limit....

(6x^3 - 13x^2 - 41x - 12) / ( x^2 - 8x + 16 )
=[(x-4)(2x+3)(3x+1)] / [(x-4)^2]
=[(2x+3)(3x+1)] / (x-4)

Let y = [(2x+3)(3x+1)] / (x-4)

For x-> 4-, (x-4) -> 0-
Then y -> -inf


For x-> 4+, (x-4) -> 0+
Then y -> inf

As lim (x-> 4-) does not equal to lim (x-> 4+) so the limit (x-> 4) does not exist.