# Computing Limits of Multivariate Functions

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous function such that partial derivatives of all orders exist. How can we compute limits of this function $\lim_{x \rightarrow x_0} f(x)$? I suppose the obvious way of computing a sequence of 1-dimensional limits should be valid under some conditions (and the order of computation of the 1d limits should not matter), i.e. $\lim_{x \rightarrow x_0} f(x)=\lim_{x_1 \rightarrow x_{0,1}} \cdots \lim_{x_n \rightarrow x_{0,n}}f(x_1,\cdots,x_n)$. What are these conditions?

Edited: How does the situation change when $f$ is not continuous?

## 2 Answers

Note that if $f$ is continuous, then $\lim_{x\to x_0}f(x)=f(x_0)$ by continuity.

When $f$ is not continuous, this approach cannot work either. For example, let $$f(x,y)=\begin{cases}0,&\text{ if } (x,y)=(0,0)\\ \ \\ \frac{x}{x+y},&\text{ if }(x,y)\ne(0,0) \end{cases}$$ Then $$\lim_{x\to0}\lim_{y\to0} f(x,y)=1,\ \ \lim_{y\to0}\lim_{x\to0} f(x,y)=0.$$

• Ok. What if $f$ is not continuous? I will also edit my question. – Manos Nov 11 '12 at 1:07
• Cannot work either. I've edited the answer. – Martin Argerami Nov 11 '12 at 1:25
• Is there any result saying that if taking the 1d limits commute, then they equal to the limit? – Manos Nov 11 '12 at 1:30
• @Manos: Take $f(x,y) = \frac{xy}{x^2+y^2}$ at $(x,y) = (0,0)$. Doing the one dimensional limits in either order gives $0$, but approaching the origin through any line $y=mx$, the limit is $\frac{m}{1+m^2}$. So the 1-d limits commute, but the limit doesn't exist. You can even make up examples where approaching through any straight line gives the same limit, but other curves don't. – Javier Nov 11 '12 at 2:33

What you want to be true is not. Unfortunately there is no simple connection between one-dimensional limits and $n$-dimensional limits. Well, not in a calculational sense. They appear the same in the $\epsilon, \delta$ terminology. Just replace an $\epsilon$-neighborhood with an $\epsilon$-ball.

$\lim_{x \rightarrow x_o}f(x) = L$ iff for each $\epsilon >0$ there exists $\delta >0$ such that $x$ with $0< ||x-x_o|| < \delta$ implies $||f(x)-L|| <\epsilon$.

For $n=1$, $||a|| = \sqrt{a^2} = |a|$. For $n > 1$, $||a|| = \sqrt{a_1^2+a_2^2+ \cdots a_n^2}$.

There are examples where $f(x,y) \rightarrow L$ along all possible lines through $(x_o,y_o)$ yet $\lim_{(x,y) \rightarrow (x_o,y_o)}f(x,y)$ does not exist. Similar examples exist for families of parabolic or cubic paths. It's subtle.

Now, if you know all possible paths give the same limit then you can conclude the multivariate limit exists. I have an animation illustrate the failure of linear and parabolic probing at http://www.supermath.info/ZooOfMathematicalCreatures.html and around page 11 of http://www.supermath.info/MultivariateCalculus2011Chapter3.pdf I show the calculational details. I also explain a few positive results about how to prove some limits do indeed exist.