1.

Evaluate `lim_(ntooo) (4^(n)+5^(n))^(1//n)`

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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