Answer
$1$
Work Step by Step
$\lim _{x\rightarrow \infty }\dfrac {x}{\sqrt {x^{2}+1}}\dfrac {=\dfrac {d}{dx}\left( x\right) }{\dfrac {d}{dx}\sqrt {x^{2}+1}}=\dfrac {1}{\dfrac {2x}{2\sqrt {x^{2}+1}}}=\dfrac {\sqrt {x^{2}+1}}{x}=\sqrt {1+\dfrac {1}{x^{2}}}=\sqrt {1+0}=1$
