barcode in ssrs report If lim an A and lim bn B 6 0, prove in .NET

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2.13. If lim an A and lim bn B 6 0, prove
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n!1 n!1
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(a) lim
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1 a A , (b) lim n . n!1 bn B B
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(a) We must show that for any given  > 0, we can nd N such that   1   1  jB bn j <  for all n > N b B jBjjbn j n
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By hypothesis, given any  > 0, we can nd N1 , such that jbn Bj < 1 B2  for all n > N1 . 2 Also, since lim bn B 6 0, we can nd N2 such that jbn j > 1 jBj for all n > N2 (see Problem 11). 2
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Then if N is the larger of N1 and N2 , we can write (1) as   1 2  1 1  jbn Bj B     for all n > N < 2 1 b  B jBjjbn j jBj 2 jBj n and the proof is complete. (b) From part (a) and Problem 2.12, we have   an 1 1 1 A lim an A lim an lim n!1 bn n!1 n!1 n!1 bn bn B B lim This can also be proved directly (see Problem 41).
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2.14. Evaluate each of the following, using theorems on limits.
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a 3n2 5n 3 5=n 3 0 3 lim n!1 5n2 2n 6 n!1 5 2=n 6=n2 5 0 0 5 lim ( ) ( ) ( ) n n 2 n3 n3 n2 2n 1 1=n 2=n2 2 lim lim n!1 n!1 n 1 n2 1 n!1 1 1=n 1 1=n2 n 1 n 1 lim 1 0 0 1 1 0 1 0
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p p p p p p n 1 n 1 lim n 1 n lim n 1 n p p lim p p 0 n!1 n!1 n 1 n n!1 n 1 n
3n2 4n 3 4=n lim n!1 2n 1 n!1 2=n 1=n2 lim
SEQUENCES
[CHAP. 2
Since the limits of the numerator and denominator are 3 and 0, respectively, the limit does not exist. 3n2 4n 3n2 3n > can be made larger than any positive number M by choosing n > N, we Since 2n 1 2 2n 2 3n 4n can write, if desired, lim 1. n!1 2n 1  e
    4 2n 3 4 2 3=n 4 2 16 lim n!1 3 7=n 2n 7 3 81 2n5 4n2 2=n2 4=n5 0 0 lim 3 10 n!1 3 1=n4 10=n7 3 n (Compare with Problem 2.5.)
n!1 3n7
n!1 5
1 2 10n 10 n 2 2 lim n n!1 5 10 n 3 3 3 10
BOUNDED MONOTONIC SEQUENCES 2.15. Prove that the sequence with nth un above, 2n 7 (a) is monotonic increasing, 3n 2 (c) is bounded below, (d) is bounded, (e) has a limit.
Now 2n 5 2n 7 A 2n 5 3n 2
(b) is bounded
(a) fun g is monotonic increasing if un 1 A un , n 1; 2; 3; . . . . 2 n 1 7 2n 7 A 3 n 1 2 3n 2 if and only if
or 2n 5 3n 2 A 2n 7 3n 5 , 6n2 11n 10 A 6n2 11n 35, i.e. 10A 35, which is true. Thus, by reversal of steps in the inequalities, we see that fun g is monotonic increasing. Actually, since 10 > 35, the sequence is strictly increasing. (b) By writing some terms of the sequence, we may guess that an upper bound is 2 (for example). To prove this we must show that un @ 2. If 2n 7 = 3n 2 @ 2 then 2n 7 @ 6n 4 or 4n < 11, which is true. Reversal of steps proves that 2 is an upper bound. (c) Since this particular sequence is monotonic increasing, the rst term 1 is a lower bound, i.e., un A 1, n 1; 2; 3; . . . . Any number less than 1 is also a lower bound. Thus, for example, we can write
(d) Since the sequence has an upper and lower bound, it is bounded. jun j @ 2 for all n. (e)
Since every bounded monotonic (increasing or decreasing) sequence has a limit, the given sequence has 2n 7 2 7=n 2 a limit. In fact, lim lim . n!1 3n 2 n!1 3 2=n 3
2.16. A sequence fun g is de ned by the recursion formula un 1 exists. (b) Find the limit in (a).
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