I have to prove that the function $f:\Bbb{R}\times\Bbb{R}\to\Bbb{R}$ defined as follows: $$f(x, y)=\frac{xy}{x^2+y^2}\text{for } (x, y)\neq (0,0)$$ $$f(x, y)=0\text{ for }(x, y)=(0,0)$$when taken as a function of $x$, is continuous.

My method:

Take any $\epsilon>0$. I have to prove that there exists a $\delta$ such that $\left|\frac{(x+\delta)y}{(x+\delta)^2+y^2}-\frac{xy}{x^2+y^2}\right|<\epsilon$. I'm getting a quadratic inequality in $\delta$.

My question is, what if the situation was more complicated? I got easily get a very messy expression for $\delta$. What should we do then? How should we prove continuity?

Thanks in advance!


marked as duplicate by Arnaud D., Cesareo, Claude Leibovici calculus Sep 20 '18 at 16:05

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If $y=0$, then $f(x,y)=0$ for all $x$, so it is continuous.

If $y\not=0$, then $x^2+y^2\ge y^2 > 0$ for all $x$. Now let $x_n\rightarrow 0$. Then

$$\frac{x_n y}{x_n^2 + y^2}\le \frac{x_n y}{y^2} =\frac{x_n}{y}\longrightarrow 0$$

as $n\rightarrow\infty$. So $f$ is continuous at $x=0$. Continuity away from $0$ is clear because it is a rational function of $x$ and the denominator has no zeros.

  • $\begingroup$ He's asking for a proof that $f$ is continuous as a function of $x$. $\endgroup$ – Justin Campbell Feb 10 '14 at 15:16
  • $\begingroup$ You have to assume that $F(x\times y)$ is a function of $x$. The book says it is continuous. $\endgroup$ – fierydemon Feb 10 '14 at 15:17
  • $\begingroup$ Oh sorry, I overread, but that's even more obvious. $\endgroup$ – J.R. Feb 10 '14 at 15:18

That's why we normally begin by proving some useful properties after having introduced a concept. For example:

Let $f,g:\Bbb R\to \Bbb R$

  • If $f$ and $g$ are continuous, then so are $f+g$, $f\cdot g$, and $\alpha f$ for $\alpha\in\Bbb R$.
  • If $f$ and $g$ are continuous, then so is $f/g$ wherever $g\ne 0$.

From these rules it follows immediately that $\frac{xy}{x^2+y^2}$ is continuous for a fixed $y\ne0$, as then the denominator is $>0$ for all $x$.

If $y=0$, then the situation is even easier. The function is constant $0$, and a constant map between topological spaces is always continuous.

  • $\begingroup$ This doesn't address continuity at $x = 0$ when $y \neq 0$, which is the interesting part about his example. $\endgroup$ – Justin Campbell Feb 10 '14 at 15:30
  • 1
    $\begingroup$ @JustinCampbell: If $y\ne 0$, then we have a quotient of continuous functions. $\endgroup$ – Stefan Hamcke Feb 10 '14 at 15:31
  • $\begingroup$ Good point! I remember there being something interesting about his example, but it's not what he asked about. It's probably that $f$ is continuous as a function of $x$ and $y$ separately, but not actually continuous at $(0,0)$. $\endgroup$ – Justin Campbell Feb 10 '14 at 15:40

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