Question

Find lim x^4 - 8x/ x - 2 as x-->2 ??

Find lim (x^4 - 8x)/ (x - 2) as x-->2 ?

Answer 1

Answer for the question :
`lim_(xto2) (x^4-8x)/(x-2)=24`

Explanation:

We know that,

`color(red)((1)a^3-b^3=(a-b)(a^2+ab+b^2)`

Let,

`L=lim_(xto2) (x^4-8x)/(x-2)`

`=>L=lim_(xto2) (x(x^3-8))/(x-2)`

`=>L=lim_(xto2) (xcolor(red)((x^3-2^3)))/((x-2))...to color(red)(Apply(1)`

`=>L=lim_(xto2)(xcolor(red)(cancel((x-2))(x^2+2x+4)))/(cancel((x-2)))...to[x!=2]`

`=>L=lim_(xto2) [x(x^2+2x+4)]`

`=>L=[2(2^2+2(2)+4)]`

`=>L=[2(4+4+4)]`

`=>L=24`

Answer 2

`lim_(x->2)(x^4-8x)/(x-2)=24`

Explanation:

We have: `lim_(x->2)(x^4-8x)/(x-2)`

Since substituting two in the place of `x` results in `0/0`, we can use the L'Hospital's Rule.

The rule states that: `lim_(x->c)f(x)/g(x)=(f'(c))/(g'(c))` if `f(c)/g(c)` results in an indeterminate form such as `0/0`.

Let's find the derivatives using the power rule: `d/dx[x^n]=nx^(n-1)` where `n` is a constant.

The numerator:

`d/dx(x^4-8x)`

`=>4x^(4-1)-8x^(1-1)`

`=>4x^(3)-8x^(0)`

`=>4x^(3)-8*1`

`=>4x^(3)-8`

The denominator:

`d/dx(x-2)`

`=>1*x^(1-1)-2*0x^(0-1)`

`=>x^(0)-0`

`=>1`

We now have:

`(4x^(3)-8)/1` Substitute 2 in the place of `x`.

`=>(4(2)^(3)-8)`

`=>24`

Therefore, `lim_(x->2)(x^4-8x)/(x-2)=24`