – y| < 8 < 82
so that
|h(x) – h(y)| = |f (x) – f(1) + g(1) – g(y)| < |f(x) – f (1)| + |g(1) – 9(y)| < €/2+ €/2 = e.
So in all cases, |h(x) – h(y)| < e, as needed.

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