Question

How do you find the limit `lim_(h->0)((2+h)^3-8)/h` ?

Answer 1

12

Explanation:

We can expand the cube:
`(2+h)^3 = 8 + 12h + 6h^2 + h^3`

Plugging this in,

`lim_(hrightarrow 0) (8+12h+6h^2+h^3-8)/h = lim_(hrightarrow 0) (12h+6h^2+h^3)/h`

`= lim_(hrightarrow 0) (12+6h+h^2) = 12`.

Answer 2

`12`

Explanation:

We know that,`color(red)(lim_(x->a)(x^n-a^n)/(x-a)=n*a^(n-1))`
`L=lim_(h->0)((2+h)^3-8)/h`,let,`2+h=xrArrhto0,then,xto2`
So,`L=lim_(x->2)((x^3-2^3)/(x-2))=3(2)^(3-1)=3*2^2=12`

Answer 3

Image reference...

Explanation:

• No intention do reply an answered answer... but as I was practicing , I added the image.