evaluate the limit of f(x)=(3x^2-6x)/(x-2) x->2

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