example(epsilonK)

# example(epsilonK) - K 1,K 2 ∈ N such that | a n-A |< ±...

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EXAMPLE OF USE OF THE DEFINITION OF LIMIT OF A SEQUENCE 1. Show that lim ( 2 n n + 2 ) = 2 . Solution. | 2 n n + 2 - 2 | = | 4 n + 2 | ≤ | 4 n | ( * ) . Fix ± > 0 . By the Archimedean property, there is N ± such that 4 ± < N ± . We thus have 4 n < ± n K := N ± . By plugging this into ( * ) , we see that, for every ± > 0 , there is a natural number K such that if n K, then | 2 n n + 2 - 2 | < ± and this means that lim ( 2 n n +2 ) = 2 . 2. Prove that if ( a n ) A and ( b n ) B, then ( a n + b n ) -→ ( A + B ) . Solution. By regrouping the terms and by the the Triangle Inequal- ity: | ( a n + b n ) - ( A + B ) | ≤ | a n - A | + | b n - B | ( * ) . Fix ± > 0 . By assumption there are
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Unformatted text preview: K 1 ,K 2 ∈ N such that | a n-A | < ± 2 , ∀ n ≥ K 1 , | b n-B | < ± 2 , ∀ n ≥ K 2 . Take K := max { K 1 ,K 2 } . We have | a n-A | + | b n-B | < ±, ∀ n ≥ K ( ** ) . By combining ( * ) with ( ** ) , we obtain that, for any ± > , there is K ∈ N such that | ( a n + b n )-( A + B ) | < ±, ∀ n ≥ K, that is, we have showed that ( a n + b n ) → ( A + B ) . 1...
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