6c For \u03f5 \u03b4 0 where x 3 8x 2x 2 3x4 \u03f5 for every 0x 2 \u03b4 Let S 1 1 Then 0x 2 \u03b4 1 1

# 6c for ϵ δ 0 where x 3 8x 2x 2 3x4 ϵ for every 0x

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6c) For ϵ > 0 δ > 0, where |x 3 -8|=|x-2||x 2 +3x+4|< ϵ , for every 0<|x-2|< δ Let S 1 =1. Then 0<|x-2|< δ 1 =1, then 1<x<3 Then |x-2||(3^2)+2(3)+4|=19|x-2||< ϵ and δ = ϵ 19 . 9a) By Theorem 5.1.10, There exists a sequence S n in D with each S n c such that S n converges to c, but f(s n ) is not convergent in R . Let f(x)= 1 x and S n = 1 n . lim n→∞ S n = 0 n but f(s n )= Lim n=+ so f(s n ) is divergent and the limit does not exist. 9b) By Theorem 5.1.10, k N Sin(2 πk + 2 ¿ = {0, if n=0 1, if n=1 0, if n=2 -1, if n=3} S n = n 2 π , for all n N , Then the limit as n approaches positive infinity of S n =0 But f(s n ) is the sequence {0,1,0,-1,0, -1, ….} which does not converge. 9c) x→ 0 + ¿ xsin 1 x lim ¿ ¿ . Let ϵ > 0 δ > 0 where ϵ = δ . Then | xsin 1 x -0|<|x| as | sin 1 x | 1 x > 0.
| xsin 1 x ≤x , x > 0. | xsin 1 x ϵ where 0 <x < δ x→ 0 + ¿ xsin 1 x = 0 lim ¿ ¿ 13) From Theorem 4.1.13, 4.4.11, and 5.1.8 we can see that a) Let S n D that converges to c with S n ≠c b) Knowing lim x→ c f ( x ) = L lim x →c h ( x ) = L, by Theorem 5.1.8, lim n→∞ f ( s n ) = L lim n→ ∞ h ( s n ) = M c) Since f(x) g(x) h(x) for every x D f(s n ) g(s n ) h(s n ) n N d) Given ϵ > 0, N 1 , such that |h(s n )-M|< ϵ n≥ N 1 e) Given ϵ > 0, N 2 , such that |f(s n )-L|< ϵ n≥ N 2 |f(s n )-L|< ϵ and |h(s n )-M|< ϵ n≥ N N 1 N 2 aremaximums f) Therefore – ϵ < f(s n )-L< ϵ and – ϵ < g(s n )-M< ϵ g) From line c, f(s n ) L≤ g(s n ) L≤ h(s n ) L n N ϵ < g(s n )-L< ϵ n N lim n→∞ g ( s n ) = L n N ****Similar can be said that lim n→∞ g ( s n ) = M
16)By theorem 5.1.2, let f:d-> and c be an accumulation point of D. Then lim as x approaches c of f(x)=L iff for each neighborhood V of L, there exists a deleted neighborhood U of c such that f(U D) V .

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• Fall '08
• Staff
• Limits, Limit of a sequence, Limit superior and limit inferior, lim sup, subsequence, Lim Sup Vn