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$$f(x,y)=\left \{ \begin{array}{lr} \dfrac{xy^3+xy\sin(2015x +2016y)}{(x^2+y^2)e^{x^2-y^2}} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{array} \right. $$

The question is whether this function is continuous at $(0,0)$ or not. I tried showing it is continuous by definition, but with no success so far...

Thanks for any help.

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    Consider polar coordinates and then $\lim_{r \to 0}$. – 2017-01-05
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    I tried that :) – 2017-01-05

4 Answers 4

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I changed to polar coordinates $x=r \cos \varphi$ and $y =r \sin \varphi$ for

$$f(r,\varphi) = {\rm e}^{(r^2-2 r^2 \cos^2 \varphi)} \left( \sin \varphi \cos \varphi \sin(2015 r \cos \varphi + 2016 r \sin \varphi) + r^2 \sin^3 \varphi \cos \varphi \right) $$

This clearly shows that $f(0,\varphi)=0$ with any divisions by zero.

Now for the continuity question. We have to look at $g(r,\varphi) = \frac{\partial}{\partial r} f(r,\varphi)$.

You can do the math (or use Wolfram) and show that

$$ g(0,\varphi) = 2015 \sin\varphi \cos^2 \varphi + 2016 \sin^2 \varphi \cos\varphi $$

The above is continuous for all $\varphi$ and since $f(0,\varphi)=0$ and $\frac{\partial}{\partial r}f(r,\varphi) |_{r=0} = 0$ it is continuous near $r=0$ also.

The 2nd order taylor expansion of $f(r,\varphi)$ is

$$f(r,\varphi) \approx r \left( (2015 \sin\varphi \cos^2 \varphi + 2016 \sin^2 \varphi \cos \varphi) + r (\sin^3 \varphi \cos \varphi) \right) $$

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    Isn't the fact that $\lim_{r\rightarrow 0} f(r,\varphi)$ is independent of $\varphi$ sufficient to establish continuity at 0? Continuity at all other points is guaranteed by the continuity of the numerator and denominator. – 2017-01-05
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    @eyeballfrog You mean independent of $r$? – 2017-01-05
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For informal exlploration, we know $\sin x \approx x$ for $x \ll 1$ and $e^{x^2-y^2}\approx 1$, so $$f(x,y)=\frac{xy^3+xy\sin(2015x +2016y)}{(x^2+y^2)e^{x^2-y^2}}\approx \frac{xy^3+2015x^2y+2016xy^2}{x^2+y^2}$$ The first term in the numerator is small compared to the others and can be ignored. The multiplicative factor in the second and third does not matter for convergence and the two have the same shape, so we will just ask if $$\frac {xy^2}{x^2+y^2}$$ is continuous at the origin. Converting to polar coordinates we have $$\frac {r^3\cos ^2 \theta \sin \theta }{r^2}$$ which goes nicely to zero with $r$

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In polar coordinates $(r, \theta)$:

$$f(r,\theta) = \left \{ \begin{array}{lr} \frac{r^4\sin^3(\theta)\cos(\theta) + r^2\sin(r(2015\cos(\theta) + 2016\sin(\theta)))}{r^2e^{r^2(\cos^2(\theta)-\sin^2(\theta))}} & r \neq 0 \\ 0 & r = 0 \end{array} \right. $$

Let: $$f(\theta) = \sin^3(\theta)\cos(\theta) \\ g(r, \theta) =\sin(r(2015\cos(\theta) + 2016\sin(\theta))) \\ h(\theta) = (\cos^2(\theta)-\sin^2(\theta))$$ Note that $-1\leq f, g, h \leq 1$. Then:

$$\lim_{r \to 0 } \frac{r^4 f(\theta) + r^2 g(r, \theta)}{r^2h(\theta)} = \lim_{r \to 0 } \frac{r^4 f(\theta)}{r^2h(\theta)} + \lim_{r \to 0 } \frac{r^2 g(r, \theta)}{r^2h(\theta)} = 0 + \lim_{r \to 0 } \frac{r^2 g(r,\theta)}{r^2h(\theta)} $$

If then $ \displaystyle \lim_{r \to 0 } \frac{r^2 g(r,\theta)}{r^2h(\theta)} = 0 $ then $f$ is continuous. Specifically, since $h$ is bounded, if $g(r, \theta)$ tends to $0$. Looking at $g$ and since $(2015\cos(\theta) + 2016\sin(\theta))$ is bounded:

$$\lim_{r \to 0} \sin(r(2015\cos(\theta) + 2016\sin(\theta))) = \sin(0) = 0$$

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$\lvert f(x,y)\lvert=\left\lvert\dfrac{xy^3+xy\sin(2015x +2016y)}{(x^2+y^2)e^{x^2-y^2}}\right\lvert=\underbrace{\dfrac{|xy|}{x^2+y^2}}_{\le \frac 12}\times\underbrace{\dfrac{1}{e^{x^2-y^2}}}_{\le 2}\times\left\lvert y^2+\sin(2015x+2016y)\right\lvert$

  • Since $\lim\limits_{u\to 0}e^u=1$ then for $(x,y)$ sufficiently small we have $e^{x^2-y^2}\ge \frac 12$

  • From $(|x|-|y|)^2=x^2+y^2-2|xy|\ge 0$ we get $\dfrac{|xy|}{x^2+y^2}\le \frac 12$

Consequently we get the simpler inequality

$$\lvert f(x,y)\lvert\le\underbrace{|y|^2}_{\to 0}+|\sin(\underbrace{2015x+2016y}_{\to 0})|\to 0$$

since $\lim\limits_{u\to 0}\sin(u)=0$