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.

  • $\begingroup$ It is enough to show that $\frac{\sin^2\theta \cos\theta}{ \cos\theta-\sin\theta}$ is a bounded function of $\theta$. $\endgroup$
    – Fin8ish
    Commented Apr 22, 2021 at 20:26
  • $\begingroup$ Hint: What your failed attempts are trying to tell you is that you should instead try to prove that the limit does not exist. $\endgroup$ Commented Apr 22, 2021 at 20:27
  • 1
    $\begingroup$ @Finish: ... which it isn't. $\endgroup$ Commented Apr 22, 2021 at 20:27
  • $\begingroup$ 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. $\endgroup$
    – Fin8ish
    Commented Apr 22, 2021 at 20:36
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    $\begingroup$ The inequality trick requires that $a,b>0$. $\endgroup$
    – user284331
    Commented Apr 22, 2021 at 20:44

3 Answers 3


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.

  • $\begingroup$ According to the wolframalpha site, the limit exists and the value is zero. See wolframalpha.com/input/… $\endgroup$
    – Fin8ish
    Commented Apr 22, 2021 at 20:42
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    $\begingroup$ What do I have to do with that? I did not write that software. By the way,$$\lim_{y\to0}\left|f\left(y+y^4,y\right)\right|=\infty.$$ $\endgroup$ Commented Apr 22, 2021 at 20:48
  • $\begingroup$ You are right. The limit does not exist. $\endgroup$
    – Fin8ish
    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 $f(x_{n},y_{n})\rightarrow\infty$.


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