2
$\begingroup$

I've been having trouble with $\epsilon$-$\delta$ proofs of multivariable limits, even simple ones like $$\lim_{ \begin{pmatrix} x \\ y \\ \end{pmatrix} \to \begin{pmatrix} 1 \\ 2 \\ \end{pmatrix} }{x^2\over x+y}={1\over3}$$ I can't move (much) beyond stating the definition $$(\forall\begin{pmatrix} x\\ y\\ \end{pmatrix})(\forall\epsilon>0)(\exists\delta>0) (\lvert \begin{pmatrix} x \\ y\\ \end{pmatrix} - \begin{pmatrix} 1\\ 2\\ \end{pmatrix} \rvert<\delta \longrightarrow \lvert{x^2\over x+y}-{1\over3}\rvert<\epsilon) $$ Are there any tips or resources on how to prove multivariable limits?

Also, I know (1,2) is in the domain of the function, which is how I found 1/3. My problem is with $\epsilon-\delta$ proofs in more than one variable.

It's my first question here so I apologise if I made anything I shouldn't have.

$\endgroup$
1
  • $\begingroup$ Hint: $|(x,y)-(1,2)| < \delta \implies |x-1| < \delta \text{ and } |y-2| < \delta$. $\endgroup$
    – TonyK
    Sep 24 '16 at 0:00
2
$\begingroup$

A general tip: inside every ball is a rectangle/every norm on the real plane is the same up to a constant you don't care about. This avoids dealing with messy squares and square roots.

Other than that, start with your epsilon bound and modify it until you have the bound in your domain, then work backwards

$\endgroup$
2
  • 1
    $\begingroup$ "A general tip: inside every ball is a rectangle/every norm on the real plane is the same up to a constant you don't care about" I feel this is a valueble tip but can you explain it in more detail? $\endgroup$ Sep 23 '16 at 20:54
  • $\begingroup$ There are lots of different notions of distance in the plane. The one you are used to is the euclidean norm, which looks like a ball or disc. However, you can use rectangles, i.e. of the form $|x-a|+|y-b|$. Draw a picture to convince yourself that you can fit a rectangle inside any circle, and then use this norm instead since it is much easier $\endgroup$ Sep 23 '16 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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