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In my second-year calculus class this term, one of the thing that the professor insisted was wrong was that the limit of a two-dimensional function as the input approached a certain point could not be calculated simply by taking the limit of the function in every direction and verifying that they were all equal.

I've taken her word for it, but why is this not true? Is there a counterexample to this proposition, and if so, what general principle does it violate?

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    You can approach with non-linear paths and get a different limit from all linear paths.2012-12-24
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    Can I get an example of that? I'm not quite sure how that would work.2012-12-24
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    I don't remember an example off-hand, but I think one is in Spivak's *Calculus on Manifolds.* There is probably one in any book on real analysis in multiple variables.2012-12-24
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    Let $f(x,y) = \begin{cases}1&\text{if }y=x^2,\\0&\text{otherwise.}\end{cases}$2012-12-24
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    On the other hand, you could calculate the limit by [using every path approaching the point](http://math.stackexchange.com/questions/203219/limit-along-a-path-equivalent-to-usual-definition-of-limit). This is hardly practical, though.2012-12-24
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    It has been awhile since multivariate calculus, but perhaps she is referring to non-euclidean geometry — http://en.wikipedia.org/wiki/Non-Euclidean_geometry.2012-12-24

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Consider the limit of the function $$f(x,y) = \begin{cases}\frac{x^4}{y^2} & \text{for } y \ne 0 \\ 0 & \text{for } y = 0\end{cases}$$ as $(x,y) \to (0,0)$. Clearly, along any line $y = ax$ passing through the origin, $$f(x,ax) = \begin{cases}\frac{x^2}{a^2} & \text{for } a \ne 0 \\ 0 & \text{for } a = 0,\end{cases}$$ and thus $\lim_{x \to 0} f(x,ax) = 0$. Indeed, $f(0,y)=0$ for all $y$ as well, so the same limit holds when approaching the origin along any line. Yet $f$ maps any open neighborhood of the origin to $[0,\infty)$, and so has no limit at the origin.

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    nice! (for those unpacking this, the idea is, on the circle of radius $\epsilon$, the function still blows up to infinity as you approach the line $y = 0$)2012-12-24