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Unformatted text preview: MAT 319/320 Correction of HW4 Exercise 1. Page 67, #6a. Proof. By the sum rule, the limit of (2 + 1/ n ) is equal to 2 . By the product rule, the limit of (2 + 1/ n ) 2 is 2.2 = 4 . square Exercise 2. Page 67, #9. Proof. One has y n = n + 1  n = 1 n + 1 + n (multiply the numerator and denominator by the conjugate quantity). But now one has lessorequalslant y n lessorequalslant 1 n . Therefore if one proves that (1/ n ) converges to zero, the squeeze theorem implies that ( y n ) converges itself to zero. Fix any &gt; , then by the archimedean property there exists a natural number K such that K &gt; 2 , but this implies that for any n greaterorequalslant K one has n &gt; 2 , implying 1 n &lt; , thus we proved that (1/ n ) converges to zero, and hence ( y n ) converges to zero. Now n y n = n n + 1 + n = n n . 1 (1 + 1/ n ) p + 1 = 1 1 + 1 + 1 n q . By the square root theorem 1 + 1 n radicalBig converges to 1 . By the quotient theorem (which applies because the limit of the denominator is....
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This note was uploaded on 01/31/2011 for the course AMS 319 taught by Professor Mark during the Fall '10 term at SUNY Stony Brook.
 Fall '10
 Mark

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