Need help on piecewise function. Please

If{ f(x)= (x^2-6x)/x for x does not =0
{f(0)=2k-1

and if f is continuous at x=0 then k=?

the other user is WRONG!

lim f(x) = lim (x²-6x)/x = lim (x-6) = 0 - 6 = -6
x->0 .......x->0.............x->0

THEREFORE, f(0) = 2k - 1 = -6 -----> 2k = -6 + 1 = -5 -----> k = -5/2

f(x) = { x² - 6x, for x ≠ 0
........{ 2k - 1 , for x = 0

Given that f(x) is continuous at x = 0, the limit as x approaches 0 of f(x) exists and is equal to f(0).
lim x→0 f(x) = f(0)
lim x→0 (x² - 6x) = 2k - 1
0 - 0 = 2k - 1
2k = 1
k = 1/2