What is the limit of the function (x^4-16)/(x-2), if x goes to 2?

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

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

use the relation a^2 - b^2 = (a - b)(a + b)

=> 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 value of lim x-->2 [(x^4-16)/(x-2)] = 32.**

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

We notice that the 1st factor is also a difference of squares that returns the product:

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

We'll re-write the function:

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

We'll simplify:

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

We'll evaluate the limit the fraction:

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

We'll replace x by 2:

lim (x+2)(x^2+4) = (2+2)(4+4) = 4*8 = 32

**The value of the limit, if x approaches to 2, is lim (x^4-16)/(x-2) = 32.**

You need to substitute 2 for x in limit of function such that:

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

You should solve the indetermination `0/0` using l'Hospital's theorem such that:

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

`lim_(x-gt2) ((x^4-16)')/((x-2)') = lim_(x-gt2) (4x^3)/1`

You need to substitute 2 for x such that:

`lim_(x-gt2) (4x^3)/1 = 4*2^3 = 32`

**Hence, evaluating the limit of function using l'Hospital's theorem yields `lim_(x-gt2) (x^4-16)/(x-2) = 32` .**

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