Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm studying for my exam and I have a bit of trouble with these kind of exercices, since I have no theory and im a bit lost:

Study the continuity in (0,0) (g(x,y) is a function depending of the exercise): $$f(x,y) = \begin{cases} g(x,y)&\text{for } (x,y)\not=(0,0)\\ 0 &\text{for }(x,y) =(0,0). \end{cases}$$ I have 2 questions:

1.What steps, or methods should I try in order to find the continuity on a function with 2 variables? (Iterated limits, etc)

2.Is there any method that can make you totally sure that a function is continous, or you can only say if its discontinous?

I really appreciate any help

Edit: This is not a real exercise its a kind of exercise.

share|improve this question

1 Answer 1

A function $f:\mathbb{R}^2\to\mathbb{R}$ is continuous at a point $(x_0,y_0)$ if and only if $$ \lim_{(x,y)\to(x_0,y_0)}f(x,y)=f(x_0,y_0). $$ This has to be true irrespective of "how" the point $(x,y)$ approaches the point $(x_0,y_0)$. In your case, the function $f$ coincides with a function $g$ on $\mathbb{R}^2\backslash\{(0,0)\}$ so is continuous at $(x_0,y_0)\neq(0,0)$ if and only if $g$ is continuous there.

For the point $(0,0)$ you will have to show that $$ \lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}g(x,y)=f(0,0)=0. $$

Since the function $g$ is not specified, not much else can be said.

share|improve this answer
    
Thanks for the reply, what methods can I use in order solve that limit with 2 variables? I read something about iterated limits and that there are more ways to solve it. –  Alejandro Jan 12 '13 at 20:44
    
@Alejandro: The method you would use would depend on the function $g$. If by iterated limits you mean performing the limits $x\to0$ and $y\to 0$ sequentially then this is not a good idea and may give a wrong result. In general, you can always resort to the $\epsilon-\delta$ definition of limits. –  Eckhard Jan 12 '13 at 20:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.