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

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The function $f(x.y)$ has the limit $L$ as $(x,y) \rightarrow (x_0, y_0)$ if $\forall \epsilon >0, \exists \delta >0$ such that $0<|x-x_0|<\delta \text{ and } 0<|y-y_0|<\delta \Rightarrow |f(x,y)-L|< \epsilon$

So, let $\epsilon >0$. We have to find a $\delta>0$ such that for $0<|x|<\delta$ and $0<|y|<\delta$ $$|\frac{x^2y^2}{\sqrt{y^2+1}-1}|=0$$

$$|\frac{x^2y^2}{\sqrt{y^2+1}-1}|=|\frac{x^2y^2(\sqrt{y^2+1}+1)}{y^2+1-1}|=|\frac{x^2y^2(\sqrt{y^2+1}+1)}{y^2}|=|x^2(\sqrt{y^2+1}+1)|<|\delta^2(\sqrt{\delta^2+1}+1)|$$

So find an appropriate $\delta$ such that $|\delta^2(\sqrt{\delta^2+1}+1)|<\epsilon$.

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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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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*}

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