2
$\begingroup$

We are proving $g(x) = x^3$ is continuous at $x_0$ arbitrary.

  • My attempt:

For all $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for an arbitrary point $x_0$ and $ \lvert x - x_0 \rvert \lt \delta \implies \lvert g(x) - g(x_0) \rvert \lt \epsilon$.

We plug in $g(x) = x^3$ to get:

$ \lvert x - x_0 \rvert \lt \delta \implies \lvert x^3 - x_0^3 \rvert \lt \epsilon$.

But we recognize $\lvert x^3 - x_0^3 \rvert = \lvert x - x_0 \rvert \lvert x^2 + x_0x + x_0^2 \rvert \lt \lvert x- x_0 \rvert \lvert x^2 + 2xx_0 + x_0^2 \rvert = \lvert x - x_0 \rvert \lvert (x+ x_0)^2 \rvert \lt \epsilon$.

So we have $ \lvert x^3 - x_0^3 \rvert \lt \lvert x-x_0 \rvert \lvert x+x_0 \rvert \lvert x+x_0 \rvert $

I'm not sure how to choose the proper relationship between $\delta$ and $\epsilon$ in this case.

Thank you!

$\endgroup$
11
  • 1
    $\begingroup$ I don't see how you can make that claim unless $x_0 = 0$ or $x = 0$. $\endgroup$
    – Jared
    May 15 '15 at 0:07
  • 3
    $\begingroup$ Try this: $|x+x_0|\le|2x_0|+|x-x_0|$ $\endgroup$ May 15 '15 at 0:09
  • 1
    $\begingroup$ Your inequality $|x-x_0||x^2+xx_0+x_0^2| < |x-x_0||x^2+2xx_0+x_0^2|$ is incorrect since $xx_0$ could be negative. You can replace it with $|x-x_0||x^2 + xx_0 + x_0^2| \leq |x-x_0||x^2 + 2|x||x_0| + x_0^2| = |x-x_0|(|x_0|+|x|)^2$. $\endgroup$
    – Michael
    May 15 '15 at 0:11
  • 1
    $\begingroup$ In other (shorter) words: Your first sentence "For all $\epsilon>0$...$|x-x_0|<\delta \implies |g(x)-g(x_0)|<\epsilon$" is what you want to prove. It does not make sense, in the course of your proof, to "plug in" to something that you want to prove, since that has not yet been proven! $\endgroup$
    – Michael
    May 15 '15 at 0:16
  • 1
    $\begingroup$ Well I will answer: "No" because what if $2x_0>\epsilon^{1/3}$? On the other hand, notice that, if we started out assuming that $|x-x_0|\leq \delta$ (hint, hint, you should start by assuming this), then we could conclude that $|2x_0| + |x-x_0| \leq |2x_0|+\delta$. $\endgroup$
    – Michael
    May 15 '15 at 0:35
7
$\begingroup$

Working from the starting point $|x^3-x_0^3|\le|x-x_0||x+x_0|^2\le\delta(2|x_0|+\delta)^2$ would suggest choosing $\delta=\min(\frac{\epsilon}{2|x_0|+1},1)$ but one of the comment points out that you made an inequality error when changing $xx_0$ to $2xx_0$. Instead, you can use $|x|\le|x_0|+|x-x_0|$ to get

$$|x^2+xx_0+x_0^2|\le(\delta+|x_0|)^2+\delta|x_0|+|x_0|^2=\delta^2+3\delta|x_0|+2|x_0|^2,$$

so $|x^3-x_0^3|\le\delta(\delta^2+3\delta|x_0|+2|x_0|^2)$ and you can pick $\delta=\min(\dfrac\epsilon{1+3|x_0|+2|x_0|^2},1)$.


Alternatively, you could just ignore the whole mess by noting that the product of continuous functions is continuous, so $x^3=x\cdot x\cdot x$ is continuous since $x$ is (choose $\delta=\epsilon$).

$\endgroup$
3
  • 1
    $\begingroup$ Just curious why you say to choose $\delta = \epsilon$. How are you able to say that so quickly without doing any work? I want to get to the point where I can make those kinds of statements immediately just by looking at the problem. $\endgroup$ May 15 '15 at 0:43
  • 1
    $\begingroup$ @DavidSouth Because it's the identity function: If I want $|x-x_0|<\delta\to|x-x_0|<\epsilon$, obviously $\delta=\epsilon$ will do the trick. But that's sort of a special case... $\endgroup$ May 15 '15 at 0:45
  • 1
    $\begingroup$ @DavidSouth In this case, you should note that the two examples I gave (for the two inequality calculations) give a general algorithm for picking $\delta$ if you can derive an inequality like $|f(x)-f(x_0)|\le\delta g(\delta)$ where $g$ is nonnegative and increasing and $g(1)>0$ - just choose $\delta=\min(\epsilon/g(1),1)$. $\endgroup$ May 15 '15 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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