$$ A_N(x) = \lim_{N\to \infty}(\sin(x)/x)^N $$
The solution to this problem is given as,
$$ A_N(x) = \exp( -Nx^2/6). $$
The problem is solved through
- Taylor series expansion for $\sin(x)$.
- And the famous limit
$$ \lim_{N\to \infty}(1+(x/N))^N = \exp(x). $$
I'm not able to figure out the disappearance of some of the higher degree terms. How exactly is this limit calculated . An insight would be great.
$\lim_{N\to \infty}(1+(x/N)+ H.O.T.)^N = exp(x)$
where H.O.T. are Higher Order Terms. So you can put the remainder of Taylor series there.
BTW your formula for $A_N(x)$ contains an error. If it's a limit as $N \to \infty$, how can the answer depend on $N$?