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How to prove $f(x, y)=\dfrac{x}{y}$ ($y$ is not $0$) is continuous using continuity definition? couldn't figure out how to define delta...

kimtahe6
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user99051
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    It's clearly not continuous at $(0, 0)$. Do you mean at a specific point, or on the plane except at $0$? Can you share your thoughts on the problem, and explain what you've tried? –  Oct 06 '13 at 03:00
  • One thing that you can do is to show that $g(x,y)=xy$ is continuous, then show that $h(y)=\frac{1}{y}$ is continuous if $y\neq 0$. Then the composition of continuous functions is continuous. – Baby Dragon Oct 06 '13 at 03:19
  • @T.Bongers It is not defined there. I makes no sense to discuss continuity in a point where the function is not defined. – Pedro Oct 07 '13 at 01:56
  • @PedroTamaroff Before the question was edited, it did not specify the domain of definition. I was inviting the OP to consider the point $(0, 0)$ to show that the function isn't continuous on the whole plane, and to restrict the question accordingly. –  Oct 07 '13 at 01:58
  • @T.Bongers The function is continuous where it is defined. It is not defined on the whole plane, thus it is meaningless to say "the function isn't continuous on the whole plane". – Pedro Oct 07 '13 at 02:01

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Hint: To show continuity at a point $(x_0, y_0)$, Pick $\delta$ so that $$ \delta = \min\left\{\frac{|y_0|^2\epsilon}{4|x_0|}, \frac{|y_0|\epsilon}{4}, \frac{|y_0|}{2}\right\} $$

($|x_0|$ could be zero, in which case just ignore the first value in the min.) Then show $$ \left| \frac{x}{y} - \frac{x_0}{y_0} \right| = \left| \frac{x_0(y_0-y)}{yy_0} + \frac{y_0(x - x_0)}{yy_0} \right| $$

From here, use triangle inequality, and note that $|y_0 - y|, |x - x_0| < \delta$. Finally, you will need to use $$ \left|\frac{1}{y} \right|= \frac{1}{|(y - y_0) + y_0|} < \frac{1}{|y_0| - \delta} \le \frac{1}{|y_0|/2} $$ Good luck!

Caleb Stanford
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  • How did you figure out to choose these three values for delta in the first place? I am okay with the later two, but not quite understand the first one...my final result is a bit different from the first value. According to my final result, I should have $2(|x_0|+|y_0|)$ for the denominator for the first delta value. – user99051 Oct 06 '13 at 20:15
  • @user99051 you're right, there's an error in my first delta value. Anyway the way you come up with the values in the minimum expression for $\delta$ is you just go through with the proof as if delta were already decided and then whenever you need delta to be less than some specific thing, add that thing to the minimum. – Caleb Stanford Oct 07 '13 at 01:48
  • @user99051 Sorry for the errors; I've fixed it now. I now use $x_0(y_0 - y) + y_0(x - x_0)$ instead of $x(y_0 - y) + y(x - x_0)$ because it's considerably cleaner. – Caleb Stanford Oct 07 '13 at 01:54