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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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    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.2012-09-24

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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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    @Primeczar: You cannot break paths up into small line segments (not even differentiable/analytical ones, as the example $y=x^2$ suggests).2012-09-24
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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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    @Primeczar If $f(x,y)=x^3\cdot y^4$ and $(x,y)\to (1,2)$, say, then $x\to 1$ *and* $y\to 2$. Since the function $g(x)=x^3$ is continuous, you get that $g(x)\to g(1)$, or that $x^3\to 1^3=1$. In the same way, $y^4 \to 2^4 =16$. Since the limit of the products is the product of the limits, you get that $x^3\cdot y^4 \to 1^3\cdot 2^4 = 16$.2012-09-24