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, andSf(x) for r€ (0, 1][9(x) for r € [1, 2) 'f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) =is uniformly continuous on (0,2).Proof:Let e >0.Since f is uniformly continuous on (0, 1], there exists d > 0 such that if r, y € (0,1] and|r – y| < d1, then |f(x) – f (y)| < e/2.Since g is uniformly continuous on [1,2), there exists d2 > 0 such that if x, y E [1,2) and|r – y| < d2, then [g(x) – g(y)| < €/2.Let 8 = min{d1, 82}.Now suppose r, y E (0, 2) with r < y and |x – y| < 8.If r, y E (0, 1], then |r – y| < ô < 8i and so |h(x) – h (y)| = |f(x) – f (y)| < €/2 < e.If x, y € [1,2), then |r – y| < d < dz and so h(x) – h(y)| = [g(x) – g(y)| < €/2 < e.If r € (0, 1) and y E (1,2), then |r – 1|< |r – y| < 8 < ôị and |1 – y| < |r – y| < 8 < 82so that|h(x) – h(y)| = |f (1) – f(1) + g(1) – g(y)| < |f(x) – f(1)| + \g(1) – g(y)| < e/2+ €/2 = e.So in all cases, h(x) – h(y)|< €, as needed.

The proof is, Let ε>0 Since f is uniformly continuous on (0,1] there exists δ1>0 and…

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 Sf(z) for 1€ (0, 1] 9(z) for r€ [1, 2) * f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) is uniformly continuous on (0,2). Proof: Let e > 0. Since f is uniformly continuous on (0, 1], there exists ô > 0 such that if r, y E (0,1] and

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 Sf(z) for 1€ (0, 1] 9(z) for r€ [1, 2) * f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) is uniformly continuous on (0,2). Proof: Let e > 0. Since f is uniformly continuous on (0, 1], there exists ô > 0 such that if r, y E (0,1] and

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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Concept explainers

Limits of Multivariable Functions

Multivariable calculus is an extension of single variable calculus. It involves several variables instead of just one. Assume there is an open set containing points (x0, y0), let f be a function defined in that open interval except for the points (x0, y0). The multivariable limit of this function as it approaches the points (x0, y0) and is denoted as:

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
Sf(x) for r€ (0, 1]
[9(x) for r € [1, 2) '
f (1) = g(1), then the function h : (0, 2) → R, defined by h(x) =
is uniformly continuous on (0,2).
Proof:
Let e >0.
Since f is uniformly continuous on (0, 1], there exists d > 0 such that if r, y € (0,1] and
|r – y| < d1, then |f(x) – f (y)| < e/2.
Since g is uniformly continuous on [1,2), there exists d2 > 0 such that if x, y E [1,2) and
|r – y| < d2, then [g(x) – g(y)| < €/2.
Let 8 = min{d1, 82}.
Now suppose r, y E (0, 2) with r < y and |x – y| < 8.
If r, y E (0, 1], then |r – y| < ô < 8i and so |h(x) – h (y)| = |f(x) – f (y)| < €/2 < e.
If x, y € [1,2), then |r – y| < d < dz and so h(x) – h(y)| = [g(x) – g(y)| < €/2 < e.
If r € (0, 1) and y E (1,2), then |r – 1|< |r – y| < 8 < ôị and |1 – y| < |r – y| < 8 < 82
so that
|h(x) – h(y)| = |f (1) – f(1) + g(1) – g(y)| < |f(x) – f(1)| + \g(1) – g(y)| < e/2+ €/2 = e.
So in all cases, h(x) – h(y)|< €, as needed.

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