How do you evaluate this limit as n approaches infinity

(1 - (1/n^2))^n

Please Show your working

Thanks, James.

log y = n log (1 - 1/n^2)
log y = log (1 - 1/n^2) / 1/n

Use LHospital's rule until you don't get the indeterminate.

(1 - 1/n^2)' = (0 + 2n^-3) = 2/n^3

2/n^3 / log(1 - 1/n^2) / 1/n
(2/n^2) /log(1 - 1/n^2)
Keep differientiating. My mind is so tired right now. I'll come back tommorrow.

It could be solved by the second important limit theorem which writes:

(1+1/n)^n -> e as n->infinity
(1-1/n)^n -> 1/e as n->infinity

Therefore, the limit is e*(1/e)=1