Limit of multivariable function $\frac{x^2 y^2}{x^2 + y^2}$

Question

I need to find the limit:

$ \lim \limits_{P \to P_.0} \frac{x^2 y^2}{x^2 + y^2}$ where $P_0 = {0,0}$

I know that I can find the limit using polar coordinates, BUT how can I find it with $\epsilon - \delta$ definition? Please explain step step so I can understand it.

Answer 1

With the AM-GM inequality: for $x,y)\neq (0,0)$ $$ 0 \leq \frac{x^2 y^2}{x^2+y^2} \leq \frac{x^2 y^2}{2\sqrt{x^2 y^2 }} = \frac{\lvert x y\rvert }{2} \xrightarrow[(x,y)\to(0,0)]{}0 $$ the last part being "easy" to quantify with $\epsilon$ and $\delta$, as $\sqrt{x^2+y^2}=\left\lVert \begin{pmatrix} x\\y\end{pmatrix}\right\rVert_2\leq \delta$, the very same AM-GM inequality implies $\sqrt{2\lvert xy \rvert }\leq \delta$.

Answer 2

Remember that $y^2$ and $x^2\leqslant x^2+y^2$, so $x^2y^2\leqslant (x^2+y^2)^2$, i.e. your function is less than $x^2+y^2$.

Using fancier terminology, if $\lVert (x,y)\rVert=\sqrt{x^2+y^2}$ is the Euclidean norm, $|x|,|y|\leqslant \lVert (x,y)\rVert$ by taking squareroots above. Your function is $$\frac{x^2y^2}{\lVert(x,y)\rVert}\leqslant\frac{\lVert (x,y)\rVert^4}{\lVert(x,y)\rVert^2}\leqslant \lVert (x,y)\rVert^2\to 0$$ when $\lVert (x,y)\rVert \to 0$, i.e. $(x,y)\to 0$.

Answer 3

Note we have,

$$|\frac{x^2y^2}{x^2+y^2}| \leq |\frac{x^2y^2}{x^2}|=|y^2| \to 0$$

Now conclude with squeeze theorem. And show $\lim_{y \to 0} |y^2|=0$.