For what values of n does the following limit exist?

$$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}$$

I've tried using the squeeze theorem but couldn't come up with anything good. Also converting it to cylindrical coordinates gave me the answer that the limit exists for all n (which I don't know if it's correct).


3 Answers 3



$$x^{2n} + y^2 = |x|^{2n} + |y|^2 \geqslant 2|x^n||y|$$


$$0 \leqslant \left|\frac{x^2y}{x^{2n} + y^2} \right| = \frac{|x|^2|y|}{x^{2n} + y^2} \leqslant \frac{|x|^{2-n}}{2}.$$

Consider the squeeze theorem to find a sufficient condition on $n$ for convergence to $0$. Then consider what happens if this condition is not satisfied.

  • $\begingroup$ So for all n in (-inf, 2] using your method? When I delete the y^2 term I get (-inf, 1]. $\endgroup$
    – Linsu Han
    Feb 6, 2017 at 5:02
  • $\begingroup$ This should give you that sufficient condition on $n$ for which it converges. Then look at the answer by@ MyGlasses. $\endgroup$
    – RRL
    Feb 6, 2017 at 5:04
  • $\begingroup$ @MyGlasses: $(|x|^n - |y|)^2 \geqslant 0 \implies |x|^{2n} - 2|x|^n|y| + |y|^2 \geqslant 0 \implies \ldots$ $\endgroup$
    – RRL
    Feb 6, 2017 at 5:14
  • $\begingroup$ @MyGlasses: Yes, if $n < 2$ this converges to $0$ and your solution shows limit does not exist if $n \geqslant2$. $\endgroup$
    – RRL
    Feb 6, 2017 at 5:39

We know with $x=0$ as $y\to0$ then $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}=\lim_{y\to0} \frac{0}{0+y^2}=0$$

For $n>2$ this limit doesn't exist. Let $y=x^n$ so $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}=\lim_{(x,y) \to (0,0)} \frac{x^{n+2}}{2x^{2n}}=\infty$$ This substituation shows for $n=2$ this limit doesn't exist. Let $y=x^2$ so $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}=\lim_{(x,y) \to (0,0)} \frac{x^4}{2x^4}=\frac12$$ For $n=1$ the limit exists. By polar coordinates $$\Big|\frac{x^2y}{x^2+y^2}\Big|=\Big|\frac{r^3\cos^2\theta\sin\theta}{r^2}\Big|=|r\cos^2\theta\sin\theta|<r$$

  • $\begingroup$ Good complement to hint +1 $\endgroup$
    – RRL
    Feb 6, 2017 at 5:05
  • $\begingroup$ What about $n \in (1,2)$? $\endgroup$ Feb 6, 2017 at 5:09

Let's eliminate first the cases $n<0$ and $n\ge2$.

  • $n<0$ converges

$$f_n(x,y)=\frac{x^2y\ .\ x^{2|n|}}{1+y^2x^{2|n|}}\sim x^{2|n|+2}y\to 0$$

  • $n>2$ diverges

$$f_n(x,x^n)=\frac{x^2x^n}{x^{2n}+x^{2n}}=\frac{1}{2}x^{2+n-2n}=\frac{1}{2}x^{2-n}\to \infty$$

  • $n=2$ has multiple limits, so doesn't converges

$$f_2(0,y)=\frac{0}{0+y^2}=0\to 0$$

$$f_2(x,x)=\frac{x^3}{x^4+x^2}\sim\frac{x^3}{x^2}\sim x\to 0$$


So let's have now $n\in[0,2[$

We loose no generality in writting $y=x^ng(x,y)$, the function $g$ being arbitrary.


According to this study : range of $\frac{x}{x^2+1}$

We can say that $|\frac{g}{1+g^2}|<1$

Thus $|f(x,y)|\le|x^{2-n}||\frac{g}{1+g^2}|<|x^{2-n}|\to 0$

$\color{teal}{Conclusion :}$ $\lim\limits_{x,y\to 0}f(x,y)$ exists for $n\in]-\infty,2[$ and this limit is $0$.

Of course we can also do this directly by setting $u=\frac{y}{x^n}$.


And since $|\frac{u}{1+u^2}|\le\frac 12$ is bounded, $f_n(x,y)\to 0$ for $n<2$.

Also since we can choose $y$ so that this expression take any desired value in $[-\frac12,\frac12]$ then $f_n(x,y)$ cannot converge for $n\ge 2$.

Note: This is because it has multiple limits. And for $n>2$ the divergence to $\infty$ is not even required, just suffice to say that $-\infty$, $0$, $+\infty$ are all possible limits.


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