# What is it called when we force a discontinuity at a point?

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)?

• Discontinuity is perhaps not the right word here (unless you assign some other value to $f(c)$): math.stackexchange.com/questions/1087623/… May 28 at 2:50
• To answer the question in the title: I'ld call it useless. May 28 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. May 28 at 14:15

"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 $$a\in\Bbb R$$, $$\lim_{x\to a}g(x)=f(a).$$
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.