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Is it possible to rewrite expression $\frac{1}{1-x}-\frac{1}{1+x}$ in order to be able to find its values near $x=1$ and $x=-1$ more precisely? This is a question in a numerical methods course. Is the problem ill posed from the start so you can't find a way to rewrite it?

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    $\begingroup$ $f$ is undefined at these points. Are you looking for a limit? $\endgroup$
    – Shahar
    Aug 21, 2014 at 23:35
  • $\begingroup$ It is clear that it is undefined at these points, I think the problem is to evaluate near $1$ and $-1$, without getting the usual, the limit doesn't exist response. $\endgroup$
    – Eesu
    Aug 21, 2014 at 23:36
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    $\begingroup$ Yeah the limits don't exist because on one side it's $\infty$ on the other it's $-\infty$ or vice versa. $\endgroup$
    – Shahar
    Aug 21, 2014 at 23:40

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$$f(x) = \frac{1}{1-x} - \frac{1}{1+x}$$ has nonremovable singularities at $x = \pm 1$. As $x$ becomes closer and closer to these points, $|f(x)|$ grows without bound. This is true no matter how you rearrange the formula for $f$. However, for values of $x$ near $1$, $\frac{1}{1+x}$ is very small in magnitude compared with $\frac{1}{1-x}$ so you could simply approximate $$f(x) \approx \frac{1}{1-x}$$ for $x$ near $1$. Similarly, if $x$ is near $-1$, then $$f(x) \approx \frac{1}{1+x}$$

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  • $\begingroup$ It's even worse than that, I'm afraid. A term becomes unbounded in either case, but its sign depends on the direction of approach. The limit, therefore, doesn't exist in either case, even if you allow $\pm \infty$ as possible limits. $\endgroup$
    – MPW
    Aug 22, 2014 at 0:36
  • $\begingroup$ @MPW: Indeed, the limit doesn't exist. It's not clear exactly what Eesu is trying to achieve, but if the goal is to evaluate $f$ near (but not at) $x = \pm 1$ then the function is well defined. The simplifications above are probably about as much as can be hoped for (which admittedly isn't much). $\endgroup$
    – user169852
    Aug 22, 2014 at 0:43

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