\displaystyle \lim_{x\to\infty}{\frac{x}{\sqrt{x^2+1}}}=
\lim_{x\to\infty} \frac{\frac{x}{x}}{\frac{\sqrt{x^2+1}}{x}}=
\lim_{x\to\infty} \frac{1}{\sqrt{1+\frac{1}{x^2}}}
=1