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Unformatted text preview: Finding Sequences to Squeeze n n + n 1 /n and Hence, to Compute lim n parenleftbigg n n + n 1 /n parenrightbigg +LHopitals Rule Does Work JO We have proven lim n n 1 /n = 1 . This means that for any > 0 there is a positive integer M such that if n M then  n 1 /n 1  . I dont need to worry about the absolute value signs, because n 1 /n 1 > , so  n 1 /n 1  = n 1 /n 1 & n 1 /n 1 . I can pick any positive value for and I am guaranteed an M (that I dont have to find) such that for n M then n 1 /n 1 =  n 1 /n 1  . 1 Out of the air, I let = 1 / 10, so there is an M 1 such that 1 n 1 /n < 11 / 10 , n M. Note: 11 / 10 = 1 + . Now I build the sequence, add n to each term and divide each term by n. 1 + n n n + n 1 /n n n + 11 / 10 n , n M. Next invert each term, recall 0 < a b then 1 /b 1 /a....
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This note was uploaded on 10/12/2009 for the course MATH calculus 3 taught by Professor Ningzhong during the Spring '08 term at University of Cincinnati.
 Spring '08
 ningzhong
 Calculus, Squeeze Theorem

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