Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\epsilon$-$\delta$?

share|cite|improve this question
First, write the definition of continuity of $f$ at $(0,0,0)$. – Asaf Karagila Jun 8 '12 at 11:00
up vote 5 down vote accepted


  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.
share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.