0

So $f(\vec{x}) =0$ if $x=0$ and equals $xyzt/(x^4+y^4+z^4+t^4)$ if $|\vec{x}|$ does not equal zero. How do I prove it is not continuous at the origin with epsilons and deltas? The whole epsilon delta thing is confusing the crap out of me to be honest. Thanks for the explanations. I don't know how the epsilon delta nut is cracked in this case and perhaps in general.

To show 'what I know' I add that I know I need to find an epsilon and show that no delta will have $|\vec{x}-0| < \delta$ and $|f(\vec{x})-0| < \epsilon$.

I also know that $|\vec{x}|=sqrt(x^2 +y^2 +z^2 +t^2)$

Andres Salas
  • 205
  • Recall that has the statement "for all $\epsilon$, there exists a $\delta$. Thus, if a function is discontinuous, we see the statement "for some $\epsilon$, there is no $\delta$". – Bib Mar 03 '15 at 23:47
  • Yes! Basic concepts of math explains this well, miraculous timing for this limit proof. I have that idea implicit in my post: "I need to find 'an' epsilon and show that 'no' delta." Thank you – Andres Salas Mar 03 '15 at 23:52
  • 2
    Hint: along any of the axes, $f$ is equal to zero. But along the line $x=y=z=t$, $f$ is equal to $1/4$. So now choose $\varepsilon=1/8$, let $\delta>0$ be arbitrary, and consider points which are on these lines and have magnitude $\delta$. – Ian Mar 03 '15 at 23:56
  • @Ian, a thousand blessings! I'll try that straightaway. – Andres Salas Mar 03 '15 at 23:59

1 Answers1

1

Let $\epsilon = \frac{1}{8}$. Given $\delta > 0$, set $x = y = z = t = \sqrt{\frac{\delta}{8}}$. Then $|\vec{x}- 0| = \frac{\delta}{2} < \delta$ and

$$|f(\vec{x}) - f(0)| = \frac{xyzt}{x^4 + y^4 + z^4 + t^4} = \dfrac{\frac{\delta^2}{64}}{4\left(\frac{\delta^2}{64}\right)} = \frac{1}{4} > \epsilon.$$

kobe
  • 41,901
  • A thousand blessings to you too kobe, I should come on here more often. A note: I used to be confused that $\vec{x}$ was limited to one multiple of \delta, because the proof requires all x's in delta to work. Could you answer how limiting x to one multiple of delta within delta is still ok as a way of proof? I may just have to look at the definition of epsilon-delta discontinuity – Andres Salas Mar 04 '15 at 00:05
  • After a look, my question is answered, the original definition says 'for all x such that ...' so the negation is 'there exists an x such that...' not for all x. – Andres Salas Mar 04 '15 at 00:11
  • @Andres correct. Glad I can help. – kobe Mar 04 '15 at 00:11