I have this piecewise function and I have to determine if it is continuous in $(0,0)$ $$f(x)=\begin{cases}\dfrac{\sqrt[]{|x|}\sin^2y}{x^2+y^2}&\text{if }(x,y)\neq (0,0)\\0&\text{if }(x,y)=(0,0)\end{cases}$$ So I have to prove if $$\lim_{(x,y)\to(0,0)}\dfrac{\sqrt[]{|x|}\sin^2y}{x^2+y^2}=0$$

Thus I used polar coordinates and got

$$\lim_{\rho\to0}\dfrac{\sqrt[]{\rho|\cos\theta|}\sin^2(\rho\sin\theta)}{\rho^2}$$ I know that $\cos x\leq1, \sin x\leq1$, then $$\frac{\sqrt[]{\rho|\cos\theta|}\sin^2(\rho\sin\theta)}{\rho^2}<\frac{\sqrt\rho\sin^2\rho}{\rho^2}=g(\rho)$$ $$\lim_{\rho\to0}g(\rho)=0\implies \lim_{(x,y)\to(0,0)}\dfrac{\sqrt[]{|x|}\sin^2y}{x^2+y^2}=0$$ However Wolfram Alpha says the limit does not exist. Why? I thought that this procedure of polar coordinates yields the final result.

Could Wolfram Alpha also consider complex $x,y$?


1 Answer 1


I think your proof is (essentially) correct (I'd say that $\cos x \le 1, \sin x \le 1$). A similar approach would be to note that $|\sin y| \le |y|$ for all real $y$, and hence for $x^2+y^2 > 0$ we have

$$\frac{\sin^2 y}{x^2+y^2} \le 1$$

and thus

$$|f(x,y)| \le \sqrt{|x|}$$

which clearly shows your limit is correct.

From the message by Wolfram Alpha about the value being dependend on the path in the complex space, maybe the reason is that you only consider real $x,y$ while Alpha considers complex values?

  • $\begingroup$ Thank you! You could be right about WA. Can I check the limit only for real x,y in WA? $\endgroup$
    – Lorenzo B.
    Oct 19, 2018 at 12:05
  • $\begingroup$ I have absolutely no experience with it, maybe somebody else can help? $\endgroup$
    – Ingix
    Oct 19, 2018 at 13:11

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