Thread: 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.

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..

Originally Posted by catcat103

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

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

ah...i forgot those useful inequalities. what a marvellous answer! thanks for your reply!