Answer
$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^2+1}} = 1$
Work Step by Step
$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^2+1}} = \frac{\infty}{\infty}$
We can try to apply L'Hospital's Rule:
$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^2+1}}$
$= \lim\limits_{x \to \infty}\frac{1}{\frac{1}{2}(x^2+1)^{-1/2}(2x)}$
$= \lim\limits_{x \to \infty}\frac{\sqrt{x^2+1}}{x} = \frac{\infty}{\infty}$
Applying L'Hospital's Rule does not help us solve the question.
We can try another method:
$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^2+1}}\cdot \frac{1/x}{1/x}$
$ = \lim\limits_{x \to \infty}\frac{x/x}{\sqrt{x^2/x^2+1/x^2}}$
$ = \lim\limits_{x \to \infty}\frac{1}{\sqrt{1+1/x^2}}$
$ = \frac{1}{\sqrt{1+0}}$
$ = 1$
Therefore:
$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^2+1}} = 1$
