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Suppose we want to know $\lim_{(x,y)\rightarrow (0,0)}{f(x,y)}$.

The epsilon-delta definition of continuity (in $\mathbb{R}^n$) implies that "all paths" to $(0, 0)$ must result in the same limit for a function to be continuous.

But for some functions (i.e. ratios of polynomials in x and y), it's easy to set $y = \lambda x$ and determine the existence of $\lambda_1$ and $\lambda_2$ such that

$\lim_{x \rightarrow 0}{f(x,\lambda_1x)} \neq \lim_{x \rightarrow 0}{f(x,\lambda_2x)}$.

Since any "path" to (0,0) can be approximated arbitrarily well (near the point (0,0)) by some line $y = \lambda x$, it looks like checking all linear paths would suffice in many circumstances.

Is there a [named] class of functions which satisfies this property (i.e. one can prove continuity by checking only linear paths)?

Is there a simple counter-example for when this scheme breaks?

Riley Murray
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  • To spot the error, state rigorously what you mean by "any path to (0,0) can be approximated arbitrarily well". Either it will be false, or it won't imply the desired conclusion. – Nate Eldredge Aug 21 '15 at 23:29

1 Answers1

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Regarding a counterexample, you can consider $$\begin{array}{l|rcl} g : & \mathbb R^2 & \longrightarrow & \mathbb R \\ & (x,y) & \longmapsto & \frac{x^2 y}{x^4+y^2} \text{ for } (x,y) \neq (0,0)\\ & (0,0) & \longmapsto & 0 \end{array}$$

For which you'll see the non-continuity along the path $x \mapsto (x,x^2)$ while $g$ is continuous at the origin along all lines.

For more insight, please have a look here.

mathcounterexamples.net
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