The value of limn→∞(1+sinan)n is equal to
limn⟶∞(1+sinan)n
As limn⟶∞sinan=limn⟶∞(an)sinanan=limn⟶∞an
So, limn⟶∞(1+sinan)n=limn⟶∞(1+an)n
Since, now undetermined form is of type 1∞
So, limn⟶∞(1+an)n=elimn⟶∞(1+an)n=ea