Question

What is `lim_(x->1) (x^3-1)/(x-1)^2` ?

Answer 1

`oo`

Explanation:

We think about factorizing `" "x^3-1` by applying the following
`" "`
polynomial identity.
`" "`
`a^3-b^3=(a-b)(a^2+ab+b^2)`
`" "`
`x^3-1=x^3-1^3=(x-1)(x^2+x+1^2)=(x-1)(x^2+x+1)`
`" "`
`lim_(x->1)(x^3-1)/(x-1)^2=((x-1)(x^2+x+1))/(x-1)^2`
`" "`
`lim_(x->1)(x^3-1)/(x-1)^2=(cancel ((x-1))(x^2+x+1))/(x-1)^cancel (2)`
`" "`
`=3/0=oo`

Answer 2

Limit does not exist.

Explanation:

`(x^3-1)/(x-1)^2`.

Plugging in 1 gives the indeterminate form: `0/0`

Using l'Hospital's rule:

`d/dx(x^3-1)=3x^2`

`d/dx (x-1)^2=2x-2`

`(3x^2)/(2x-2)`

Plugging in 1:

`(3(1)^2)/(2(1)-2) =3/0`

Limit does not exist.

Answer 3

`lim_(x->1) (x^3-1)/(x-1)^2` does not exist since the left and right limits disagree.

Explanation:

Note that:

`(x^3-1)/(x-1)^2 = (color(red)(cancel(color(black)((x-1))))(x^2+x+1))/(color(red)(cancel(color(black)((x-1))))(x-1)) = (x^2+x+1)/(x-1)`

Note that the numerator is always positive. So the sign of `(x^2+x+1)/(x-1)` is determined by the sign of the denominator.

So we find:

`lim_(x->1-) (x^2+x+1)/(x-1) = -oo`

`lim_(x->1+) (x^2+x+1)/(x+1) = +oo`

Since the left and right limits disagree, the two-sided limit does not exist.

graph{(x^3-1)/(x-1)^2 [-9.92, 10.08, -60, 60]}