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I'm trying to get rid of a removable discontinuity at $x=0$ in the function: $$\frac{1}{2}\left(1-\frac{|x|}{x}\right)\left|\sqrt[3]{x^3-c}-x\right|+\frac{1}{2}\left(1+\frac{|x|}{x}\right)\left(\sqrt[3]{x^3+c}-x\right)$$ where $c$ is a constant greater than $0$.

I'm only working with the reals. No matter what I do, I can't get rid of the $x$'s in the denominators. I've tried WolframAlpha but that didn't work. Any help or suggestions would be appreciated.

For context, I am trying to turn a continuous piecewise function into a single elementary function. I have done this kind of thing with many other piecewise functions, but this is the first one that I just can't make work.

Blue
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Matt MacKinnon
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3 Answers3

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Just take the limit of your function as $x \to 0$ and define the value at $x=0$ to be that limit. When $x \gt 0$ the first set of parentheses evaluates to $0$ and you are left with the second term, where the $\left(1+\frac{|x|}{x}\right)=2$. When $x \lt 0$ the first term is what contributes. The limits from the two sides have to be equal for the discontinuity to be removable.

Ross Millikan
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  • Sorry, I wasn't clear about what I was trying to do. I know how to find the value to "fix" the function, but I'm trying to find a function equivalent to the initial function except without discontinuity. – Matt MacKinnon Oct 03 '22 at 19:11
  • There is no guarantee that a simple expression exists. I would find the function clearer if the $\frac{1}{2}\left(1\pm \frac{|x|}{x}\right)$s were removed and we had cases for positive and negative $x$, then a third case for $x=0$ of $\sqrt[3]c$. – Ross Millikan Oct 03 '22 at 19:17
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I'm not sure what you're trying to achieve here. You already have an expression that is a single elementary function. A piecewise expression for this function would be

$$f(x) = \begin{cases} \sqrt[3]{x^3+c} - x, & x \ge 0 \\ \sqrt[3]{x^3-c} - x, & x < 0. \end{cases}$$

We can clearly see that $$\lim_{x \to 0} f(x) = \sqrt[3]{c}.$$ I suppose you could write $$f(x) = \sqrt[3]{x^3 + \frac{|x|}{x} c} - x,$$ but you seem to imply that you don't want to use expressions like $|x|/x$.

heropup
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    That function fails the "defined at 0" test, still have a division by $x$ – Alan Oct 03 '22 at 19:18
  • @Alan Other than using something like the "sign" function $$\operatorname{sgn}(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x < 0 \end{cases}$$ there is no elementary way of getting a function that does what you want. – heropup Oct 03 '22 at 19:21
  • @heropup Why is that when it works for something like: $$f(x) = \begin{cases} \left|\sqrt{x^2+c} + x\right|, & x \leq 0 \ \sqrt{x^2+c} - x, & x > 0 \end{cases}$$ which works out to $$\sqrt{x^2+c}-|x|$$. – Matt MacKinnon Oct 03 '22 at 19:26
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$$\frac{1}{2}\left(1-\frac{|x|}{x}\right)\left|\sqrt[3]{x^3-c}-x\right|+\frac{1}{2}\left(1+\frac{|x|}{x}\right)\left(\sqrt[3]{x^3+c}-x\right) \\\\ =\frac1{2x} \left[(x-|x|) \left|\sqrt[3]{x^3-c} - x\right| + (x+|x|) \left(\sqrt[3]{x^3+c} - x\right)\right] \\\\ = \frac1{2x} \left[x \left(\left|\sqrt[3]{x^3-c} - x\right| + \sqrt[3]{x^3+c} - x\right) + |x| \left(\sqrt[3]{x^3+c} - x - \left|\sqrt[3]{x^3-c} - x\right|\right)\right]$$

The discontinuity at $x=0$ vanishes since $\frac xx\to1$ and $\frac{|x|}x\to\pm1$. You would be left with

$$\frac12 \left[\left(\left|\sqrt[3]{x^3-c} - x\right| + \sqrt[3]{x^3+c} - x\right) \pm \left(\sqrt[3]{x^3+c} - x - \left|\sqrt[3]{x^3-c} - x\right|\right)\right] \\\\ = \begin{cases}\sqrt[3]{x^3+c}-x &\text{if }x>0 \\ \left|\sqrt[3]{x^3-c}-x\right|&\text{if }x<0\end{cases}$$

which could probably be rewritten further if you know something more about $c$.

user170231
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  • I am trying to come up with a non-piecewise function to generate a "perfect" $\delta$ for a given $c$ and $\varepsilon$ for $\lim_{x\rightarrow c}{x^3}$. Ironically, the $x$ in my the equation in my question is actually $c$ and the $c$ in my question is $\varepsilon$. For $\lim_{x\rightarrow c}{x^2}$ I got $\delta=\sqrt{c^2+\varepsilon}-|c|$, which is very nice. But despite $x^3$ having a nicer inverse than $x^2$, I can't find a function like this for $x^3$. – Matt MacKinnon Oct 03 '22 at 19:41
  • It kinda sounds like you're just trying to prove the limit $\lim_{x\to c}x^3=c^3$, in which case I don't know where you're getting the function in your question from. Per the limit definition, you want to find $\delta$ such that for any $\varepsilon>0$, you have$$|x-c|<\delta\implies|x^3-c^3|<\varepsilon$$Well,$$|x^3-c^3|=|x-c||x^2+cx+c^2|$$and you just need to impose a bound on the quadratic expression to write up the proof. Note that $x^2+cx+c^2=\left(x+\frac c2\right)^2+\frac{3c^2}4\ge\frac{3c^2}4$, so you need to pick an appropriate upper bound. – user170231 Oct 03 '22 at 20:18