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This is a generalization of this question

A quick and easy was to prove that a 2 dimensional limit like $$\lim\limits_{(x,y)\to0}\frac{xy}{x^2+y^2}$$ is to try 2 different linear paths and prove that they aren't equal or that the limit depends on the path i.e choosing $y=mx$

$$\lim\limits_{x\to0}\frac{mx^2}{x^2(1+m^2)}=\frac{m}{1+m^2}$$

But that doesn't necessarily mean that the limit exist if mean the limit exist at every possible linear path that the limit exist as shown in the answer of the linked question

However let $y=\sum\limits_{k=1}^na_k x^k$ does the existence of a limit on all polynomial paths that tend to zero (and all of them are equal) enough for the limit to exist ?

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  • $\begingroup$ @MathIsAlwaysRight it doesn't even work for n=1 $\endgroup$ Commented Nov 16 at 19:17
  • $\begingroup$ @MathIsAlwaysRight any polynomial bath that tend to 0 as x tend to 0 $\endgroup$ Commented Nov 16 at 19:21
  • $\begingroup$ Oh, I'm sorry. Can't believe I've had such a brainfart. My point is that your polynomial paths dont include vertical or near-vertical paths. Here is a counterexample that hopefully works now: $$f(x,y) = \begin{cases} 0, & x \neq 0, \\ 1, & x = 0. \end{cases}$$ Now the limit along vertical (not polynomial) path is 1, but the limit on all the polynomial paths is 0 $\endgroup$ Commented Nov 16 at 19:27
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    $\begingroup$ Let $$f(x,y) = \begin{cases} 1 & y = e^{-1/x^2}, \\ 0 & \text{ otherwise.} \end{cases}$$ $\endgroup$ Commented Nov 16 at 19:27
  • $\begingroup$ @MathIsAlwaysRight the factorial of x is never zero $\endgroup$ Commented Nov 16 at 19:28

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I assume you know that the function \begin{equation} h \colon x \mapsto \begin{cases} \qquad 0 & \text{ if } x = 0 \\ \exp(-1/x^2) & \text{ if } x \neq 0 \end{cases} \end{equation} is smooth with $h^{(k)}(0) = 0$ for all $k \in \mathbb{N}$. Thus for every polynomial — more, every real analytic — $p$ with $p(0) = 0$ there is $\varepsilon > 0$ such that $p(t) \neq h(t)$ for all $t \in (-\varepsilon, \varepsilon) \setminus \{0\}$.

Hence for \begin{equation} f(x,y) = \begin{cases} 1 & \text{ if } y = h(x) \\ 0 & \text{ if } y \neq h(x) \end{cases} \end{equation} we have \begin{equation} \lim_{\substack{(x,y) \to (0,0) \\ \text{analytic}}} f(x,y) = 0\,, \end{equation} but of course every neighbourhood of the origin contains points $(x,h(x))$, so the unconstrained limit doesn't exist.

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