# Finding the double limit of a function: $\lim_{(x,y)\rightarrow(0,0)} \frac{x^2y}{x-y}$

I'm trying to determine the limit of the following function as it approaches (0,0): $$\lim_{(x,y)\rightarrow(0,0)} \frac{x^2y}{x-y}$$

The function is not defined for x=y, {(0,0)}. Just like with one-variable functions, where I could find one-sided limits, here I can examine all paths leading to (0,0), apart from y=x. I've tried writing the limit in polar coordinates, eventually getting at: $$\lim_{r\rightarrow0} \frac{r^2sin^2\theta cos\theta}{cos\theta-sin\theta}$$ I can't prove that the limit approaches 0, since the part that is dependent on $$\theta$$ in the denominator can vary and be as small as I want while $$\theta$$ approaches $$\frac{\pi}{4}$$, meaning I can't bound it.

I've tried finding an upper bound function that approaches 0 in order to use the "Sandwich" theorem without success, as well.

I would appreciate some help.

• It is enough to show that $\frac{\sin^2\theta \cos\theta}{ \cos\theta-\sin\theta}$ is a bounded function of $\theta$. Commented Apr 22, 2021 at 20:26
• Hint: What your failed attempts are trying to tell you is that you should instead try to prove that the limit does not exist. Commented Apr 22, 2021 at 20:27
• @Finish: ... which it isn't. Commented Apr 22, 2021 at 20:27
• Use the inequality $\dfrac{2ab}{a+b}=\dfrac{2}{\frac{1}{a}+\frac{1}{b}} \leq \dfrac{a+b}{2}$ to show that $\dfrac{-xy}{x-y} \leq \dfrac{x-y}{4}$. Then, $\dfrac{x^2y}{x-y} \leq (-x) \left( \frac{x-y}{2} \right)$. Hence the your limit is zero. Commented Apr 22, 2021 at 20:36
• The inequality trick requires that $a,b>0$. Commented Apr 22, 2021 at 20:44

If $$y=0$$, then $$f(x,y)=0$$ and therefore the limit, if it exists, is $$0$$. But$$f(y+y^3,y)=\frac{(y+y^3)^2y}{y^3}=(1+y^2)^2,$$ and therefore the limit, if it exists, is $$1$$. So, there is no limit.

• According to the wolframalpha site, the limit exists and the value is zero. See wolframalpha.com/input/… Commented Apr 22, 2021 at 20:42
• What do I have to do with that? I did not write that software. By the way,$$\lim_{y\to0}\left|f\left(y+y^4,y\right)\right|=\infty.$$ Commented Apr 22, 2021 at 20:48
• You are right. The limit does not exist. Commented Apr 22, 2021 at 20:56

If you consider the path $$y= x-x^4$$ you will see that along this path the limit is $$+\infty$$.

How to see that? The denominator $$x-y$$ suggests that if you get fast enough close to the line $$x=y$$ while approaching $$(0,0)$$, then you can drive the value of the expression to $$+\infty$$. So, a standard trick in such a case is considering

$$y= x-x^n\Rightarrow \frac{x^2y}{x-y}=\frac{x^2(x-x^n)}{x^n}=\frac{x^3}{x^n}-x^2$$ Now, you see immediately that for $$n>3$$ approaching $$(0,0)$$ along $$(x,x-x^n)$$ gives $$+\infty$$.

Let $$x_{n}=r_{n}\cos^{2}\theta_{n}$$ and $$y_{n}=r_{n}\sin^{2}\theta_{n}$$, where $$r_{n}$$ and $$\theta_{n}$$ will be determined later, then \begin{align*} f(x_{n},y_{n})=r_{n}^{2}\dfrac{(\cos^{4}\theta_{n})(\sin^{2}\theta_{n})}{\cos(2\theta_{n})}. \end{align*} Now we choose $$\theta_{n}=\pi/4-1/n$$, we have \begin{align*} f(x_{n},y_{n})=r_{n}^{2}\dfrac{(\cos^{4}\theta_{n})(\sin^{2}\theta_{n})}{\sin(2/n)}=r_{n}^{2}\dfrac{(\cos^{4}\theta_{n})(\sin^{2}\theta_{n})(2/n)}{\sin(2/n)}\dfrac{n}{2}. \end{align*} Let $$r_{n}=1/\sqrt[4]{n}$$ and use the fact that $$\dfrac{2/n}{\sin(2/n)}\rightarrow 1$$, it is easy to see that $$f(x_{n},y_{n})\rightarrow\infty$$.