<s>[INST] prove the limit using definition Prove the limit by using the definition
$$\lim_{(x,y) \rightarrow (0,0)} \frac{x^{2}y^{2}}{ \sqrt{y^2 + 1}-1 } = 0 $$
 [/INST] Hint: for all $y\neq0$,
$$\frac{x^{2}y^{2}}{\sqrt{y^2+1}-1}=\frac{x^{2}y^{2}}{\sqrt{y^2+1}-1}\cdot\frac{\sqrt{y^2+1}+1}{\sqrt{y^2+1}+1}\\
=\frac{x^2y^2(\sqrt{y^2+1}+1)}{y^2+1-1}\\
=\frac{x^2y^2(\sqrt{y^2+1}+1)}{y^2}\\
=x^2\left(\sqrt{y^2+1}+1\right).$$
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