9. Let (x) and (yn) be sequences of positive numbers such that lim (xn/Yn) = +∞, a. Show that if lim (yn) = +∞, then lim (xn) = +∞. b. Show that if (xn) is bounded, then lim (yn) = 0. Solution 9. (a) Since x/yn → ∞, there exists K₁ such that if n> K₁, then xnyn. Now apply Theorem 3.6.4(a). (b) Let 0 < x < M. If (y) does not converge to o, there exist & > 0 and a subsequence (yn) such that E0Yn. Since lim(x/yn) = ∞, there exists K such that if k > K, then M/ɛ0 < xn/Ynk, which is a contradiction. Hint >>
9. Let (x) and (yn) be sequences of positive numbers such that lim (xn/Yn) = +∞, a. Show that if lim (yn) = +∞, then lim (xn) = +∞. b. Show that if (xn) is bounded, then lim (yn) = 0. Solution 9. (a) Since x/yn → ∞, there exists K₁ such that if n> K₁, then xnyn. Now apply Theorem 3.6.4(a). (b) Let 0 < x < M. If (y) does not converge to o, there exist & > 0 and a subsequence (yn) such that E0Yn. Since lim(x/yn) = ∞, there exists K such that if k > K, then M/ɛ0 < xn/Ynk, which is a contradiction. Hint >>
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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Transcribed Image Text:9. Let (x) and (yn) be sequences of positive numbers such that lim (xn/Yn) = +∞,
a. Show that if lim (yn) = +∞, then lim (xn) = +∞.
b. Show that if (xn) is bounded, then lim (yn) = 0.
Solution
9.
(a) Since x/yn → ∞, there exists K₁ such that if n>
K₁, then xnyn. Now apply Theorem 3.6.4(a).
(b) Let 0 < x < M. If (y) does not converge to o,
there exist & > 0 and a subsequence (yn) such that
E0Yn. Since lim(x/yn) = ∞, there exists K such
that if k > K, then M/ɛ0 < xn/Ynk, which is a
contradiction.
Hint >>
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