Quiz3-Solution

Math 3000 Fall 2020 Quiz 3 For each sequence, use the " - N " language to prove the existing of the limit (if the limit is not given, you should first find the limit). 1. (7 marks) lim n →∞ n ( p n 4 + 2 n + 2 - n 2 ) =? ANS: We can calculate its limit as follows: lim n →∞ n ( p n 4 + 2 n + 2 - n 2 ) = lim n →∞ n ( √ n 4 + 2 n + 2 - n 2 )( √ n 4 + 2 n + 2 + n 2 ) ( √ n 4 + 2 n + 2 + n 2 ) = lim n →∞ 2 n 2 + 2 n √ n 4 + 2 n + 2 + n 2 = lim n →∞ 2 + 2 n q 1 + 2 n 3 + 2 n 4 + 1 = 1 . For all > 0, consider the following inequality 2 n 2 + 2 n √ n 4 + 2 n + 2 + n 2 - 1 = n 2 + 2 n - √ n 4 + 2 n + 2 √ n 4 + 2 n + 2 + n 2 < . In fact, one has 2 n 2 + 2 n √ n 4 + 2 n + 2 + n 2 - 1 = n 2 + 2 n - √ n 4 + 2 n + 2 √ n 4 + 2 n + 2 + n 2 = ( n 2 + 2 n - √ n 4 + 2 n + 2)( n 2 + 2 n + √ n 4 + 2 n + 2) ( √ n 4 + 2 n + 2 + n 2 )( n 2 + 2 n + √ n 4 + 2 n + 2) < 4 n 3 + 4 n 2 - 2 n - 2 ( √ n 4 + 2 n + 2 + n 2 ) 2 < 8 n 3 4 n 4 = 2 n . We can then let 2 n < ⇐⇒ n > 2 . Now let N ( ) = [2 / ]. Then, for all > 0 , ∃ N ( ) = [2 / ], such that, for all n > N ( ) (or n ≥ N ( ) + 1), one has 2 n 2 + 2 n √ n 4 + 2 n + 2 + n 2 - 1 < 2 n ≤ 2 N ( ) + 1 < 2 2 / = . Hence, by definition, one gets lim n →∞ n ( √ n 4 + 2 n + 2 - n 2 ) = 1 . 2. (8 marks) lim n →∞ ( n 2 + 3 n + 1) 1 /n = 1 . ANS: For all > 0, consider the following inequality ( n 2 + 3 n + 1) 1 /n - 1 < . 1

Math 3000 Fall 2020 Quiz 3 Then, one has n 2 + 3 n + 1 < (1 + ) n . (1) Note that the Binomial formula implies (1 + ) n = 1 + n + n ( n - 1) 2 2 + · · · . In order to let (1) hold true, we can let n 2 + 3 n + 1 < 1 + n ( n - 1) 2 2 , which implies that n > 6 + 2 2 - 2 . Now let N ( ) = [ 6+ 2 2 - 2 ]. Then, for all > 0 , ∃ N ( ) = [ 6+ 2 2 - 2 ], such that, for all n > N ( ) (or n ≥ N ( ) + 1), one has ( n 2 + 3 n + 1) 1 /n - 1 < 1 + n ( n - 1) 2 2 1 /n - 1 < 1 + N ( )( N ( ) - 1) 2 2 1 /n - 1 = . Hence, by definition, one gets lim n →∞ ( n 2 + 3 n + 1) 1 /n = 1 . 2