prove that y$x^3$ + x$y^3$ has a limit as x,y approach 0,0

Question

I don't know how to prove this with epsilon delta. So far all i can do is express |$yx^3$ + $xy^3$| <= |$(y^2)x^3$ + $(x^2)y^3$| = $(x^2)(y^2)(|x+y|)$ <= $(x^2)(y^2)(|x|+|y|)$

How would i go on to prove that o is a limit.

Answer 1

Your work so far is not actually correct -- it is not true that $\lvert yx^3+xy^3\rvert\leq\lvert y^2x^3+x^2y^3\rvert$. Take for instance $x=y=\frac{1}{2}$.

But, not all is lost: I'd start out by noting that $$ \lvert yx^3+xy^3\rvert=\lvert x\rvert\,\lvert y\rvert\,\lvert x^2+y^2\rvert. $$

Now, your eventual goal is to show that if $(x,y)$ is close to $(0,0)$, this quantity is small (because the limit should clearly be $0$ if it exists). So, you need to find a way of bounding this using the distance between $(x,y)$ and $(0,0$) -- namely, $\sqrt{x^2+y^2}$.

A common trick to do this is to note that $$ \lvert x\rvert=\sqrt{x^2}\leq\sqrt{x^2+y^2} $$ and similarly $\lvert y\rvert\leq\sqrt{x^2+y^2}$; finally, $\lvert x^2+y^2\rvert=(\sqrt{x^2+y^2})^2$.

So, in all, $$ \lvert yx^3+xy^3\rvert\leq\left(\sqrt{x^2+y^2}\right)^4. $$ Can you figure out how to use this to fashion an $\epsilon-\delta$ argument?

Answer 2

For every $\epsilon >0$ you have to fing $\delta>0$ such that $\sqrt{x^2+y^2}< \delta \Longrightarrow |yx^3+xy^3|< \epsilon$

You have that $x^2+y^2< \delta^2$

$|yx^3+xy^3|=|xy|(x^2+y^2)\leqslant \frac{x^2+y^2}{2}(x^2+y^2)<\frac{\delta^4}{2}$

Take $\delta=\sqrt[4]{2 \epsilon}$ and you are done.

In one step i used the inequality in which $ab \leqslant\frac{a^2+b^2}{2}$

Answer 3

def: $d((x,y),(a,b)) = \max(|x-a|,|y-b|)$

If you feel the need, prove that this is a valid metric.

$d((x,y),(0,0)) <\delta \implies |x|<\delta$ and $|y|<\delta$ and

$|x^3y+y^3x| < 2\delta^4$

for any $\epsilon$ let $\delta = \sqrt[4]{\frac \epsilon2}$ and $d((x,y),(0,0))<\delta \implies |f(x,y)|<\epsilon$