You can use an alternate formula for the large values:
T = (1 + 1./exp(x)) ./ (1 - 1./exp(x))
Note that once x gets large enough (19), the ratio calculation becomes equivalent to the "first order" approximation 1 + 2/exp(x) because of the limits of double precision arithmetic.
A general answer is probably not possible. It will depend on the calculations being done. First one would look to see if the ratio has a finite limit. Simple inspection or algebraic manipulation of your particular example easily shows that it does. Then either look for a simple rearrangement of terms so that the numerator & denominator converge instead of diverge, or divide things out to reduce it to a series that converges rapidly enough to be of practical use. I did both above. E.g., just doing a long divide results in this:
T = 1 + 2/exp(x) + 2/exp(x)^2 + 2/exp(x)^3 + ...
Those 2/exp(x)^n terms are going to rapidly become insignificant compared to 1 for double precision arithmetic as x gets large. It only takes x=19 before the entire ratio calculation becomes equivalent to only the first term approximation of 1 + 2/exp(x), and when x=38 the 2/exp(x) part becomes smaller than eps(1) and no longer has any effect on the double precision result.
If you use the symbolic toolbox, you could
syms x x0 T = (exp(x) + 1)/(exp(x) - 1); Tt = taylor(T, x, x0, 'Order', 20) Tt160 = simplify(subs(Tt, x0, 160)); fplot(Tt160, [120 200]);
and do other x0 values for higher accuracy in local ranges.
... Looks like Order 100 is pretty hard for MATLAB to process... I'm going to have to shoot that session it looks like!
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