We have to find lim x-->2 [(x^4-16)/(x-2)]

lim x-->2 [(x^4-16)/(x-2)]

=> lim x-->2 [(x^2 - 4)(x^2 + 4)/(x-2)]

=> lim x-->2 [(x - 2)(x + 2)(x^2 + 4)/(x-2)]

=> lim x-->2 [(x + 2)(x^2 + 4)]

Substitute x = 2

=> (2 + 2)(4 + 4) = 4*8 = 32

**The required limit is 32**

We calculate the limit substituting x by 2

lim (x^4-16)/(x-2) = (16-16)/(2-2) = 0/0

We notice that we've get an indetermination case

We'll solve using L'Hospital rule:

lim (x^4-16)/(x-2) = lim (x^4-16)'/(x-2)'

lim (x^4-16)/(x-2) = lim 4x^3/1

We'll substitute x by 2:

lim 4x^3 = 4*2^3

lim 4x^3 = 32

The limit of the function is:

lim (x^4-16)/(x-2) = 32

We also could write the numerator as a difference of squares:

x^4-16 = (x^2 - 2^2)(x^2 + 4)

But x^2 - 2^2 = (x-2)(x+2)

lim (x^4-16)/(x-2) = lim (x-2)(x+2)(x^2 + 4)/(x-2)

We'll simplify and we'll get:

lim (x+2)(x^2 + 4) = (2+2)(4+4)

lim (x+2)(x^2 + 4) = 4*8

lim (x+2)(x^2 + 4) = 32