`lim_(ntooo) {((n)/(n+1))^(alpha)+"sin"(1)/(n)}^(n)` (where `alphainQ`) is equal to

Options

A. `a=1//3,b=1`
B. `a=1,b=-1`
C. `a=9,b=-9`
D. `a=2,b=2//3`

Answer

C

Explanation

`1^(oo)" form"`
`L= e^(underset(ntooo)limn(((n)/(n+1))^(a)+sin(1)/(n)-1))=e^(underset(ntooo)limn"sin"(1)/(n)+underset(ntooo)limn(((n)/(n+1))^(a)-1)`
`"Consider "^(underset(ntooo)limn(((n)/(n+1))^(a)-1)=underset(ntooo)limn(((1)/(1+1//n))^(a)-1))`
Put `n=(1)/(y)`. Then
`underset(yto0)lim(1)/(y)(((1)/(1+y))^(a)-1)=underset(yto0)lim(1-(1+y)a)/(y)=-a`
`:. L=e^(1-a)`