## Calculus (3rd Edition)

$1$
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, $\lim\limits_{n \to \infty} a_n =e^{\lim\limits_{x \to \infty}\frac{1}{x}} \\ =e^0 \\=1$