# Thread: Finding the points where a function is continous

1. ## Finding the points where a function is continous

Find the set of all points of continuity for:

$\displaystyle f(x,y) = ln(\frac{x-y}{x^2+y^2})$

i know that the domain is whenever ln is positive, so its whenever x-y>0 , x>y

is it continuous on all points of the domain?
also is it considered continuous on the edges of the surface?

thanks

2. ## function is continuous everywhere

Originally Posted by travelalfred
Find the set of all points of continuity for:

$\displaystyle f(x,y) = ln(\frac{x-y}{x^2+y^2})$

i know that the domain is whenever ln is positive, so its whenever x-y>0 , x>y

is it continuous on all points of the domain?
also is it considered continuous on the edges of the surface?

thanks
f(x) is said to be continuous when lim(x->a+)f(x)=lim(x->a-)f(X)=f(a).
where a is any point in the domain.
following a similar analogy for your 2 variable function i think f(x,y)is continuous everywhere as f(a,a)=0=lim(x->a+-)lim(y->a+-)f(x,y) in the domain which means that all the points of continuity will lie in the
domain. and the domain is given by whenever x>y for all reals.

3. Originally Posted by travelalfred
Find the set of all points of continuity for:

$\displaystyle f(x,y) = ln(\frac{x-y}{x^2+y^2})$

i know that the domain is whenever ln is positive, so its whenever x-y>0 , x>y

is it continuous on all points of the domain?
also is it considered continuous on the edges of the surface?

thanks
Since the "edge of the surface", x= y, is NOT in the domain, the question of whether the function is continuous there is moot- it is not even defined threre!

4. Originally Posted by travelalfred
Find the set of all points of continuity for:

$\displaystyle f(x,y) = ln(\frac{x-y}{x^2+y^2})$

i know that the domain is whenever ln is positive, so its whenever x-y>0 , x>y

is it continuous on all points of the domain?
also is it considered continuous on the edges of the surface?

thanks

Re-write it as $\displaystyle ln(x-y) \, - \, ln(x^2+y^2)$.

Looks easier, Right?