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