`lim_(x->2) (3x^2-4x-4)/(x-2)`

To solve, first plug-in x=2 to the function.

`lim_(x->2) (3x^2-4x-4)/(x-2) = (3(2)^2-4(2)-4)/(2-2)=0/0`

Since the result is indeterminate, to get its limit, we have to use other methods. One of the methods is to simplify the function.

To simplify the function, factor the numerator.

`lim _(x->2)(3x^2-4x-4)/(x-2)`

`= lim_(x->2)((x-2)(3x+2))/(x-2)`

Then, cancel the common factor between the numerator and denominator, which is (x - 2). So, the simplified form of the function is:

`=lim_(x->2) 3x + 2`

Now that the function is in simplified form, plug-in x=2.

`= 3(2)+2`

`=6+2`

`=8`

**Hence, `lim_(x->2) (3x^2-4x-4)/(x-2) = 8` .**

The limit `lim_(x->2) (3x^2 - 4x - 4)/(x - 2)` has to be determined.

Notice that substituting x = 2 gives the numerator as 3*2^2 - 4*2 - 4 = 12 - 8 - 4 = 0 and the denominator is also 2 - 2 = 0. This is an indeterminate form of the type `0/0` . In this case it is possible to use l'Hospital's rule to arrive at the value of the limit. Replace the numerator and the denominator by their derivatives.

`lim_(x->0) (6x - 4)/1`

Substituting x = 2 now gives 12 - 4 = 8

**The limit **`lim_(x->2) (3x^2 - 4x - 4)/(x - 2) = 8`