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I am a high school student and I am slightly confused regarding certain aspects of continuity in my calculus class. Rational functions are often given as examples of functions which possess so-called removable and asymptotic discontinuities. For example, consider the function $f: I \rightarrow \mathbb{R}$ where $f(x) = \frac{x^2 - 16}{x^2 + 5x + 4}$. This function is said to be discontinuous at $x=-1$ and $x=-4$. However, I find this question is sort of trivial in a sense---it's like asking if $f(x) = \ln(x)$ is continuous at $x=-16$. The value is not even in its domain. Why bother discussing it?

It gets particularly hairy if we're expected to classify $f(x)$ as being a continuous or discontinuous function, which, as it so happens, we sometimes are. The "correct" answer is that it is not continuous. However, I disagree; I think $f(x)$ is continuous. This is evident from the definition of continuity. A function is said to be continuous at some point $c$ in its domain if $\forall \epsilon > 0 \; \exists \; \delta > 0$ such that for all $x$ within the domain of the function, $ |x-c|<\delta \Longrightarrow |f(x) - f(c)|<\epsilon$. $f$ is said to be a continuous function if this holds for all numbers $c$ within the domain of $f$. It should not have to hold for all $c \in \mathbb{R}$; otherwise a number of functions such as $\ln(x), \arcsin(x), x^{\frac{1}{2}}$, etcetera would be classified discontinuous.

What am I to make of this? I'm just...confused. Is my argument incorrect? Is this a definitional issue? Or is this a pedagogic issue?

SupInf123
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  • Maybe this is interesting for you. – drhab Dec 19 '15 at 09:06
  • An important question: what is $I$? It makes sense only to speak of a function $f$ being continuous or discontinuous at points of its domain — here, $I$. I'd say, put aside considerations of whether $f$ has a removable discontinuity at a point: in the case of your rational function $f$, no discontinuities have been removed, and the function is simply not defined at $x = -1, -4$. That is, $-1, -4\notin I$: it's not continuous at either point, it isn't even defined there. – BrianO Dec 19 '15 at 09:22

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It is a matter of definitions. In higher analysis and topology, continuity is defined at points of the domain. In elementary calculus, instructors tend to slightly generalize the concept of discontinuity by saying (roughly)

Definition. A function $f$ is discontinuous at a point $c$ if $c$ is either a point of the domain of $f$ or an accumulation point of the domain of $f$, and furthermore $f$ cannot be extended to a continuous function at $c$.

This way, $f(x)=\sin x/ x$ is considered to be continuous at $x=0$, while $f(x)=1/x$ is not.

Just be patient: in your future courses this will disappear ;-)

Siminore
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