Limits with assumptions

Question

I have an example where Mathematica is not able to compute limits in a function restricted to assume that the input parameter is Real. I don't understand why this is so. 1. Could someone explain why this is a problem? 2. Is there a way to get around it, other than removing the restrictions on my function?

I would appreciate any help.

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Here's the code:

blah3[x_] := -(1/32) x^4 Log[x^2/E^5.4076];
blah4[x_ /; x \[Element] Reals] := -(1/
    32) x^4 Log[x^2/E^5.4076];
Limit[-(1/32) x^4 Log[x^2/^E^5.4076], x -> 0]
Limit[blah3[Sin[x]], x -> 0]
Limit[blah4[Sin[x]], x -> 0]

The first two work, while the third limit evaluation does not.

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Here's the context for the problem:

I'm trying to compare $$\int_0^\infty x^2 Log[1-E^{-\sqrt{\xi^2+x^2}}]$$ with its Taylor expansion, where the first few terms are given by $$-\frac{\pi ^4}{45}+\frac{\pi ^2 \zeta ^2}{12}-\frac{\pi \zeta ^3}{6}-\frac{1}{32} \zeta ^4 \log \left(\frac{\zeta ^2}{a_b}\right) + \ldots$$

To do this, I'm evaluating the former numerically and comparing against the latter approximation (of which blah3 and blah4 capture the part that is not easily calculable near $\xi = 0$). Just to ensure that I don't blindly trust results where things could go wrong (eg: $\xi^2 < 0$) I put in a condition that $\xi \in \mathbb{R}$ in my function definitions.

Answer 1

A conditional definition (using /;, that is) is not the same thing as a "Limit with assumptions", so far as Mathematica is concerned. To achieve that one does e.g. Limit[...,Assumptions->...]. To un derstand why, notice that, as written, blah4[Sin[x]] does not evaluate. So Limit can do nothing with it.

For your example, below is a form that Mathematica will be able to work with.

Limit[blah3[Sin[x]], x -> 0, Assumptions -> Element[x, Reals]]