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The first example I remember seeing for this was the sum of all Real numbers.

\begin{equation} \int_\mathbb{R} x\ dx = \int^{\infty}_{-\infty} x\ dx = \lim_{a\rightarrow\infty^-} \int^{a}_{0} x\ dx + \lim_{b\rightarrow-\infty^+} \int^{0}_{b} x\ dx \end{equation}

We can't evaluate that sum because we have no control over how fast each limit approaches infinity and we can't assume that they approach infinity at the same speed (which would result in $0$, if we could).

A more useful example is a problem I have actually encountered. The following function R is used to represent the progress per day given a devotion score of $x$ and a distraction score of $y$ in a simulation I was running a while back.

\begin{equation} R(x, y) = \frac{\lim_{a\rightarrow x} \ \ln{\frac{100 - a}{100}}}{\lim_{b\rightarrow y}\ \ln{\frac{100 - b}{100}}} \quad \text{for } x,y \in [0, 100] \end{equation}

Not sure about the notation here for dividing the limits, but this is essentially the function I have in my code, but with added limits. This code is very old so I don't remember why I came to believe these scores contributed in this way. Any analysis of the validity of this function for that purpose will be outside the scope of this question.

The value of $R(100, 100)$ would be how much progress a person who was $100\%$ devoted but also $100\%$ distracted would accomplish. Is $R(100, 100)$ incalculable for the same reason? Are there any techniques for finding a better way to represent this than a double limit that would lead to a meaningful answer?

Axoren
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1 Answers1

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As you've written it, $R(100,100)$ is undefined because it is $\frac{-\infty}{-\infty}$.

Your function $R(x,y)$ is actually equal to $\frac{\ln\left(1 - \frac{x}{100}\right)}{\ln\left(1 - \frac{y}{100}\right)}$ for all points $(x,y)$ where $0 \leq x < 100$ and $0 < y < 100$, so it is probably better to think of the problem in these terms: you have a function $R(x,y)$ defined on the square $[0,100) \times (0,100)$, and you would like to know how to extend this in a sensible way to the remaining three sides of that square, so that the function is defined throughout $[0,100] \times [0,100]$.

This is really a question about limits of functions of two variables. In this context, it makes sense to define $R(a,b)$ for $(a,b)$ on one of the edges by the following limit: $$R(a,b)= \lim_{\substack{(x,y) \to (a,b) \\ (x,y) \in [0,100) \times (0,100)}} R(x,y).$$

In this case, we find: for $0 < a \leq 100$, $R(a,0) = +\infty$; for $ 0 \leq a < 100$, $R(a,100) = 0$; for $0 \leq b < 100$, $R(100,b) = +\infty$. However, the limits defining $R(0,0)$ and $R(100,100)$ do not exist. For example, you get different limits at $(0,0)$ if you follow different paths: $$\lim_{\substack{(x,y) \to (0,0) \\ y = x}} R(x,y) = \lim_{t \to 0} R(t,t) = 1,$$ while $$\lim_{\substack{(x,y) \to (0,0) \\ y = 2x}} R(x,y) = \lim_{t \to 0} R(t,2t) = 1/2.$$ A similar phenomenon will occur at $(100,100)$ for $y = x$ on the one hand and $1-y/100 = (1 - x/100)^2$ on the other.

As these examples show, the limits at $(0,0)$ and $(100,100)$ may well exist if $(x,y)$ is restricted to belong to a certain subset of $[0,100) \times (0,100)$ as it tends towards those points. If there is such a restriction on $x$ and $y$ that would represent a meaningful condition in the real-world context of your problem, then it would make sense to determine whether that restricted limit exists.

I believe the difficulty here may have arisen because you tried to approach a problem that was really about a function of two variables as one in terms of single-variable limits. Looking at the problem within the context of limits and continuity of two-variable functions clarifies the issues.

user187373
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  • This is a really good explanation of what I was worried about. That there's no one general answer to how to resolve this. On a case-by-case basis, we need to decide a path for the limit to take that makes sense. In the case of my function, I wouldn't know which path would make sense generally. – Axoren Nov 02 '14 at 01:10
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    In the case of your function, if there are positive constants $m$ and $M$ such that, in a neighbourhood of $(100,100)$, we only look at points $(x,y)$ such that $m(1-x/100) \leq 1 - y/100 \leq M(1 - x/100)$, then the limit at $(100,100)$ will be $1$. This region represents some angle centred at $(100,100)$ and extending into the square, but held away by at least some small angle from the sides of the square. The only problems are paths that get very close to one of the two edges on the way to $(100,100)$. – user187373 Nov 02 '14 at 05:09