# Limit and Continuity (multivariable calculus) [help]

• Jun 13th 2008, 08:10 AM
catcat103
Limit and Continuity (multivariable calculus) [help]
Question : For f(x,y)=xy, show that f(x,y) is continuous everywhere.
i.e. for every point (a,b) in the plane and for every epsilon>0, find delta>0 depending on a,b and epsilon such that whenever
0<|(x,y)-(a,b)|<delta, |f(x,y)-f(a,b)|<epsilon

How to find an epsilon in terms of delta? I am troubled in expressing epsilon as a function of delta.

Many thanks.
• Jun 13th 2008, 08:14 PM
kalagota
Quote:

Originally Posted by catcat103
Question : For f(x,y)=xy, show that f(x,y) is continuous everywhere.
i.e. for every point (a,b) in the plane and for every epsilon>0, find delta>0 depending on a,b and epsilon such that whenever
0<|(x,y)-(a,b)|<delta, |f(x,y)-f(a,b)|<epsilon

How to find an epsilon in terms of delta? I am troubled in expressing epsilon as a function of delta.

Many thanks.

epsilon is an arbitrary number.. you should find delta in terms of epsilon and not the other way around..
• Jun 14th 2008, 07:41 PM
ThePerfectHacker
Quote:

Originally Posted by catcat103
Question : For f(x,y)=xy, show that f(x,y) is continuous everywhere.

We shall prove $f$ is continous at $(a,b)$. Let $\sqrt{(x-a)^2+(y-b)^2} < \delta$. Then $|x-a| = \sqrt{(x-a)^2}\leq \sqrt{(x-a)^2+(y-b)^2} < \delta$. Similarly $|y-b| < \delta$. Now use the inequality $||y|-|b|| \leq |y-b| < \delta \implies - \delta < |y| - |b| < \delta$. Thus, $|y| < \delta + |b|$. With these facts we can prove continuity. We need to show $|xy-ab|$ can be made arbitrary small. Add and subtact, $|y(x-a) + a(y-b)| \leq |y||x-a| + |a||y-b| < (\delta + |b|)\delta + |a| \delta < \delta (1+|a|+|b|)$ if $0<\delta \leq 1$.
This tells us that for any $\epsilon > 0$ choose $\delta = \min \left\{ 1 , \tfrac{\epsilon}{1+|a|+|b|} \right\}$.
• Jun 17th 2008, 05:37 AM
catcat103