Solve Showing Function f Continuous on R^n & f(x)=c.x Confused

[mr_coffee]

Hello everyone, I'm so lost on these 2 questions...they are:
http://img437.imageshack.us/img437/5193/q30pi.jpg
if that link sucks try this one:
http://show.imagehosting.us/show/789841/0/nouser_789/T0_-1_789841.jpg

Question 2:
http://img437.imageshack.us/img437/8540/q47ni.jpg
mirrior:
http://show.imagehosting.us/show/789846/0/nouser_789/T0_-1_789846.jpg

I know a function f of two variables is called continuous at (a,b) if
lim (x,y)->(a,b) f(x,y) = f(a,b);
We say f is continuous on D if f is continuous at every point (a,b) in D.

Also If f is defined on a subset D of R^n, then lim x->a f(x) = L means that for every number E > 0 there is a corresponding number delta > 0 such that |f(x)-L| < E wherever x is a subset of D and 0 < |x-a| < delta. But I'm so lost on how I'm suppose to even start these 2. thanks.
Note: the BoLDED variables means they are vectors. and E is suppose to be epsalon or however u spell it

 

[kurt.physics]

.For the first question, you need to determine whether the given function is continuous at (0,0). To do this, you need to evaluate the limit of the function as (x,y) approaches (0,0). This means that you need to calculate:lim (x,y)->(0,0) f(x,y).To do this, you need to substitute x=0 and y=0 into the given expression for f(x,y). This gives you:lim (x,y)->(0,0) f(x,y) = lim (x,y)->(0,0) (x^2 + y^2)/(x^2 - y^2)You can then apply the limit rules to evaluate the limit. The result is:lim (x,y)->(0,0) f(x,y) = 1Since the limit of the function is equal to 1, the function is continuous at (0,0). For the second question, you need to determine whether the given function is continuous on the given set D. To do this, you need to check that the function is continuous at all points (x,y) in D. To check continuity at a point (x,y), you need to calculate:lim (x,y)->(x,y) f(x,y).To do this, you need to substitute x and y into the given expression for f(x,y). This gives you:lim (x,y)->(x,y) f(x,y) = lim (x,y)->(x,y) (x^4 + y^4)/(x^2 + y^2)You can then apply the limit rules to evaluate the limit. The result will depend on the particular point (x,y) being considered. Therefore, you need to check each point (x,y) in turn to see if the limit is equal to the value of the function at that point. If the limit is equal to the value of the function at each point (x,y) in D, then the function is continuous on D.