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Evaluate `lim_(ntooo) (4^(n)+5^(n))^(1//n)` |
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Answer» We have, `underset(ntooo)lim(4^(n)+5^(n))^(1//n)" "(oo^(0)"from")` `=underset(ntooo)lim5(1+((4)/(5))^(n))^(1//n)` `=5" "(because((4)/(5))^(n)to0" as n"tooo)` |
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