Answer
The fraction of the original amount $m(0)$ remaining after 68 million years is $\frac{1}{e^{8228}}~m(0)$
The maximum age of a fossil that could be dated using carbon-14 is 57,089 years.
Work Step by Step
We can find the value of $k$ when we use carbon dating:
$m(t) = m(0)e^{kt}$
$m(5730) = m(0)e^{5730k} = 0.5~m(0)$
$e^{5730k} = 0.5$
$5730~k = ln(0.5)$
$k = \frac{ln(0.5)}{5730}$
$k = -0.000121$
Then:
$m(t) = m(0)~e^{-0.000121~t}$
We can find the amount remaining after 68 million years:
$m(t) = m(0)~e^{-0.000121~t}$
$m(65,000,000) = m(0)~e^{(-0.000121)~(68,000,000)}$
$m(65,000,000) = \frac{1}{e^{8228}}~m(0)$
This number is an extremely small fraction and it is even less than the number $10^{-2500}$
We can find the time $t$ when only 0.1% remains:
$m(t) = m(0)~e^{-0.000121~t} = 0.001~m(0)$
$e^{-0.000121~t} = 0.001$
$(-0.000121)~t = ln(0.001)$
$t = \frac{ln(0.001)}{-0.000121}$
$t = 57,089~years$
The maximum age of a fossil that could be dated using carbon-14 is 57,089 years.
