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For each sequence a, find a number k such that nkan has a finite non-zero limit. This is of use, because by the limit compari

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A) 6+4n7 an (6+4m)7 Takingn 7 am (6+4m)7 G+4n) m7 4 Donoma Common m (a b)= a +4 +4) m an (7 +4 0 47 is Comuragent, Simce omvB) am Lot an 7 x m Taking Comnon n the demounat gel- we an 7. m 7 =7キ0 m On ( m-00 -Zan amd conuenge togethon lmit compatikonam am 7m t 4 Lot m highert dagnea tenm in haghent daas team m Dr 5 Tn+S 91+7nt 4 () (s+uL+G anTnt4 +) +4 n8 t um am #0 9+1+ 42 5m+7n+6 10 +7m +4m Let tvghost dogree term m N 1o aloqee tem in D 210 10 70 an 10 5 m2-+7ntb 9m+7n+4 1 10 10 5m2+77+6 70 +7

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