hi, i hae a question in calculus regarding functions with three variables and continuity?

Question

for what values of the number r is the function

f(x,y,z) = (x+y+z)^r/(x^2 + y^2 + z^2) if (x,y,z) not equal to 0
= 0 if (x,y,z) equal 0

thanks!

Answer 1

Set x and y = 0, and let z vary.
Set x and z = 0, and let y vary.
Set y and z = 0, and let x vary.

In each case you'll get the same answer, namely r >2.

So that's your clue.

For an epsilon/delta PROOF that the limit is 0, I'll give you a hint -- replace x + y + z by the sum of their absolute values, and then by 3 times the max of their absolute values, and you should have a useful estimate for f.

Answer 2

In these kinds of problems, one thing to do is to try approaching 0 along a line or a curve. The trick is finding the line or curve. In this case, if you go to 0 along the line y = 0, z = 0, (this is, of course, the x-axis), then your function is f(x) = x^r / x^2, which reduces to x^(r-2). That WON’T be continuous at (0,0,0) if r < 2, and it WILL be continuous if r >= 2. My guess is that this also works in three dimensions, but you will have to prove that.

Answer 3

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