Why is the input Limit[IntegerPart[Sin[x]/x], x -> 0] not being evaluated?

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A closely related problem was treated in this question I asked some time ago that received a beautiful answer from @acl. The documentation for
Nevertheless, the symbolic part of the assertion has been proved false in the aforementioned question. Just take a look at what Mma thinks of its derivative:
So, What does
Ha! it does nothing! That doesn't mean that Mma can't calculate limits for analytic non-symbolically treatable functions. It can:
Of course this is useless in your case since So, there is a deadlock: Mma does not know how to treat Sorry :) EditThis is a cheater using PiecewiseExpand:
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Limit[IntegerPart[Sinc[x]], x -> 0]works, though. – J. M.♦ May 16 '12 at 17:07IntegerPartrequires infinite precision, which may be a little much to ask :-). Nevertheless MMA should be able to figure this one out from the series, and in fact it quickly gets the correct answer with, say,Limit[IntegerPart[Series[Sin[z]/z, {z, 0, 4}]], z -> 0]. But this is a snare: replacing the4with1gets a wrong answer. It does not seem to realize that $1+O[z]^2$ can be less than $1$! – whuber May 16 '12 at 19:43Limit[IntegerPart[Sin[x]/x], x -> 10^-80]returns 0, so it must be a precision thing. – Guillochon May 16 '12 at 22:250forLimit[IntegerPart[Sinc[x]], x -> 0]. Which version are you running? I'm using v8.0.4, but it does the same on v7. – rcollyer May 17 '12 at 1:45