Let epsilon1 0 f x g x f x g x g x f x M f x for all x U D Since f x 0 as x c

# Let epsilon1 0 f x g x f x g x g x f x m f x for all

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Unformatted text preview: Let epsilon1 > 0. | f ( x ) g ( x )- | = | f ( x ) g ( x ) | = | g ( x ) || f ( x ) | < M | f ( x ) | for all x ∈ U ∩ D. Since f ( x ) → 0 as x → c There exists a neighborhood V of c such that | f ( x )- | = | f ( x ) | < epsilon1/M for all x ∈ V ∩ D. Let δ 2 be the radius of V . Let δ = min { δ 1 , δ 2 } . Then | f ( x ) g ( x )- | < M | f ( x ) | < M epsilon1 M = epsilon1 whenever | x- c | < δ. Therefore, lim x → c f ( x ) g ( x ) = 0. 16. Suppose that lim x → c f ( x ) = L > 0. Let epsilon1 = L . Then there exists a deleted neighborhood U = N * ( c, δ ) of c such that | f ( x )- L | < L whenever x ∈ U ∩ D. This implies that- L < f ( x )- L < L whenever x ∈ U ∩ D and it follows that 0 < f ( x ) < 2 L . In particular, f ( x ) > 0 for all x ∈ U ∩ D . 18. Suppose lim x → c f ( x ) = L . Let epsilon1 = 1. There exists a neighborhood U of c such that | f ( x )- L | < 1 for all x ∈ U ∩ D . This implies that- 1 < f ( x )- L < 1 and L- 1 < f ( x ) < L + 1 for all x ∈ U ∩ D Let max {| L- 1 | , | L + 1 |} . Then | f ( x ) | < M for all x ∈ U ∩ D . 4 Additional Problems 1. For each of the following sequences, determine: (1) Whether the sequence converges; if it does, give the limit. (2) If the sequence does not converge, find the set of subsequential limits (if any) and give lim sup and lim inf. (a) s n = 1 n + 1 , n odd 3 2 n- 1 , n even s n → 0. (b) s n = 1 n , n = 3 , 6 , 9 , ··· n n + 1 , n = 1 , 4 , 7 , 10 , ··· n 2 , n = 2 , 5 , 8 , 11 , ··· ( s n ) does not converge. Subsequential limits: S = { , 1 , ∞} , lim sup s n = ∞ , lim inf s n = 0. (c) s k = k sin 2 parenleftbigg kπ 4 parenrightbigg ( s k ) does not converge. Subsequential limits: S = { , ∞} , lim sup s n = ∞ , lim inf s n = 0. (d) s n = 1 2 n + cos parenleftBig nπ 6 parenrightBig ( s n ) does not converge. Subsequential limits: S = { 1 , √ 3 / 2 , 1 / 2 , ,- 1 / 2 ,- √ 3 / 2 ,- 1 } , lim sup s n = 1, lim inf s n =- 1. 2. Let f be some function for which you know only that if < | x- 3 | < 1 , then | f ( x )- 4 | < . 1 . Which of the following statements are necessarily true? I. If | x- 3 | < . 1, then | f ( x )- 4 | < . 01. II. If | x- 2 . 6 | < . 3, then | f ( x )- 4 | < . 1. III. If 0 < | x- 3 | < . 5, then | f ( x )- 4 | < . 1. IV. lim x → 3 f ( x ) = 4. 5 (a) II and IV (b) IV only (c) I and II (d) II and III (e) II, III and IV Answer: (d) II and III. 3. Use the epsilon1- δ definition of the limit to prove that lim x → 2 (3 x + 4) = 10. Let epsilon1 > 0. | 3 x + 4- 10 | = | 3 x- 6 | = 3 | x- 2 | . Choose δ = epsilon1/ 3. If | x- 2 | < δ = epsilon1/ 3, then | 3 x + 4- 10 | = 3 | x- 2 | < 3 δ = 3 · epsilon1 3 = epsilon1 Therefore lim x → 2 (3 x + 4) = 10. 6...
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• Fall '08
• Staff
• Limits, lim, Limit of a function, Limit of a sequence, lim sup, subsequence

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