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Hello im studying calculus at the university and I dont know how to solve the following exercise: Study the continuity of the next function: $$f(x,y) = \begin{cases} \frac{x^2-xy}{x+y}&\text{for } x+y\ne0\\ 0 &\text{for }(x,y) =(0,0). \end{cases}$$

I've tried to resolve it with iterated limits and directional limits, but im sure if its correct.

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6  
Note that the function is not defined on the line $x+y=0$ except at the one point $\langle 0,0\rangle$. –  Brian M. Scott Dec 21 '12 at 12:05

5 Answers 5

up vote 11 down vote accepted
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Note that $f(0,t) = 0$ (for $t \neq 0$), but $$f(t,-t+t^2) = \frac{t^2+t^2-t^3}{t^2} = 2-t,$$ again for $t \neq 0$. What happens when $t \to 0$? This shows that $f$ is not continuous at $(0,0)$.

On the other hand, $f$ is continuous everywhere else where $f$ is defined, since the numerator and denominator clearly are continuous.

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For the function to be continuous at a point $(x_0,y_0)$, you need to prove that

$$ \lim f(x,y) = f(x_0,y_0)\quad \mathrm{as}\quad (x,y)\to (x_0,y_0). $$

To find the limit of the function at the point $(0,0)$, use the polar coordinates $x=r\cos(\theta), y= r\sin(\theta)$ and consider taking the limit as $r\to 0.$

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By using polar coordinates we can find the limit, if it exists, but not show that it actually exists, only that it is the same along any line through the point $(x_0,y_0)$. However, since the limit does not exist in this case, there is enough to find two cases in which the results of the limit differ, which can be done by using polar coordinates... Using polar coordinates in this case and letting $t$ tend to 0 would get the limit 0, which is clearly not the limit along all curves as shown by mrf in one (the) other answer to this post. –  malin Dec 28 '12 at 23:08

mrf's solution is simple and correct. Here I want to show an alternative one. The only question is $\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$ or not. $$ f(x,y)=\frac{x^2-xy}{x+y}=x\frac{x+y-2y}{x+y}=x-2\frac{1}{\frac{1}{x}+\frac{1}{y}}. $$ Obviously $\lim_{(x,y)\to(0,0)}x=0$. If we can choose $y:=y(x)$, ($y(x)\neq -x$), such that $\lim_{x\to 0}\left(\frac{1}{x}+\frac{1}{y(x)} \right)\neq\pm\infty$, then $f$ is not continuous in $(0,0)$. Choose, for example, $\frac{1}{x}+\frac{1}{y(x)}=1$, that is, $y(x):=\frac{x}{x-1}$, where we may assume that $x\neq 1$ because $x\to 0$. ($y(x)\neq -x$ for any $x\neq 0$.)

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$$ \text{We choose a path like} \hspace{20mm} y=mx$$ $$ f=\frac{x^2-xy}{x+y}=\frac{x^2-x^2m}{x+my}=\frac{x^2(1-m)}{x(1+m)}$$ $$ \lim_{x \rightarrow \alpha} \frac{x(1-m)}{(1+m)}=\frac{\alpha (1-m)}{(1+m)}$$ $$\text{answer of the limit is depend on path (m) , so } \mathcal{f} \text{ is not continuous} $$

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The limit doesn't depend on $m$ when $\alpha = 0$ (and this is the interesting case). –  mrf Jan 1 '13 at 16:45

Yet another way to show non-continuity in $(0,0)$ is to show that $f$ is unbounded in every neighborhood of $(0,0)$:

Let $\epsilon > 0 $ and $x\in\mathbb{R}$ such that $0<x<\frac{\epsilon}{2}$. Let $\delta\in\mathbb{R}_+$. We have $$\|(x, -x + \delta)\|_2 = \sqrt{x^2 + (-x+\delta)^2} = \sqrt{2x^2 - 2x\delta + \delta^2}\\ \leq \sqrt{2x^2 + \delta^2} \leq \sqrt{\frac{\epsilon^2}{2} + \delta^2} < \epsilon $$ for sufficiently small $\delta$, so $(x, -x + \delta)\in B_{\epsilon}(0,0)$. However, $$ f(x, -x + \delta) = \frac{x^2- x(-x+\delta)}{\delta} = \frac{2x^2}{\delta} - x$$ and we see that $$ \lim_{\delta\rightarrow 0}\; \left| f(x,-x+\delta) \right| = \infty, $$ i. e. $f$ is unbounded on $B_{\epsilon}(0,0)$.

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