numeric_limits and constexpr

Hello,

I notice that the default numeric_limits in C++0X reads:
static constexpr T min() throw() { return T(); }

I assume this fails if T doesn't have a constexpr default constructor
(T isn't a literal type).

It's never been clear whether users were supposed to provide a
specialization of numeric_limits for their own number types, but this
sounds like one has to be careful using numeric_limits.

If I write something like:
if(numeric_limits<T>::is_bounded) {
  ... use numeric_limits<T>::min() ...
}
do I have the guarantee that the compiler won't try to instantiate min
for wrong types? (I could test std::is_builtin<T>::value (can't
remember the real name) but the issue is the same) Or do I need to
write 2 functions?

Since there is no has_epsilon, for a type that has
numeric_limits<T>::is_specialized, I would check epsilon()==T() which
is a bad idea. Should I assume that only float, double and long double
have epsilon?

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