prove the limit using definition Prove the limit by using the definition
$$\lim_{(x,y) \rightarrow (0,0)} \frac{x^{2}y^{2}}{ \sqrt{y^2 + 1}-1 } = 0 $$

First multiply the fraction by $\frac{\sqrt{y^2+1} + 1}{\sqrt{y^2+1} + 1}$ in order to simplify the denominator then switch to polar coordinates $x\mapsto r\cos \theta, \; y \mapsto r\sin \theta$ and take the limit $r \to 0$.
In other words,
\begin{eqnarray*}
\lim_{(x,y)\to 0} \frac{x^2y^2}{\sqrt{y^2+1}-1} =& \lim_{(x,y)\to 0} \frac{x^2y^2}{\sqrt{y^2+1}-1} \frac{\sqrt{y^2+1} + 1}{\sqrt{y^2+1} + 1} \\
=&  \lim_{(x,y)\to 0} \frac{x^2y^2(\sqrt{y^2+1} + 1)}{y^2+1-1} \\
=& \lim_{(x,y)\to 0} \frac{x^2y^2(\sqrt{y^2+1} + 1)}{y^2}  \\
=& \lim_{r \downarrow 0} \frac{r^4(\cos^2 \theta \sin^2 \theta)(\sqrt{r^2 \sin^2 \theta+1} + 1)}{r^2\sin^2 \theta}  \\
=& \lim_{r \downarrow 0} r^2\cos^2 \theta (\sqrt{r^2 \sin^2 \theta+1} + 1)  \\
=& 0.
\end{eqnarray*}
