Answer
converges to $1$
Work Step by Step
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})^n$
But $ \lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})=\lim\limits_{n \to \infty} \dfrac{\ln (1-\dfrac{1}{n^2})}{(1/n)}=\dfrac{0}{0}$
This shows that the limit has an Indeterminate form, so we will use L-Hospital's rule.
This implies that $\lim\limits_{n \to \infty} (1-\dfrac{1}{n^2})=\lim\limits_{n \to \infty} \dfrac{-2n}{n^2-1}=0$
Thus, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (e)^{(n \ln n) (1-\dfrac{1}{n^2})}=e^0=1$
Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} converges to $1$.
