I was looking at Peter Tamaroff's answer to this question. The question was to show that the following limit did not exist: $$\mathbf{\lim_{(h_1, h_2)\to (0,0)}\frac{\sqrt {|h_1.h_2|}}{\sqrt {h_1^2+ h_2^2}}}$$

In the answer it was explained that since if you let $h_1=h_2$ you would get a different limit than if you let $h_1=2h_2$ , the limit of the original does not exist. Why are you allowed to choose different "speeds" at which the variables approach $0$? Doesn't $\mathbf{\lim_{(h_1, h_2)\to (0,0)}}$ imply that both variables approach at the same "speed" so the only valid solution is $h_1=h_2$? Thanks.

P.S. I know this might be terribly wrong because I've never done multivariable limits, so do you also have a recommendation for an online source for me to learn this?


No, the notation $\lim_{(x,y) \rightarrow (a,b)}$ means you are allowed to approach the point $(a,b)$ in EVERY possible way. In 2-dimensions (in the plane) $lim_{(t,t):t \rightarrow (a,a)}$ will mean approaching the point $(a,a)$ at equal speed and $lim_{(t,2t):t \rightarrow (a,2a)}$ will mean approaching $(a,2a)$ along a line of slope $2$.

You should think that approaching a point along any one dimensional continuous curve (like a line, a parabola etc) in two or higher dimensions is not very far from working in one dimensions (because everything is constrained to move in one dimension). This is called parametrization.

In other words, a two dimensional space is inherently two dimensional and hence cannot (and shouldn't) be continuously parametrized by one dimension.


The formal definition of a limit as the point $\mathbf{x} = (x_1,...,x_n)$ in $n$ dimensions approaches some point $\mathbf{p_0}$ of a function $f(\mathbf{x})$ is that for some $L$ to be the limit, then for all $\varepsilon > 0$, there exists a $\delta > 0$ such that for all points $\mathbf{p}$ in the open ball of radius $\delta$ about $\mathbf{p_0}$, $|f(\mathbf{p}) - L| < \varepsilon$.

The different variables $x_1,...,x_n$ can be 'approaching $\mathbf{p_0}$ at different speeds', that is, that $\mathbf{x}$ approaches $\mathbf{p_0}$ along some vector whose components are different, since in that ball, for each vector $\mathbf{v}$, there exists a point $\mathbf{y}$ such that $\mathbf{p_0}$ is in the direction of $\mathbf{v}$ from $\mathbf{y}$, and the inequality given above must hold for all $\mathbf{p}$ in that ball, including each of these $\mathbf{y}$.

  • $\begingroup$ I don't think this answer is appropriate for the OP's level... $\endgroup$ – Hawk Jul 9 '13 at 5:54
  • $\begingroup$ Neither do I, really, but I'm not sure how to explain the why properly without getting into the definition. $\endgroup$ – qaphla Jul 9 '13 at 5:57

It has nothing to do with speeds, it is paths of approach that matter.

Replace the first example by $h_1=h_2=\frac{h}{\sqrt{2}}$.

Replace the second by $h_1=\frac{h}{\sqrt{5}}$ and $h_2=\frac{2h}{\sqrt{5}}$.

As $h$ approaches $0$, each of the two points approaches the origin at the same speed. But if you calculate, you will find that the limits of the ratios differ.

All we have done in changing the example is some scaling. The scaling does not affect the ratio in this case.

Note that in general we must consider all paths. It could be that all approaches along any line give the same limit, but an approach along a parabolic path gives a different limit.


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