Let  P = _{[n ][ ]} [tan{(pie-4)/4+(1+1/n)}^a]ⁿ

 

Let t = 1 / n When n tends to infinity, t tends to 0

 

ln P = log [ tan { pie/4 - 1 + ( 1 + t ) } ^ a ] / t

 

Put t = 0 and the above limit turns out to be in 0/0 form. So apply L.Hospital in order to evaluate the limit.

 

ln P

=_{[t ][0]} [sec ² { pie/4 +  t } ^ a   .a ( pie / 4 + t ) ^ (a-1)]   / [ tan { pie/4 + t} ^ a ]

 

Put t = 0 to find ln P and hence evaluate P