You should substitute 2 for x in `(x^4-16)/(x-2)` such that:

`lim_(x-gt2) (x^4-16)/(x-2) = (16-16)/(2-2) = 0/0`

Since the limit is indeterminate and you should not use l'Hospital's theorem, then you should evaluate the limit using the formula of difference of squares such that:

`lim_(x-gt2) (x^4-16)/(x-2) = lim_(x-gt2) ((x^2 - 4)(x^2 + 4))/(x - 2)`

You may use again the formula of difference of squares such that:

`lim_(x-gt2) ((x^2 - 4)(x^2 + 4))/(x - 2) =lim_(x-gt2) ((x-2)(x+2)(x^2 + 4))/(x - 2)`

You should reduce like terms such that:

`lim_(x-gt2) ((x-2)(x+2)(x^2 + 4))/(x - 2) = lim_(x-gt2) (x+2)(x^2 + 4)`

You should substitute 2 for x in `lim_(x-gt2) (x+2)(x^2 + 4)` such that:

`lim_(x-gt2) (x+2)(x^2 + 4) = (2+2)(4 + 4) = 32`

**Hence, evaluating the limit yields `lim_(x-gt2) (x^4-16)/(x-2) = 32.` **