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iii. is continuous at a provided g(a) # 0.

iii. is continuous at a provided g(a) # 0.

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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2.19 † Prove part (iii) of Theorem 2.2.2. The domain D is that subset of D;ND, consisting of points x for which g(x) + 0. For Reference Theorem 2.2.2 Suppose both f and g are continuous at a e D;n Dg. Then i. f+gis continuous at a. ii. fg is continuous at a. iii. is continuous at a provided g(a) # 0. iv. If, moreover, h is continuous at f(a), then the composition ho f is continuous at a, where ho f(x) = h(f(x)). а, Proof: We will apply the Sequential Criterion for Continuity (Corollary 2.2.1) throughout. i. For the first case, bear in mind that Df±g = Df n Dg. If xn → a with all an E D; n Dg, then f(n) → f(a) and g(an) → g(a). We conclude that lim,-a(f + g)(x) exists and equals lim f(x) + lim g(x) = f(a) ± g(a). ii. This part has a nearly identical proof. iii. This part is left to the reader in Exercise 2.19. iv. We note that Dhof = {x € Df | f(x) E Dh}. Now let rn E Dhof such that In - a. Then letting yn = f(In), we have lim h(f(rn)) = lim h(yn) = h(f(a)) I| n-00 n+00 since yn -- f(a) as n -→ o since f and h are continuous at a and f (a), respectively.
Transcribed Image Text:2.19 † Prove part (iii) of Theorem 2.2.2. The domain D is that subset of D;ND, consisting of points x for which g(x) + 0. For Reference Theorem 2.2.2 Suppose both f and g are continuous at a e D;n Dg. Then i. f+gis continuous at a. ii. fg is continuous at a. iii. is continuous at a provided g(a) # 0. iv. If, moreover, h is continuous at f(a), then the composition ho f is continuous at a, where ho f(x) = h(f(x)). а, Proof: We will apply the Sequential Criterion for Continuity (Corollary 2.2.1) throughout. i. For the first case, bear in mind that Df±g = Df n Dg. If xn → a with all an E D; n Dg, then f(n) → f(a) and g(an) → g(a). We conclude that lim,-a(f + g)(x) exists and equals lim f(x) + lim g(x) = f(a) ± g(a). ii. This part has a nearly identical proof. iii. This part is left to the reader in Exercise 2.19. iv. We note that Dhof = {x € Df | f(x) E Dh}. Now let rn E Dhof such that In - a. Then letting yn = f(In), we have lim h(f(rn)) = lim h(yn) = h(f(a)) I| n-00 n+00 since yn -- f(a) as n -→ o since f and h are continuous at a and f (a), respectively.
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