Answer
$+\infty$
(does not exist)
Work Step by Step
Note that
$\sqrt{x^{2}+2}=\sqrt{x^{2}(1+\frac{2}{x^{2}})}=\sqrt{x^{2}}\cdot\sqrt{1+\frac{2}{x^{2}}}=|x|\sqrt{1+\frac{2}{x^{2}}}$
Since $ x\rightarrow\infty$ it is positive, so $|x|=x.$
$\displaystyle \lim_{x\rightarrow\infty}\frac{x^{3}}{\sqrt{x^{2}+2}}= \lim_{x\rightarrow\infty}\frac{x^{3}}{x\sqrt{1+\frac{2}{x^{2}}}}$
$= \displaystyle \lim_{x\rightarrow\infty}\frac{x^{2}}{\sqrt{1+\frac{2}{x^{2}}}}$
When $ x\rightarrow\infty$, the term $\displaystyle \frac{2}{x^{2}}$ approaches zero, and the
denominator approaches 1.
The numerator, $x^{2}$, becomes arbitrarily large.
The limit does not exist.
We write
$\displaystyle \lim_{x\rightarrow\infty}\frac{x^{3}}{\sqrt{x^{2}+2}}=+\infty$
