Find multivariable limit $\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{\sqrt{x^2y^2+(x-y)^2}}$

Question

I am trying to find the limit

$$\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{\sqrt{x^2y^2+(x-y)^2}}\text{.}$$

I tried the following approach

$$\begin{align*} \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{\sqrt{x^2y^2+(x-y)^2}}&=\lim_{(x,y)\to(0,0)}\sqrt{\frac{x^4y^4}{x^2y^2+(x-y)^2}}\text{,}\\ \end{align*}$$

now I just need to find $$\begin{align*} \lim_{(x,y)\to(0,0)}\frac{x^4y^4}{x^2y^2+(x-y^2)}&=\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{1+\frac{(x-y)^2}{x^2y^2}}\\ \end{align*}$$

and that's where I got lost. I have no idea how to work with

$$\lim_{(x,y)\to(0,0)}\frac{(x-y)^2}{x^2y^2}\text{.}$$

Expanding the numerator didn't help me at all.

Answer 1

Hint. Note that if $xy\not=0$ (otherwise the ratio is zero), then $$0\leq \frac{x^2y^2}{\sqrt{x^2y^2+(x-y)^2}}\leq \frac{x^2y^2}{\sqrt{x^2y^2}}=|xy|.$$

Answer 2

By polar coordinates we have

$$\frac{x^2y^2}{\sqrt{x^2y^2+(x-y)^2}}=r^3\frac{\cos^2\theta\sin^2\theta}{\sqrt{\cos^2\theta\sin^2\theta+(\cos \theta -\sin\theta)^2}}\to 0$$

indeed

$$\cos^2\theta\sin^2\theta+(\cos \theta -\sin\theta)^2\ge m>0$$