Answer
$\lim _{x\rightarrow \infty }\dfrac {x^{2}+1}{x^{4}-3x+6}=0$
Work Step by Step
$\lim _{x\rightarrow \infty }\dfrac {x^{2}+1}{x^{4}-3x+6}=\lim _{x\rightarrow \infty }\dfrac {1+\dfrac {1}{x^{2}}}{x^{2}-\dfrac {3}{x}+\dfrac {6}{x^{2}}}=\dfrac {1+\dfrac {1}{\infty }}{\infty -\dfrac {3}{\infty }+\dfrac {6}{\infty }}=\dfrac {1}{\infty }=0$
