Lim as x approaches 2 of (x^2-kx+4)/(x^3-8)

Answer 1

if k = -2, there's no problem. The fraction is just 1/(x-2), so the limit is ∞

Otherwise, the numerator is only ever zero if k>4, at

x = 1/2 (k±√(k^2-16))

In any case, as x->2, the numerator is not zero, the the limit is ±∞