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In the real world, that is, in engineering, computer science, or whatever, whenever we get undefined or indeterminate expressions, do we just take the limit, if it exists? Does this work in the real world?

Example:

$$\frac{x^2-1}{x-1}$$

The above is undefined when $x=1$, but it does have a limiting value when $x\to1$, which is $2$. So, in the real world, do we regularly use the limiting values whenever we get undefined forms like these?

tryingtobeastoic
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2 Answers2

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Physics uses mathematics to model the real world. Those models may behave very well in some ranges, and they may do unusual things in other ranges. Sometimes, it's fine to consider how those models behave in the limit approaching various singularities and that provides insight into what's happening in the real world, and sometimes it's necessary to find a different model that behaves better in that limit, and possibly look for ways to reconcile the difference between the two models.

ConMan
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  • I edited the question. Kind sir, would you like to change your answer or keep it how it is? – tryingtobeastoic Oct 05 '21 at 14:41
  • It's still basically the same thing - if my mathematical model for something has a singularity like $\frac{x^2-1}{x-1}$, then my first guess is that it will behave like the limit, but it's only in observation that I know for sure. Black holes are an example of a singularity that (to some approximation) behave just like the model predicts, even though that involves a division by zero. – ConMan Oct 05 '21 at 22:09
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Just a references and a summary to think about:

Instead of the OP's $(x^2-1)/(x-1)$ another function is considered, namely the Sinc function, which has important physical applications in for example spectrography and signal processing: $$ \operatorname{sinc}(x) = \begin{cases} \sin(x)/x & \text{for } x \ne 0 \\ 1 & \text{for } x =0 \end{cases} $$ It is argued in the reference that nobody in physics would ever think of a definition with a discontinuity at $x=0$ such as: $$ \operatorname{suck}(x) = \begin{cases} \sin(x)/x & \text{for } x \ne 0 \\ 0 & \text{for } x =0 \end{cases} $$ The basic reason is that, from a physics viewpoint, continuity is very similar to equality. In order to appreciate this, you might have to notice an equivalence between the $(\epsilon,\delta)$ formalism in analysis and the processing of errors in physics. The continuum in physics is always error prone. There is no other equality there than approximately equal. That's why we have the following sequence of abstractions: $$ |x-a| \lt \delta \quad \Longrightarrow \quad |f(x)-f(a)| \lt \epsilon \\ x \approx a \quad \Longrightarrow \quad f(x) \approx f(a) \\ x=a \quad \Longrightarrow \quad f(x)=f(a) $$ With other words: functions defined on a continuum in the real world can be only functions if they are themselves continuous as well.
All this in a nutshell. More details are in the reference as mentioned. Alas, there is much more to tell that doesn't fit into the margins of MSE :-(
As far as the OP's function is concerned, indeed: $(x^2-1)/(x-1)=x+1$ is always true in the real world. No black holes are going to change that fact.
Han de Bruijn
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