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What can we do with this function, so the function will be continuous in $(0,0)$?

$f:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto \frac{x^2+y^2-x^3y^3}{x^2+y^2}$

What I think we should do, is:

approximate $(0,0)$ via the line $y=x$, so substitute $y=x$ and take the limit of that function, i.e. $\lim_{x\rightarrow0}$:

$\lim_{x\rightarrow0} 1 - \frac{x^6}{2x^2} = \lim_{x\rightarrow0}1-\frac{x^4}{2} = 1$

So the new function, that is continuous in $(0,0)$ is defined by:

$F:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto F(x) = \begin{cases} 1 & \mbox{if } (x,y) = (0,0) \\ f(x) & \mbox{if } (x,y) \neq (0,0) \end{cases}$

I'm sorry for my english, but if you understand my question, could you say if I'm right?

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First of all, $f$ is not a function of a single variable... –  Siminore Aug 30 at 15:52
    
yes, I know, so ... ? –  Louis Aug 30 at 15:55
    
So you can't write $f \colon \mathbb{R} \to$ –  Siminore Aug 30 at 15:56
    
Ow, ok, yes, Ill change that, with my LaTeX code, I forgot it, sorry :) –  Louis Aug 30 at 15:57

2 Answers 2

up vote 1 down vote accepted

Your intuition is correct, however you can't prove that $F$ is continuous that way. You must prove that $$\lim_{(x,y) \to (0,0)} F(x,y)=1.$$ You only computed the limit $$\lim_{x \to 0} F(x,x)=1,$$ and this is not enough. However, $$ F(x,y)=1+\frac{x^3 y^3}{x^2+y^2} = 1 + x^2 y^2 \frac{xy}{x^2+y^2}. $$ Now, $$ \left| \frac{xy}{x^2+y^2} \right| \leq \frac{1}{2}, $$ and $x^2 y^2 \to 0$ as $x \to 0$, $y \to 0$. Hence $F(x,y) \to 1$ as $x \to 0$, $y \to 0$.

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I dont understand why that absolute value is always smaller or equal to 0,5, could you explain that? –  Louis Aug 30 at 16:01
1  
Because $2xy \leq x^2+y^2$, as you can see by expanding the trivial inequality $0 \leq (x-y)^2$. –  Siminore Aug 30 at 16:01
    
ah, ok, now I see, thanks for that! –  Louis Aug 30 at 16:02

Yes, it is going to work, but you need to justify it better.

In fact you only demonstrated that the limit along the line $y=x$ equals $1$, but to ensure $f$ continuos you have to be sure that the limit holds from whichever direction one takes.

In particular, you have to set $x = \rho \cos \theta$, $y = \rho \sin \theta$, and as you can see if $\rho \to 0$, then the $f \to 1$

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and that works always? –  Louis Aug 30 at 15:57
    
@Louis certainly in a lot of cases. Sometimes it does not, for example if you arrive to an expression in the form $\frac \rho {\cos \theta}$. In this case you can't conclude anything since the denominator is $0$ along some directions. –  Ant Aug 30 at 16:00
    
Ok, thanks very much! –  Louis Aug 30 at 16:02

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