ok you need to replace the x, y, z cartesian coordinates by their corresponding rho, phi, theta coordinates...
I have this one in my textbook.
It is really hard!!!
I can't do anything.
For what values of the number r is the function
continuous on R ?
the problem here is that the question did not specify r, is it natural or real .. etc
f(x(ρ,Φ,θ),y(ρ,Φ,θ),z(ρ,Φ,θ)) = (ρsinΦcosθ + ρsinΦcosθ + ρcosΦ )^r / ρ^2
= [ρ^(r)](sinΦcosθ + sinΦcosθ + cosΦ) / ρ^2
note that the numerator is bounded because its sin terms and cosine terms are all bounded by -1 and 1. Also note that the function is piecewise continuous at (x,y,z) = (0,0,0) or extended by continuity. all we need is to equate the limit of f as ---> ρ <--- approches zero to zero. that is:
ρ^(r) > ρ^2 <=> ρ^(r) / ρ^2 > 1 <=> ρ^(r - 2) > 1 or r - 2 > 0 <=> r > 2
Hope this convinces you ( and is 100% correct)
OK, here's another way to look at it (btw i did a minor algebra error, its supposed to be rho^r and not ^1/r)...
(x + y + z)^r / (x2 + y2 + z2) must approach zero thats the numerator is muuuuuuch smaller than the denominator... This is equivelent to saying that (x2 + y2 + z2) / (x + y + z)^r goes to infinity as (x,y,z) tend to (0,0,0)... u know that x2 + y2 + z2 = (x+y+z)2 - 2xy - 2yz - 2xz so:
(x2 + y2 + z2) / (x + y + z)^r = (x+y+z)2 / (x+y+z)r -2(xy + xz +yz)/(x + y + z)^r
the 1st term is muuuuch greater than the second term (x,y and z are all of order 2). So:
(x+y+z)r / (x+y+z)2 ---> +inf so (x+y+z)^2 / (x+y+z)^r goes to zero...
implies that (x+y+z)^r is "greater" than (x+y+z)^2 or simply r>2
QED??!! LETS HOPE