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Let $$f(x) = \frac{1}{\log(\frac{x}{c})}$$ where $c$ is some constant number. Consider the variable $x$ in the large regime where $x \gg c$ and small regime where $x \ll c$. How would $f(x)$ depend on $x$ in those regimes? Would one use some kind of Taylor expansion?

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In the examination of the two limits of your function, we should probably start with the identity $\log{\frac x c} = \log x - \log c$. We can then identify the largest of these two terms, and make a series expansion in the ratio of the smaller over the larger term. Assuming for convenience that $x > 1$ and $c > 1$, this procedure yields:

$x >> c: f(x) = \frac 1 {\log x - \log c} = \frac 1 {\log x} \frac 1{1 - \log c/\log x} $

Expanding the term between square brackets gives

$f(x) = \frac 1{\log x} \left[1 + \frac{\log c}{\log x} + \left(\frac{\log c}{\log x}\right)^2 +\cdots \right]$

$x << c: f(x) = \frac 1{\log x - \log c} = -\frac 1 {\log c} \frac 1 {1 - \log x/\log c}$

Once again we expand the term between square brackets.

$f(x) = -\frac 1 {\log c} \left[ 1 + \frac{\log x}{\log c} + \left( \frac{\log x}{\log c} \right)^2 + \cdots \right] $

So it is indeed possible to expand your function in the given limits. However, I doubt the result has much practical use. In particular since the original function is very simple.

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