Answer
$1$
Work Step by Step
When $\lim\limits_{x \to \infty} f(x)$ exists, then the sequence $a_n=f(n)$ converges to the same limit, so we have:
$ \lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} f(x)$
Now, $ \lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} (1+\dfrac{1}{x^2})^x \\=\lim\limits_{x \to \infty} (1+\dfrac{1}{x^2})^{x^2 \times \dfrac{1}{x}}$
Since, $\lim\limits_{x \to \infty} (1+\dfrac{1}{x^2})^{x^2}=e$
Thus, $ \lim\limits_{n \to \infty} a_n =e^{\lim\limits_{x \to \infty}\frac{1}{x}} \\ =e^0 \\=1$
