Answer
$\lim\limits_{n \to \infty} \sqrt {n^2 + 1} - n = 0$
Work Step by Step
$\lim\limits_{n \to \infty} \sqrt {n^2 + 1} - n = \lim\limits_{n \to \infty} [(\sqrt {n^2 + 1} - n) \times \frac{\sqrt {n^2 + 1} + n}{\sqrt {n^2 + 1} + n}] =\lim\limits_{n \to \infty} \frac{1}{\sqrt {n^2+1} + n} = \frac{1}{\infty + \infty} = 0$
