The limit has the form 0/0, so we could use L'Hopital's Rule. The limit can also be evaluated algebraically as follows:
limx→1[(x2-kx+k-1)/(x-1)]
= limx→1[((x2-1) + (k-kx))/(x-1)]
= limx→1[((x+1)(x-1) -k(x-1))/(x-1)]
= limx→1[((x-1)(x+1-k))/(x-1)]
= limx→1(x+1-k) = 1+1-k = 2-k