Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You can approach with non-linear paths and get a different limit from all linear paths. – Potato Dec 24 '12 at 1:25
Can I get an example of that? I'm not quite sure how that would work. – Joe Z. Dec 24 '12 at 1:25
It has been awhile since multivariate calculus, but perhaps she is referring to non-euclidean geometry — – David542 Dec 24 '12 at 1:27
Let $f(x,y) = \begin{cases}1&\text{if }y=x^2,\\0&\text{otherwise.}\end{cases}$ – Rahul Dec 24 '12 at 1:36
On the other hand, you could calculate the limit by using every path approaching the point. This is hardly practical, though. – user53153 Dec 24 '12 at 1:59
up vote 12 down vote accepted

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.

share|cite|improve this answer
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 $y = 0$) – uncookedfalcon Dec 24 '12 at 1:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.