So, when you have a removable discontinuity, that means the function is defined at every point but the point in question. In this problem, the discontinuity can be found by factoring the numerator and denoinator and simplyfing the rational function. [tex] \frac{x^3-3x-4}{x^2-5x-4}=\frac{(x-4)(x+1)}{(x-1)(x-4)}=\frac{x+1}{x-1} [/tex]
with a removable discontinuity at [tex] (4,\frac{5}{3}) [/tex]
So the answer is none of the above.
