Evaluate the limit (x^4-16)/(x-2), x-->2.

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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**

The expression (x^4-16)/(x-2) at x = 2 assumes (2^4-16)/(2-2) = 0/0 form.

So clearly both numerator and denominators have the factor x-2 by reainder theorem

Therefore we divide both numerator and denominator by x-2 and then take the limit.

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

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

Therefore Lt x-->2 (x^4-16)/(x-2) = Ltx-->2 (x+2)(x^2+4) = (2+2)(2^2+4) = 4*8 = 32.

Therefore Lt (x^4-16)/(x-2) = 32.

Hi.

Just for a graphical appreciation of the problem.

limit (x^4-16)/(x-2), x-->2.

First we will try substituting to find the limit.

We will substitute with x= 2:

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

==> Now we will factor the numerator:

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

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

Now we will reduce similar:

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

Now substitute with x=2:

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

= 4*8 = 32

**Then, lim (x^4-16)/(x-2) when x--> 2 is 32.**

We have to evaluate limit (x^4-16)/(x-2), x-->2.

Now (x^4 - 16)/(x-2)

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

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

cancel (x-2)

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

Therefore limit (x^4 - 16)/(x-2), x-->2

=> limit ( x + 2) (x^2 + 4), x-->2

=> (2+2)(2^2 + 4)

=> 4 ( 4 + 4)

=>4 * 8

=> 32

**Therefore limit (x^4-16)/(x-2), x-->2 = 32**

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