To determine the value of the given limit, we'll substitute x by 2 in the expression of the function.

lim f(x) = lim (3x^2-6x)/(x-2)

lim (3x^2-6x)/(x-2) = (12-12)/(2-2) = 0/0

We notice that we've obtained an indetermination case.

We could solve the problem in 2 ways, at least.

The first method is to factor the numerator. Since x = 2 has cancelled the numerator, then x = 2 is one of it's roots. The other root we'll calculate it using Viete's relations.

x1 + x2 = 6/3

2 + x2 = 2

x2 = 0

The factored numerator is:

3x^2-6x = 3x(x-2)

We'll re-write the limit of the function, having the numerator factored:

lim f(x) = lim 3x(x-2)/(x-2)

We'll simplify:

lim 3x(x-2)/(x-2) = lim 3x

We'll substitute x by 2:

lim 3x = 3*2

**lim f(x) = 6, for x->2**

Another method is to use L'Hospital rule, since we've get an indeterminacy "0/0".

lim (3x^2-6x)/(x-2) = lim (3x^2-6x)'/(x-2)'

lim (3x^2-6x)'/(x-2)' = lim (6x-6)/1

lim (3x^2-6x)/(x-2) = lim (6x-6)

We'll substitute x by 2:

lim (6x-6) = 6*2 - 6

**lim (6x-6) = 6**