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Prove that $f(x,y,z)=x^4+y^4+z^4$ is continuous on point $(x,y,z)=(0,0,0)$ with epsilon-delta

I prove this so:

if $$\lim_{x,y,z \to 0,0,0} f(x,y,z) = f(0,0,0)$$ then that function is continuous

$$\lim_{x,y,z \to 0,0,0} x^4+y^4+z^4 = 0^4+0^4+0^4=0$$

But how to prove this with $\epsilon$-$\delta$?

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First, write the definition of continuity of $f$ at $(0,0,0)$. –  Asaf Karagila Jun 8 '12 at 11:00
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2 Answers

up vote 5 down vote accepted

HINTs

  1. $g(x) = x^4$ is continuous at $0$. In fact, if we choose $\delta < \epsilon^{1/4}$, we're set.
  2. The triangle inequality is really useful.
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If A, B positive, we have: $A^{2}+B^{2}\leq (A+B)^{2}$. Since $x^2,y^2,z^2$ are positive, then: $x^4+y^4+z^4\leq (x^2+y^2+z^2)^2$. And, with spherical coordinates:

$(x^2+y^2+z^2)^2\leq \rho^4< \epsilon$, where $\rho$ is the neighbourhood's radius. I hope I have understood your request.

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