Complete proof about the limits of a multivariable function

Question

I just started learning about multivariable functions and I have a question about the complete proof of the existence of a limit of a function. So let's take for example the function $$f(x,y)=\frac{x^3y}{x^2+y^2}$$ This function has a limit in $(x,y)=(0,0)$ and I'll have to prove it

So according to my understanding, I'll have to examine the behavior of the function while it approaches $(0,0)$ from every possible path. First I examined the trivial limits when $x=0$ and $y=0$ cases which both wielded the result $0$ as expected. Then I tried for $t \in (0,1]$ the approach from every different line of origin $y=kx$ by setting $x=at$ and $y=bt$ (with a and b never both equal to $0$) to test the limit $$\lim_{t\to0} f(x(t),y(t))$$ which also gave $0$. At this point, I don't think that I've covered every single approach of $f$ to the point $(x,y)=(0,0)$, and even if I have, I think that in order to complete the proof I'll have to also mention the definition of limits, for which I am completely unfamiliar in multivariable functions, but I think that for this example it goes like this:

Since for every $ε>0$ there is a $δ>0$ so that for every $(x,y)$ with $0 < ||(x,y)-(0,0)|| < δ$ then $|f(x,y)-0|<ε$ and the limit exists.

Any thoughts or insight would be really helpful.

Answer 1

Let me first give you a definition for the limit of a function. I'll try to bring some generality with it, so let $(M,d_M),(N,d_N)$ be metric spaces (if you like, you can think $\mathbb{R}^2$ or $\mathbb{R}$ with the usual metrics induced by the norms $||\cdot||_2$ and $|\cdot|$):

Definition: Let $f:M\to N$, $x_0$ a limit point of $M$, $a\in N$. Then $$\lim_{x\to x_0}f(x)=a$$ if $\lim_{n\to\infty}f(x_n)=a$ for every sequence $(x_n)_{n\in\mathbb{N}}$ in $M\setminus\{x_0\}$ s.t. $\lim_{n\to\infty}x_n=x_0$.

This is an alternate formulation of the $\epsilon$-$\delta$-criterion of limits, i.e. equivalent with such. Another thing of use is a simplification of convergence of sequences in the spaces $\mathbb{R}^n$:

Theorem: Let $v_k=(x_1^{(k)},\dots,x_n^{(k)})^\top\in\mathbb{R}^n$. $(v_k)_{k\in\mathbb{N}}$ converges iff every $(x_i^{(k)})_{k\in\mathbb{N}}$ converges and $$\lim_{k\to\infty}v_k=\begin{pmatrix}\lim_{k\to\infty}x_1^{(k)}\\\vdots\\\lim_{k\to\infty}x_n^{(k)}\end{pmatrix}$$

In words, a sequence in $\mathbb{R}^n$ converges iff every coordinate sequence converges and the limit is computable as such. (I will not prove this now, but if you want to try it yourself, an easy access is first showing that every norm over $\mathbb{R}^n$ is equivalent and then choosing a suitable one for this problem(Hint: $\infty$-norm))

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Now, lets tackle your problem and evaluate $$\lim_{(x,y)\to (0,0)}f(x,y)=\lim_{(x,y)\to (0,0)}\frac{x^3y}{x^2+y^2}$$

We have to show, that for every sequence $(v_n)_{n\in\mathbb{N}}$ in $\mathbb{R}^2\setminus\{(0,0)\}$ convergent to $(0,0)$, the real sequence $(f(v_n))_{n\in\mathbb{N}}$ converges to the same limit. This limit is then the limit of the function at that point.

Let $(v_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}^2\setminus\{(0,0)\}$ s.t. $\lim_{n\to\infty}v_n=(0,0)$. Then as above, for $v_n=(x_n,y_n)$, $\lim_{n\to\infty}x_n=0$ and $\lim_{n\to\infty}y_n=0$. Note, that these limits now take place over $\mathbb{R}$.

Now, for any $n\in\mathbb{N}$ s.t. $x_n\neq 0$, we have

$$|f(v_n)|=|\frac{x_n^3y_n}{x_n^2+y_n^2}|=\frac{|x_n|x_n^2|y_n|}{x_n^2+y_n^2}\leq\frac{|x_n|x_n^2|y_n|}{x_n^2}=|x_ny_n|$$

If $x_n=0$, then anyway $|f(v_n)|=0\leq |x_ny_n|$. Thus $|f(v_n)|\leq |x_ny_n|$ f.a. $n\in\mathbb{N}$. Since $\lim_{n\to\infty}y_n=0$ and $\lim_{n\to\infty}x_n=0$, we have $\lim_{n\to\infty}|x_ny_n|=0$, and thus via the Sandwich-Theorem and

$$0\leq |f(v_n)|\leq |x_ny_n|$$

we derive $\lim_{n\to\infty}|f(v_n)|=0$, i.e. $\lim_{n\to\infty}f(v_n)=0$.

As this sequence $(v_n)_{n\in\mathbb{N}}$ was arbitrary, we have that $\lim_{(x,y)\to(0,0)}f(x,y)=0$.

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At last, I want to note that this approach via considering every possible convergent sequence to a certain point corresponds very well to your intuition about considering every possible path towards this point, especially in higher dimension where you really have the freedom to walk a path with a sequence.

EDIT: Note, that in the evaluation of your limit, I used a lot of notions from classical, one-dimensional real analysis, as we were looking at a function going into $\mathbb{R}$. E.g. I used that $$\lim_{n\to\infty}a_n=0\text{ iff }\lim_{n\to\infty}|a_n|=0$$ for a real sequence $(a_n)$ as well as the Sandwhich-Theorem, i.e.

Theorem: If $(a_n)_{n\in\mathbb{N}},(b_n)_{n\in\mathbb{N}},(c_n)_{n\in\mathbb{N}}$ are sequences in $\mathbb{R}$ s.t. $a_n\leq b_n\leq c_n$ f.a. $n\in\mathbb{N}$, $\lim_{n\to\infty}a_n$ and $\lim_{n\to\infty}c_n$ exists and $\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n$, then $$\lim_{n\to\infty}b_n=\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n$$

which of course strongly relies on the order of the reals. But even in general, I personally find that a reduction to sequences is a comprehensible way to go. Note, that DinosaurEgg also has nice way to approach this problem, where should definitely take a look at.

EDIT2: As Mark Viola pointed out, every possible path is a very general statement not imposing a relationship between the $x_n$ and $y_n$ at all. You can specifically formally observe this in the fact, that for convergence of $((x_n,y_n))_{n\in\mathbb{N}}$, it is necessary and sufficient that the coordinates converge separately.

Answer 2

In problems like this, I've found that one of the easiest ways to approach them is the following:

1) Rewrite the function in spherical coordinates of the corresponding dimension around the point of interest. In your case we have 2 variables, so we can write the function in terms of polar coordinates $x=r\cos\theta, y=r\sin\theta$ and the function becomes $f(x=r\cos\theta,y=r\sin\theta)=g(r,\theta)=r^2\sin\theta\cos^3\theta$.

The reason why it is useful to do this is because the definition of the limit in more than two dimensions requires the epsilontic definition to hold in a ball around the point of interest. The parametrization in terms of the spherical coordinates does the wonderful trick of reducing the limit to proving the statement:

"for every $ε>0$ there is a $δ>0$ so that for every $(x,y)$ with $0 < r < δ$ then $|f(r\cos\theta,r\sin\theta)-0|<ε$"

which can be tackled using one-dimensional limit arguments if the limit exists.

2) In your case the proof can be completed as follows:

Suppose $0<r<\sqrt{\epsilon}$.

Then $|g(r, \theta)|=r^2|\sin\theta||\cos\theta|^3<\epsilon$, since$|\sin\theta|,|\cos\theta|<1$

and thus: $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=\lim_{r\rightarrow0}g(r, \theta)=0.$