Show limit of function $\frac{xy(x+y)}{x^2-xy+y^2}$ for $(x,y)\to(0,0)$

Question

Show $$\lim_{(x,y)\to (0,0)} xy \frac{(x+y)}{x^2-xy+y^2}=0$$

If I approach from $y=\pm x$, I get $0$. Is that sufficient?

Answer 1

From $\bigl(|x|-|y|\bigr)^2\geq0$ we get $2|xy|\leq x^2+y^2$ and then $$|x|+|y|\leq\sqrt{2(x^2+y^2)},\qquad x^2-xy+y^2\geq{1\over2}(x^2+y^2)\ .$$ This implies $$\bigl|f(x,y)\bigr|=\left|{xy(x+y)\over x^2-xy+y^2}\right|\leq{{1\over2}(x^2+y^2)\sqrt{2(x^2+y^2)}\over{1\over2}(x^2+y^2)}=\sqrt{2(x^2+y^2)}\ .$$ Therefore, given an $\epsilon>0$, we have $\bigl|f(x,y)\bigr|<\epsilon$ as soon as $\sqrt{x^2+y^2}<\delta:={\displaystyle{\epsilon\over\sqrt{2}}}$.

Answer 2

Let $t=\frac yx$. (Note the we can do this only for $x\ne0$. But $x=0$ should not be a problem, since the whole expression is zero in that case.)

Then we have $$\frac{xy(x+y)}{x^2-xy+y^2}=\frac{x^3}{x^2} \cdot \frac{t(1+t)}{1-t+t^2}$$ If we can show that the expression $$g(t)=\left|\frac{t(1+t)}{1-t+t^2}\right|$$ is bounded, then this is sufficient to show that the limit is zero.

For example:

• Can you show that for $t\le\frac12$ you have $|t^2-t| \le t^2-t+1$, which means that in this case we have $g(t)<1$?
• Can you show that for $t>\frac12$ the value $g(t)$ is positive and $t^2+t<3(t^2-t+1)$, which means that $g(t)<3$?

Or simply try some other way to show that this expression is bounded.

If will add plot form WA:

Answer 3

No, that is not sufficient. That can only show that if the limit exists, then it should equal $0$. But it doesn't show that the limit exists.

Change to polar coordinate system, so that

$$ \lim_{(x,y)\to (0,0)} \frac{xy(x+y)}{x^2-xy+y^2} = \lim_{r\to 0} \frac{ (r^2 \sin \theta \cos \theta) (r\sin \theta + r\cos \theta)}{r^2-r^2\sin \theta \cos \theta}= \lim_{r\to 0} \frac{r\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta} $$

(Check that two limits are equilvalent from the ($\varepsilon$-$\delta$) definition of the limits)

But notice that $|1-\sin \theta \cos \theta|=|1-\sin (2\theta) /2|\ge 1/2$.

Hence $\displaystyle \frac{\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}$ is bounded for all $\theta\in \mathbb{R}$, and so

$$ \lim_{r\to 0} \frac{r\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}=0 $$