Solve!!!!

lim_{nâ†’âˆž}(n^Î±sin²n!/n+1), where 0<Î±<1

2 Answers

oh sorry.. its... [(n^Î±sin²n!)/(n+1)

sin² can have maximum mod of 1

put n= 1/t

so tâ†’0+

n^a/n+1 = t^1-a/1+t = 0 as tâ†’0 (note 0<1-a )

so its 0 X something bounded by mod1 = 0

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