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I don't get the idea of trying "all" paths to know the value of the limit. The way I see it is I have to "scan the area" around the point, and that would be enought. for example, if all the straight lines whose equation is (y = mx) lead to the same value, then any other curvy path is just a "jump" from a straight line path to another. So, Can you explain why this procedure isn't enough? A counterexample would be great.

Samuel Shokry
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    In my answer to Multivariable limits I gave an example of very simply defined function such that not only all straight line approaches to a certain point, but even all polynomial curve approaches to a certain point, give the same limit, but the limit at the point does not exist. – Dave L. Renfro Oct 10 '18 at 16:33
  • @DaveL.Renfro: nice example. +1 for your comment. – Paramanand Singh Oct 10 '18 at 21:31
  • @DaveL.Renfro: You might also be interested in this post, which raises the general question, "can we say $f(\mathbf{x})$ has property $P$ at $\mathbf{x}=\mathbf{a}$ if $f\circ\gamma$ has property $P$ at $t=0$ for all curves $\gamma(t)$ with $\gamma(0)=\mathbf{a}$?" https://calculus7.org/2012/07/24/how-much-multivariable-calculus-can-be-done-along-curves/ – symplectomorphic Oct 10 '18 at 21:47
  • Moreover, this result -- "Continuously Differentiable Curves Detect Limits of Functions of Two Variables" -- shows that Dave Renfro's example is about as bad as it gets, in a certain sense: there is no counterexample where the limit agrees along all continuously differentiable curves but the limit does not exist. This is not too surprising. – symplectomorphic Oct 10 '18 at 21:54

2 Answers2

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The idea of a multivariable limit is not exactly related to approach via multiple paths. If the multivariable limit exists then limits via various paths also exist and are equal. But this is one way implication.

The meaning of the limit is similar to one dimensional case ie values of function are near the limit for all values of its arguments near the point under consideration.

Consider the case mentioned in question where you think that all straight lines passing through the point under consideration cover the area near the point. This is accepted but the problem is something else. Consider two such lines. It might be possible that for a given $\epsilon$ the $\delta$ corresponding to these lines is different. If you want to cover all possible lines then you may have to deal with all the possible $\delta$'s and choose a minimum out of them. And it is quite possible that there is no desired minimum available here.

Thus even though you can find nearby points on each line to fit the definition of limit, yet you may not be able to find a neighborhood of point which fits the definition of multivariable limit. And this can happen no matter how many paths of various nature you choose to approach the point under consideration.

You should also observe that the multiple path approach is almost always used to prove the non-existence of a limit instead of existence of limit.

Paramanand Singh
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It is easy to construct such a counterexample.

Just take the indicator function of the easiest nonlinear curve, the parabola $y=x^2$:

$$f(x,y)=\begin{cases} 1 & y=x^2 \\ 0 & y\neq x^2\end{cases}$$

If you approach $(0,0)$ along any line, the limit is zero. (To see this, note that any line through the origin eventually misses the parabola, as you move sufficiently close to the origin.) But if you approach along $y=x^2$, the limit is one.

Dave Renfro's comment gives a much more radical example, where the limit agrees along all polynomial paths, but still fails to exist.

symplectomorphic
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