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Consider a function $f(x,y)$ of two variables $x$ and $y$. Let us consider a point $(a,b)$ in $\mathbb{R}^2$. Now the limit of the function as $(x,y)$ tends to $(a,b)$ is said to be existing if and only if it has the same value for each and every path through which $(x,y)$ approaches $(a,b)$. Now suppose we are required to prove that a given value $L$ is the limit of the function $f$ as $(x,y)$ tends to $(a,b)$. So is it sufficient to prove that the limit of the function is $L$ whenever $(x,y)$ approaches $(a,b)$ through any of the straight lines passing through $(a,b)$, as any other path can be broken up into several smaller straight line segments?

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You've been here for a year, and only accepted one answer out of ten questions? I won't read this, personally, until you go back and give people who have helped you in the past some credit for their efforts. You should also know that you can put dollar signs around your mathematical expressions, write R as $\mathbb{R},$ etc. to make it easier to read your question. –  Kevin Carlson Sep 24 '12 at 8:33
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@KevinCarlson I promise to mark out the answers to those questions as well as to the questions that will come up in future. I hope (optimistically) that I am going to understand this one too with the help of the dedicated members of MSE. –  Primeczar Sep 24 '12 at 9:32
    
Thanks for your promise, but why not just go do it? It's not as if it takes any time, unless you've kept asking questions without actually reading the answers. –  Kevin Carlson Sep 24 '12 at 10:30

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Please consider the function $f \colon \mathbb{R}^2 \to \mathbb{R}$ defined by $$ f(x,y)= \begin{cases} 1 &\text{if $y=x^2$ and $(x,y) \neq (0,0)$}\\ 0 &\text{otherwise.} \end{cases} $$ Does your idea apply?

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@siminore- In this case I don't think the limit attains the same value for all straight line paths. Could you please elaborate your answer . I mean could you suggest where the idea of proving the constancy of limit for straight line paths is wrong. –  Primeczar Sep 24 '12 at 9:22
    
@AlexanderThumm I am not concerned about continuity here, limit of the function is my main concern . –  Primeczar Sep 24 '12 at 9:23
    
Yes, the value at $(0,0)$ is irrelevant. –  Siminore Sep 24 '12 at 9:23
    
$f(x,mx)=0$ in any neighborhood of $(0,0)$. But you cannot conclude that $\lim_{(x,y)\to (0,0)} f(x,y)=0$. –  Siminore Sep 24 '12 at 9:25
    
@Primeczar: You cannot break paths up into small line segments (not even differentiable/analytical ones, as the example $y=x^2$ suggests). –  Alexander Thumm Sep 24 '12 at 9:28

Your suggestion is not correct. Consider the function defined by $$f(x,y)={x^2y\over x^4+y^2}, \qquad (x,y)\neq (0,0)$$ A straight line through $(0,0)$ is either given by $x=0$ or by $y=ax$ for some $a$. In either case, $f(x,y)\to 0$ when $(x,y)\to(0,0)$ along the straight line.

However, if $(x,y)\to (0,0)$ along the parabola given by $y=x^2$, then $f(x,y)\to {1\over 2}$.

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; Now suppose I need to prove that a value L is the limit of the function. Then what should I do? As it seems I need to show that the value remains independent of the function chosen(path). So how to how this independence? –  Primeczar Sep 24 '12 at 9:28
    
You cannot add any functional constraint between the variables. You should move them independently towards $(a,b)$. –  Siminore Sep 24 '12 at 9:30
    
You need to show that $|f(x,y)-L|$ can be made as small as you wish by requiring $\sqrt{(x-a)^2+(y-b)^2}$ to be small. The last expression here is the distance between the points $(x,y)$ and $(a,b)$. –  Per Manne Sep 24 '12 at 9:34
    
@Siminore Could you answer the question that I have written in reply to Per Manne ? –  Primeczar Sep 24 '12 at 9:35
    
@PerManne I hope you won't mind illustrating it in case of f(x,y)= x^3.y^4 as it tends to a definite point. –  Primeczar Sep 24 '12 at 9:39

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