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Consider this rational function:

$$ f(x) = \frac{x^2 - 2x - 24}{x^2 + 10x + 24} $$

I have been taught that to solve for a removable discontinuity, I find the $x$ values such that both the numerator and denominator are equal to $0$; and to solve for vertical asymptotes, I find the $x$ values that make just the denominator equal to zero. So:

$$ f(x) = \frac{(x-6)(x+4)}{(x+6)(x+4)} = \frac{x-6}{x+6}, \quad x \neq -4 $$

What this means is that we have a vertical asymptote at $x = -6$ and a removable discontinuity at $x = -4$. Great. I can compute these.

But why?

We have two kinds of undefined here, $f(x_0) = \frac{0}{0}$ and $f(x_0) = \frac{g(x)}{0}$. Why do these result in different types of undefined behavior?

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  • $\begingroup$ As you said in $-4$ the discontinuity can be removed, which is not the case in $-6$. That makes all the difference. $\endgroup$ – Pjonin Aug 6 '18 at 14:52
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Let's be a bit careful with the term undefined, since it has a bit more of a precise meaning than the one you've used in your question.

You correctly observe that $f(-4) = \frac 0 0$ and $f(-6) = \frac{g(x)}0$. Why are they different? The reality is that $\frac 0 0 $ is actually indeterminate, meaning that it does not have an established value, instead of being undefined, which would imply having no value.

Hence, the case of $\frac 0 0 $ is interesting because it implies that there could exist a whole range of values that a function that approaches such a fraction could approach, since it all depends on how fast the numerator and denominator approach 0. In other words, removing the terms that take the function to zero in essence is us recognizing that these terms approach zero at the same rate, and at all points near -4, we can ignore them.

On the other hand, approaching a fraction of $\frac {g(x_0)}0$ where $g(x_0)$ is non-zero means that the function must approach either $\pm\infty$, since there are no ways for any value to exist and be equal to that fraction.

This is some rough intuition, but should shine some light on the difference between removable discontinuities and asymptotes.

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We have that

  • at $x=-4$ the function is not defined but we don't have any vertical asymptote since for $x\to -4$ we have $f(x) \to -5$

  • at $x=-6$ also the function is not defined but we have a vertical asymptote since for $x\to -6$ we have $f(x) \to \pm \infty$

Note that in the case of $x=-4$ we can remove the discontinuity defining $f(-4)=-5$ but for $x=-6$ we can't since $|f(x)| \to \infty$.

Note also that not always the discontinuity in the form $f(x_0) = \frac{0}{0}$ can be removed as for example for

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

at $x=0$ since $|f(x)| \to \infty$.

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    $\begingroup$ eh, I don't think this is really enough detail for an answer $\endgroup$ – Don Thousand Aug 6 '18 at 14:51
  • $\begingroup$ @RushabhMehta Maybe the OP can say whether or not it is useful $\endgroup$ – user Aug 6 '18 at 14:52

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