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For instance consider the function

$$f(x) = \frac{x-2}{(x-2)(x+2)}$$

I don't understand why some discontinuities are considered "removable". Yes, we can cancel out the $x-2$'s but isn't this invalid at $x=2$ because it's like canceling out $\frac{0}{0}$? I don't understand why we're allowed to do this.

Anyway if we "remove" that piece we get:

$$f(x) = \frac{1}{x+2}$$

Which has a discontinuity at $x=-2$ but we can't "remove" that.

So we end up with two discontinuities -- do we define the domain to include or exclude these? Is the domain all real numbers? All except $2$? All except $-2$? All except $-2, 2$?

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  • $\begingroup$ Because limit . $\endgroup$ – user202729 Jan 29 '18 at 14:39
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The discontinuity at $x=2$ can be removed because $\lim_{x\to2}f(x)=1/4$. When 'removing' the discontinuity, we define a new function $f^\prime$ that takes on the values of $f$ at all points in the domain of $f$ and $f^\prime(2)=1/4$, thereby extending the domain.

The discontinuity at $x=-2$ cannot be removed because the limit there is improper: $\lim_{x\to-2}f(x)=\infty$.

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  • $\begingroup$ How do you know the limit at each point? Is there an analytic way? Is the only difference between removable and nonremovable discontinuities whether or not the limit is a number or not? $\endgroup$ – user525966 Jan 29 '18 at 14:57
  • $\begingroup$ When the limit at a certain point is improper, it cannot be represented by a real number. Only proper limits can be used to remove discontinuities. $\endgroup$ – Jeroen van Riel Jan 29 '18 at 15:03
  • $\begingroup$ What then would be the domain of $f$? All reals except $-2,2$? $\endgroup$ – user525966 Jan 29 '18 at 15:04
  • $\begingroup$ Yes, as @57Jimmy already pointed out in his answer. $\endgroup$ – Jeroen van Riel Jan 29 '18 at 15:13
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You clearly understand the algebra, and the idea behind "removable singularity". The definition of the domain is a little subtle (and usually not particularly important, given your understanding).

The expressions on the right in

$$f(x) = \frac{x-2}{(x-2)(x+2)}$$

and

$$g(x) = \frac{1}{x+2}$$

define the same function where both make sense. That function has an unremovable singularity at $x=-2$.

The domain of $g$ contains the point $x=2$; the domain of $f$ does not, so strictly speaking they are not the same function. But the limit of $f$ at $x=2$ does exist, and has value $g(2) = 1/4$. That's exactly what we mean when we say the singularity is removable.

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  • $\begingroup$ So a removable discontinuity is an undefined point that has a defined limit, whereas a nonremovable discontinuity has a limit that does not exist or is infinity? $\endgroup$ – user525966 Jan 29 '18 at 14:52
  • $\begingroup$ @user525966 Yes. $\endgroup$ – Ethan Bolker Jan 29 '18 at 15:33
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The domain of $f$ is $\mathbb{R} \backslash \{-2,2\}$, because at this two points $f$ is not defined. Having said that, some singularities are better behaved than others: $f$ goes to infinity as $x \to -2$, whereas $\underset{x \to 2}{\lim} f(x) = \frac{1}{4}$ is well-defined. This is exactly the difference between poles and removable singularities.

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