limn→∞ nn+1α+sin1nn, where α ∈ Q is equal to
e-α
-α
e1-α
e1+α
limn→c nn+1α+sin1nn
=limn→∞ 1+1n−α+sin1nn
=limn→∞ 1−αn+α(α+1)2!⋅1n2⋯+1n−13!n3+15!n5−⋯n=limn→∞ 1+1−αn+α(α+1)2!n2⋯n
=elima→∞1-αn+α(α+1)2!n2+....n=e1−α