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If I wanted to work out the range of values for something like

$$\frac{xy}{x + y},$$

then what do I take into consideration first? Clearly, if $x = y = 0$, then the denominator is $0$ which is "not allowed", but at the same time, the numerator is $0$ so doesn't that make the whole fraction $0$ before we look at the denominator?

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    $\begingroup$ Why are people looking at the limit in the origin? That's not what's he's asking about. $\endgroup$ – Git Gud Mar 23 '13 at 15:16
  • $\begingroup$ @GitGud: because he says he in interested in $x=y=0$ $\endgroup$ – Ross Millikan Mar 23 '13 at 15:55
  • $\begingroup$ @RossMillikan That was just an example to highlight my question $\endgroup$ – Kaish Mar 23 '13 at 16:48
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Take $y=kx$ then $\displaystyle \frac{xy}{x+y}=\frac{kx}{(1+k)}$

If $k\ne -1$ is fixed then then the expression can be anything between $-\infty$ and $+\infty$ by varying $x$ appropriately. So the range of this function is $(-\infty,\infty)$

Geometric idea: You choose any curve in $R^2$ and show that as you vary your points on that curve to get any desired value for the expression . In this case i chose a line in $R^2$.

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If the denominator is zero, that is not allowed. It is possible the hole is removable. If you look at $y=\dfrac {x^2-4}{x-2}$ it has no value at $x=2$ but it does have a limit of $4$ as you approach $x=2$. In fact its graph is just the line $y=x+2$ with the point $(2,4)$ missing. Sometimes it makes sense to redefine the function to add that point.

In your example, there is no limit at $x=y=0$ If you approach $(0,0)$ along either axis the value of the fraction is nicely $0$. But it has no value at all along the line $x=-y$ and if you approach along a path that approaches that it will blow up. Try $y=x+x^2$, for example.

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    $\begingroup$ That's not what the question is asking. $\endgroup$ – Javier Mar 23 '13 at 15:29

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