$\lim_{n\rightarrow \infty }{(1 - 1/n^2)}^n$
This is of $1^\infty$,
for this we need to use $\lim_{n\rightarrow \infty}{f(x)}^{g(x)} = e^{ \lim_{n\rightarrow \infty}(f(x)-1)*g(x)}$
$f(x) = (1 - 1/n^2)$
$g(x) = n$
$= e^{ \lim_{n\rightarrow \infty}(1 - 1/n^2-1)*n}$
$= e^{ \lim_{n\rightarrow \infty} - 1/n}$
$= e^{ - 1/\infty }$
$= e^{ 0 }$
$= 1$