I'm trying to check whether the function $$ f(x,y)=\begin{cases} \dfrac{1}{x^2+y^2} & \text{for $(x,y)\ne(0,0)$}\\[6px] 0 & \text{for $(x,y)=(0,0)$} \end{cases} $$ is continuous. My problem is with trying to check if $$ \lim_{(x,y)\rightarrow(0,0)}\frac{1}{x^2+y^2} = 0, $$ Any clues on how to do it?
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It should be really obvious that the limit at the origin is not zero. – Jun 23 '16 at 19:56
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1What do you mean by continuous? The function is not defined at $(0,0)$ (division by $0$ is undefined). That point is outside the function's domain. For continuity or discontinuity points need to be in the function's domain. You can only talk about continuity or dicontinuity when you've assigned some value to that point. – Aritra Das Jun 23 '16 at 19:58
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1Try with $x=y=1/n$ and do the limit as $n\to\infty$: what do you get? Besides, there is no point in computing the limit, unless you're given that $f(0,0)=0$ and $f(x,y)=1/(x^2+y^2)$ for $(x,y)\ne(0,0)$. – egreg Jun 23 '16 at 20:01
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@egreg yes, f(0,0) = 0. I'll edit the question. – Pedro Destri Jun 23 '16 at 20:10
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The function is clearly continuous everywhere, except at $(0,0)$.
To see why it's not continuous at $(0,0)$, consider the restriction of the function to a line through the origin, for instance $y=0$. Since $$ \lim_{x\to0}f(x,0)=\lim_{x\to0}\frac{1}{x^2}=\infty $$ you have the answer.
egreg
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