2

Is this scratch work for the proof of $\lim_{x\rightarrow c}x^3=c^3$ correct?

We have that $|x^3-c^3|=|x-c||x^2+xc+c^2|$.

If $|x-c|<|c|$ then $|x|=|x-c+c|\leq |x-c|+|c|<2|c|$, it follows that $|x^2+xc+c^2|\leq |x|^2+|x||c|+|c|^2< 4|c|^2+2|c|^2+|c|^2=7|c|^2$.

So if $|x-c|<|c|$ and $|x-c|<\frac{\epsilon}{7|c|^2}$ we have that $|x^3-c^3|=|x-c||x^2+xc+c^2|<\frac{\epsilon 7|c|^2}{7|c|^2}=\epsilon$.

So if we let $\epsilon>0$ be given and $\delta=\min(|c|,\frac{\epsilon}{7|c|^2})$ then $|x^3-c^3|<\epsilon$ if $0<|x-c|<\delta$.

Gary
  • 31,845
per persson
  • 745
  • 7
    This looks well-written and excellent — except for the case $c=0$. – Ted Shifrin Oct 31 '22 at 21:19
  • Now do a similar proof for exponent $m$ instead of $3$. – marty cohen Oct 31 '22 at 21:20
  • @TedShifrin Is it because if $c=0$ then $|x-c|<|c|\iff |x|<0$. – per persson Oct 31 '22 at 21:39
  • 1
    But then you are taking $\delta = \min(0, \frac \epsilon{0})=0$. Not a valid choice. You will need to do a separate case where $c=0$. (Which is trivially easy...) – fleablood Oct 31 '22 at 21:47
  • 1
    Well, so this procedure fails for many reasons. You need a separate argument in this case. – Ted Shifrin Oct 31 '22 at 21:48
  • 1
    Your proof assumes that $|c|$ is a legitimate value from which we can chose an $x$ where $|x-c| \le |c|$. This is true if and only if $|c| \ne 0$. If $|c|=0$ then $|x-c| < |c|=0$ is simply not possible and we must find another choice of $\delta$. I don't see any easy argument that would combine the two but I don't see any great need to look hard for one either. Just say. If $c=0$ then $|x-c| =|x| < \epsilon^{\frac 13}\implies \epsilon > |x^3| = |x^3 - 0^3|=|x^3 -c^2|$ so let $\delta = \epsilon^{\frac 13}$. (Letting $\delta = \min(\epsilon, 1)$ works as well and may be more in the spirit.) – fleablood Oct 31 '22 at 21:59
  • @fleablood Thanks I understand. But I have a little more difficult question here https://math.stackexchange.com/questions/4563392/problem-with-understanding-proof-if-two-different-subsequences-have-different-li . You're $\epsilon$-$\delta$ game seems really good. Answer would be appreciated :). – per persson Oct 31 '22 at 22:15

1 Answers1

1

(After reading the comments.) With a little more care, you can include the case $c=0$:

If $|x-c|\le\delta\le1$ then $|x|=|x-c+c|\leq |x-c|+|c|\le1+|c|,$ so

$|x^3-c^3|\le\delta\left(3(1+|c|)^2\right),$

which is $\le\epsilon$ if $\delta$ is not only $\le1$ but also $\le\frac\epsilon{3(1+|c|)^2}.$

Anne Bauval
  • 34,650
  • This is the textbook slick answer. :P – Ted Shifrin Nov 01 '22 at 00:46
  • I saw this in Bartles book. First I had difficulty understanding why $|x-c|<1 \implies |x|<|c|+1$ but it became easier when visualizing the distances on the real line. – per persson Nov 01 '22 at 05:45
  • 1
    I don't know that textbook. I only slightly improved you own proof (I just inserted your argument for $|x-c|<1 \implies |x|<|c|+1$). – Anne Bauval Nov 01 '22 at 06:46