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I want to show that the function $f : \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ defined by

$ f(x,y)= \begin{cases} \frac{xy}{x^{2}+y^{2}},& \text{if } (x,y)\neq (0,0)\\ 0, & \text{otherwise} \end{cases} $

is continuous on $\mathbb{R} \times \mathbb{R} \setminus{(0,0)}$.

Here is my approach so far:

Fix $(x_{0},y_{0}) \in \mathbb{R} \times \mathbb{R} \setminus{(0,0)}$. We will show $f$ is continuous here, that is:

$\forall$ $\epsilon > 0$, $\exists \ \delta > 0$ such that

$d((x,y),(x_{0},y_{0})) < \delta \implies |f(x,y) - f(x_{0},y_{0})| < \epsilon$

I'm not sure which metric to use for the delta condition yet, but let's say I use $d((x,y),(x_{0},y_{0})) = |x-x_0| + |y - y_{0}|$.

the epsilon conditions holds if and only if

$$|\frac{xy}{x^{2}+y^{2}}-\frac{x_{0}y_{0}}{x_{0}^{2}+y_{0}^{2}}| < \epsilon$$ $$ \iff |\frac{xx_{0}(yx_{0}-xy_{0})+yy_{0}(xy_{0}-yx_{0})}{x^{2}+y^{2}}| < \epsilon|x_{0}^2 + y_{0}^2| := \epsilon'$$

Now, it's relatively easy to bound the terms in the numerator, given $|x-x_0| + |y - y_{0}| < \delta$. But I'm having a lot of trouble trying to show that $|x^{2}+y^{2}|> \eta$, for some $\eta > 0$. Surely this must be true, since if $(x,y)$ is close to $(x_{0},y_{0})$, then $x^{2}+y^{2}$ must be close to $x_{0}^{2} + y_{0}^{2}$.

Is this the right direction to head with this proof? Can I find a lower bound for $|x^{2}+y^{2}|$ using the triangle inequality?

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    This function has no limit at $(0,0)$. At other points, polynomial functions are continuous as well as quotients thereof, i.e. rational functions. – 2017-02-12
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    If a derivative is defined then a function must be continuous in that domain. Though it is not the necessary condition but it is sufficient to conclude continuity. – 2017-02-12
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    Your question is very ambigus. You want to show that $f$ is continuous on $\mathbb R^2\backslash \{(0,0)\}$ ? If $h,g$ are continuous and $g\neq 0$, then $h/g$ is continuous... there is noting to prove here... – 2017-02-12
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    @Surb So you can just use the fact that the quotient of continuous functions is continuous without proof? – 2017-02-12
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    @GitGud: You can (more) easily show it in a general way and use it in this specific example ;-) – 2017-02-12

2 Answers 2

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Using the property of the lengths of a triangle vertices $A = (0,0), B = (x_0,y_0), C = (x,y)$ we have: $AC > AB - BC \implies \sqrt{x^2+y^2} > \sqrt{x_0^2+y_0^2} - BC > \sqrt{x_0^2+y_0^2} - \delta > \sqrt{x_0^2+y_0^2} - f(\epsilon)$

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    Why is $\delta < f(\epsilon)$? and can we also be sure that $\sqrt{x_{0}^{2}+y_{0}^{2}}-f(\epsilon) > 0$? – 2017-02-12
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    ah right, we can just take $\delta < min\{ f(\epsilon),..$whatever else I need$.,\}$. Thanks! – 2017-02-12
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I think your method is better than polar coordinates, but that polar is easier : $$|f(r,\theta)-f(r_0,\theta_0)|=\frac12|\sin2\theta-\sin2\theta_0|\leq|\theta-\theta_0|$$