[Legal counting forum game] Counting with four fours

I think you should be able to use limits as long as you use lim(x -> infinity), so for like asymptotes. That still leaves room for creativity but it doesn't make place for cheasy tactics like that one .

88 = 44 * 4 / sqrt(4)

(gamma (4))! / (4 * sqrt (4)) - gamma (sqrt (4)) = 89

4! * 4 - 4 - (gamma (√4)) = 91

4! * 4 - sqrt (4) * sqrt (4) = 92
(gamma (4))! / (4 * sqrt (4)) + sqrt (4) = 92 (as well...)
(4! - sqrt (4)) * 4 + 4 = 92 (another one...)
(4! - 4 / 4) * 4 = 92 (there seems to be a lot of ways)
44 * sqrt (4) +4 = 92

93 = ((4!/4)!+4!)/(4!!)

95 = 4! * 4 - 4/4

96 = 4!*4*4/4

97 = 4! * 4 + 4/4

(4 / .4) ^ sqrt (4) – sqrt (4) = 98

"(4-1)!" = "gamma (4)"

102 = 4(4!+sqrt(4))-sqrt(4)

I know 102 has already been done, but I personally quite like this variation of 102:

(4! + sqrt (4)) * 4 – gamma (sqrt (4)) = 92

Actually that's not 92, but 103.