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For example, if we want to "force" a discontinuity at $x = c$ of the function $f(x) = x^2$ we change it to $$f(x) = \frac{x^3-cx^2}{x-c}$$ Then, if we want the $\lim\limits_{x \to c} f(x)$, we factor the denominator to recover $f(x) = x^2$. Does this process have a name?

To what extent, can we say that $x^2 = \dfrac{x^3-cx^2}{x-c}$ in calculus (limits)?

Rodrigo de Azevedo
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isedgar
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    Discontinuity is perhaps not the right word here (unless you assign some other value to $f(c)$): https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark May 28 '23 at 02:50
  • To answer the question in the title: I'ld call it useless. – Michael Hoppe May 28 '23 at 13:39
  • That function is continuous on its domain. It just isn't defined at $c$. You haven't forced a discontinuity at $c$, you've just restricted the domain of $x^2$. This is a problem with the high school definition of "continuity" not being the same as the working mathematician definition. – Charles Hudgins May 28 '23 at 14:15

2 Answers2

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"Forcing a discontinuity" is a reasonable name for it. The method of multiplying by $\frac{x-c}{x-c}$ isn't really useful, though. It's just a way to produce example functions with removable discontinuities, usually to motivate studying limits.

Be careful when saying $x^2=x^2\frac{x-c}{x-c}$. The following statements are true:

  • If $x\ne c$, then $x^2=x^2\frac{x-c}{x-c}$. If $x=c$, the expression $\frac{x-c}{x-c}$ is undefined, so "$x^2=x^2\frac{x-c}{x-c}$" can't be evaluated.
  • $x^2=x^2\frac{x-c}{x-c}$ whenever both sides of the equation are defined.
  • If $f$ and $g$ are both functions with domain $\Bbb R\setminus\{c\}$, defined by $f(x)=x^2$ and $g(x)=x^2\frac{x-c}{x-c}$, then $f=g$. (In this case $f$ is not defined at $c$ because $c$ is not in the domain.)
  • If $f:\Bbb R\to\Bbb R$ and $g:\Bbb R\setminus\{c\}\to\Bbb R$ are defined by $f(x)=x^2$ and $g(x)=x^2\frac{x-c}{x-c}$, then $f\ne g$ because the domains differ. However, $f$ and $g$ have all the same limits on $\Bbb R$. $f$ is the unique continuous extension of $g$ to the domain $\Bbb R$. For every $a\in\Bbb R$, $\lim_{x\to a}g(x)=f(a).$
Karl
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Looking at the function $g(x) = \frac{x^3-cx^2}{x-c}$ (I used $g(x)$ so there's no confusion with $f(x)=x^2$) we can see that $\lim\limits_{x\rightarrow c^{-}}g(x)=\lim\limits_{x\rightarrow c^{-}}g(x)=\lim\limits_{x\rightarrow c}g(x)=c^2$, but $g(c)\neq c^2$. So both left and right-hand limits are equal, but the function isn't defined at $c$. This means that the function $g(x)$ has a discontinuity of first kind, therefore this discontinuity is also called removable discontinuity (basically the process of what you've asked for). To my knowledge forcefully creating a discontinuity does not necesarilly have a name.

Regarding your second question considering that $g(x)$ has a removable discontinuity in most use cases disconinuity would be removed by modifying the function. But to answer your question $f(x)=g(x), \forall x\in\mathbb{R}\backslash\{c\}$, also $\lim\limits_{x\rightarrow c}g(x)=\lim\limits_{x\rightarrow c}f(x)=c^2$, but at that specific point $c$, $g(x)$ is not defined, hence you'd either need to remove this discontinuity or incoroporate it in all further calculations you are making.

bb_823
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