f(x,y,z) = (x+y+z)^r/(x^2+y^2+z^2) if (x,y,z) is not equal to (0,0,0), and 0 if otherwise.
Suppose that r is less than or equal to 2. Show that the lim of f(x,y,z) at (x,y,z) -> (0,0,0) does not exist.
I have a general idea of how to do this, but the fact that r is less than or equal to 2, I'm a little tripped up. Hope someone could help :)
Suppose that r is less than or equal to 2. Show that the lim of f(x,y,z) at (x,y,z) -> (0,0,0) does not exist.
I have a general idea of how to do this, but the fact that r is less than or equal to 2, I'm a little tripped up. Hope someone could help :)
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Letting y = z = 0 with x → 0, we obtain
lim(x→0) (x + 0 + 0)^r / (x^2 + 0 + 0)
= lim(x→0) x^(r - 2).
If r < 2, then this is infinite (or does not exist), and we are done.
However, if r = 2, then this equals 1.
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Now, we need to show that this has no limit when r = 2.
Letting y = x and z = mx (for some constant m):
lim(x→0) (x + x + mx)^2 / (x^2 + x^2 + (mx)^2)
= lim(x→0) (2 + m)^2 x^2 / [(2 + m^2) x^2]
= (2 + m)^2 / (2 + m^2).
Since this is dependent on the value of m, the limit does not exist.
I hope this helps!
lim(x→0) (x + 0 + 0)^r / (x^2 + 0 + 0)
= lim(x→0) x^(r - 2).
If r < 2, then this is infinite (or does not exist), and we are done.
However, if r = 2, then this equals 1.
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Now, we need to show that this has no limit when r = 2.
Letting y = x and z = mx (for some constant m):
lim(x→0) (x + x + mx)^2 / (x^2 + x^2 + (mx)^2)
= lim(x→0) (2 + m)^2 x^2 / [(2 + m^2) x^2]
= (2 + m)^2 / (2 + m^2).
Since this is dependent on the value of m, the limit does not exist.
I hope this helps!
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Use l'hopitals rule, sorry I can't really solve it, I know this rule will help as I'm doing the same thing. This rule is a big help.