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Consider the function $ f:\mathbb{R^2} \rightarrow \mathbb{R}$ given by

$ f(x,y) = \left\{ \begin{array}{l l} \frac{xy}{x^2 + y^2} & \quad \text{if (x,y) $\neq$ (0,0)}\\ 0 & \quad \text{if (x,y) = (0,0)} \end{array} \right.$

Show that f is continuous along every horizontal and every vertical line (i.e. for every $x_0, y_0 \epsilon \mathbb{R}$, the functions $g,h:\mathbb{R}\rightarrow \mathbb{R}$ given by $g(t) = f(x_0,t)$ and $h(t)=f(t,y_0)$ are continuous).

I know that this function is not continuous at $(0,0)$, but the horizontal/vertical line issue is proving difficult to work with. Any help/hints appreciated!

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    $\begingroup$ If the line does not passes through the origin, then as $f$ is a quotient of two polynomials, it is continuous. If the line does pass through the origin, then it is $x=0$ or $y=0$. In this case the function is $0$, thus continuous. $\endgroup$ – user99914 Nov 26 '13 at 1:31
  • $\begingroup$ Actually, $f$ is continuous when restricted to lines of the form $ax+ by = 0$, not only vertical and horizontal one. $\endgroup$ – user99914 Nov 26 '13 at 1:34
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    $\begingroup$ @John: False for this function. $\endgroup$ – Ted Shifrin Nov 26 '13 at 1:40
  • $\begingroup$ @Ted Shifrin: Oh you are right. So used to "that"function. $\endgroup$ – user99914 Nov 26 '13 at 1:53
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Along a fixed horizontal line $y = c$ (constant). Put $y = c$ in the function and get $f(x,c) = \frac{xc}{x^2 + c^2}$ when $x \neq 0$ , which is a function of $x$ only and see it is a continuous one variable function. Do the same for a verticle line.

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  • $\begingroup$ Don't forget to check the special cases, i.e. what happens when $c=0$? What happens when $(x,c)\rightarrow(0,0)$? Also note that in this case the answer does not depend on $c$ $\endgroup$ – Octania Dec 23 '14 at 10:19

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