Question

What is `lim_(xrarr4) ( x^4 -256) / (x^3 - 64)`?

Answer 1

`lim_(xrarr4)(x^4-256)/(x^3-64) = 16/3`

Explanation:

Either using polynomial long division or synthetic division we can discover:
`color(white)("XXX")(x^4-256) = (x-4)(x^3+4x^2+16x+64)`
and
`color(white)("XXX")(x^3-64)=(x-4)(x^2+4x+16)`

As long as `x!=4`
`color(white)("XXX")(x^4-256)/(x^3-64) = (x^3+4x^2+16x+64)/(x^2+4x+16)`

`lim_(xrarr4) (x^4-256)/(x^3-64) = ((4)^3+4(4)^2+16(4)+64)/((4)^2+4(4)+16)`

`color(white)("XXXXXXXXX")= (4(64))/(3(16)) = 16/3`

Answer 2

`16/3`

Explanation:

`lim_(x→4) (x^4 - 256)/(x^3-64)`

`=lim_(x→4) (x^4 - 4^4)/(x^3-4^3)`

i'm multiplying both numerator and denominator by `(x-4)`.
the reason for why i came up with this idea is that there is a theorem in limits as follows,
`lim_(x→a) (x^n - a^n)/(x-a) = n a^(n-1)`

so,
`=lim_(x→4) (x^4 - 4^4)/(x-4)*(x-4)/(x^3-4^3)`

`=lim_(x→4) (x^4 - 4^4)/(x-4)*lim_(x→4) (x-4)/(x^3-4^3)`

`= ( 4*4 ^ 3) /( 3 *4^2)`

`= 4^2/3`

`=16/3`

Answer 3

`16/3`

Explanation: