You may also use the following alternative method, such that:

`lim_(x->2) (3x^2-6x)/(x-2) = (3*2^2 - 6*2)/(2 - 2)`

`lim_(x->2) (3x^2-6x)/(x-2) = 0/0`

The indetermination `0/0` allows you to use L'Hospital's theorem, such that:

`lim_(x->2) (3x^2-6x)/(x-2) = lim_(x->2) ((3x^2-6x)')/((x-2)') `

`lim_(x->2) ((3x^2-6x)')/((x-2)') = lim_(x->2) (6x - 6)/1`

`lim_(x->2) (6x - 6)/1 = 6*2 - 6 = 6`

**Hence, evaluating the limit, using L'Hospital's theorem, yields **`lim_(x->2) (3x^2-6x)/(x-2) = 6.`

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

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

We've get an indetermination case.

Since x = 2 has cancelled the numerator, then x = 2 is one of it's roots. We'll find the 2nd root 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

**Therefore, the limit of the given function, if x approaches to 2, is lim f(x) = 6.**