Answer
When applying the l'Hospital's rule $n$ times the indeterminate form $\frac{\infty}{\infty}$ still exists.
The limit is equal to $1$.
Work Step by Step
Using the l'Hospital's rule it follows:
$$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^{2}+1}}=\lim\limits_{x \to \infty}\frac{1}{\frac{2x}{2\sqrt{x^{2}+1}}}=\lim\limits_{x \to \infty}\frac{\sqrt{x^{2}+1}}{x}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule for second time it follows:
$$\lim\limits_{x \to \infty}\frac{\sqrt{x^{2}+1}}{x}=\lim\limits_{x \to \infty}\frac{\frac{2x}{2\sqrt{x^{2}+1}}}{1}=\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^{2}+1}}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule for third time it follows:
$$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^{2}+1}}=\lim\limits_{x \to \infty}\frac{1}{\frac{2x}{2\sqrt{x^{2}+1}}}=\lim\limits_{x \to \infty}\frac{\sqrt{x^{2}+1}}{x}=\frac{\infty}{\infty}$$
$$\vdots$$
If we apply the l'Hospital's rule for $n$ times the indeterminate form $\frac{\infty}{\infty}$ still exists.
$$\lim\limits_{x \to \infty}\frac{x}{\sqrt{x^{2}+1}}=\lim\limits_{x \to \infty}\frac{x}{x\sqrt{1+\frac{1}{x^{2}}}}=\lim\limits_{x \to \infty}\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}=\frac{1}{\sqrt{1+0}}=1$$
