1.

Evaluate `lim_(ntooo) nsin(2pisqrt(1+n^(2))),(n inN).`

Answer» L= `underset(ntooo)limnsin(2pisqrt(1+n^(2)))`
`underset(ntooo)limnsin(2pisqrt(1+n^(2))-2npi)`
`underset(ntooo)limnsin{(2pi(sqrt(1+n^(2))-n))/((sqrt(1+n^(2))+n))(sqrt(1+n^(2))+n)}`
`=underset(ntooo)lim{(nsin((2pi)/(sqrt(1+n^(2))+n)))/(((2npi)/(sqrt(1+n^(2))+n)))((2pi)/(sqrt(1+n^(2))+n))}`
`underset(ntooo)lim(2npi)/(n(sqrt(1+(1)/(n^(2)))+1))`
`=(2pi)/(2)=pi`


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