The value of (e^x - 1) is representable as a long double when x <=
11356.5234, but expm1l(x) overflows to infinity when x > 11356.1768
(approximately).

Steps to Reproduce:
Compile and run the following program:

#include <stdio.h>
#include <math.h>

main ()
{
  long double a, b;
  const long double x = 11356.35L;

  a = expl (x) - 1.0L;
  b = expm1l (x);
  if (a == b) {
    printf ("e^%.2Lf - 1 = %Lg\n", b);
    return 0;
  } else {
    printf ("e^%.2Lf - 1 = %Lg; but expm1l(%.2Lf) returned %Lg\n",
	    x, a, x, b);
    return 1;
  }
}


Actual Results:  e^11356.35 - 1 = 1.00032e+4932; but expm1l(11356.35)
returned inf


Expected Results:  e^11356.35 - 1 = 1.00032e+4932


Additional info:

This bug probably isn't even worth reporting, except for completeness. 
expm1l is designed to compute e^x-1 more accurately when x is negative, so
[at least the i386 version] uses a slightly different algorithm than expl. 
This algorithm overflows for large x where expl would not.  But then, expm1
really isn't intended for large x anyway.

In my own port of the math library, I circumvent the problem by calling
__ieee754_expl when x >= 64, because at that magnitude, -1 is too small to
affect the result.  This is probably overkill, but it works.