The only way that f(x,y) is undefined is when both x and y are 0, since
that is the only way that the denominator, namely x² + y², is = 0. So,
the domain is (x,y) ∈ R² / (0,0); also known as all points in the x-y
plane (where both x and y are real numbers) except the origin.
To find the range, let's first assume that y =/= 0, so that we
can divide both the top and the bottom of our function f(x,y)
by y² to get that f(x,y) = (xy + y²) / (x² + y²), so f(x,y)
= (xy/y² + y²/y²) / (x²/y² + y²/y²) = ([x/y] + 1) / ([x/y]² + 1),
so f(x,y) = (z + 1) / (z² + 1), where z = x/y, and since both
x and y are real numbers, then z can be any real number.
Therefore, in order to find the range of f(x,y), we must then
find z in terms of f and then find the domain of z(f) = z:
f = (z + 1) / (z² + 1), so (z² + 1) f = z + 1, fz² + f = z + 1, so
fz² - z + f-1 = 0, so z = ( 1 +- √[ 1 - 4*(f)*(f-1) ] ) / (2*f)
Therefore, for z to be defined, f must be such that:
f =/= 0 and 1 - 4f(f-1) >= 0, so 1 >= 4f(f-1), so 4f(f-1) <= 1,
so f(f-1) <= 1/4, so f² - f <= 1/4, so f² - f - 1/4 <= 0,
so f is between ( 1 - √[ 1 - 4*(1)*(-1/4) ] ) / (2*1) and
( 1 + √[ 1 - 4*(1)*(-1/4) ] ) / (2*1), so if we simplify this
we get that f is between ( 1 - √[ 1 - (-1) ] ) / 2 and
( 1 + √[ 1 - (-1) ] ) / 2, so f is between (1 - √2) / 2
and (1 + √2) / 2. This includes f = 0 since (1 - √2) / 2 =
- (√2 - 1) / 2, which is < 0; (1 + √2) / 2 = (√2 + 1) / 2,
which is > 0; and f = f(z) = (z + 1) / (z² + 1) = ([x/y] + 1)
/ ([x/y]² + 1) = f(x,y) = 0 when z = x/y = -1. So, the range
of f(x,y) is all real numbers f ∈ [-(√2-1)/2, (√2+1)/2].
Since when y = x, f(x,y) = f(x,x) = ([x/x]+1) / ([x/x]²+1),
which = (1+1) / (1²+1) = 2 / (1+1) = 2/2 = 1; and when y = -x,
f(x,y) = f(x,-x) = ([x/-x]+1) / ([x/-x]²+1), which = (-1+1) /
([-1]²+1) = 0 / (1+1) = 0/2 = 0; we can conclude that the
limit as (x,y) goes to (0,0) of f(x,y) does not exist.
