Question

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

Answer 1

See below

Explanation:

`lim_(x to oo) x^3 / (x-1) - x^2`

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

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

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

Using Binomial Series

`= lim_(x to oo) x^2 ( (1+1/x + mathcal O (1/x)^2 ) - 1 )`

`= lim_(x to oo) x^2 ( (1/x + mathcal O (1/x)^2 ) )`

`= lim_(x to oo) x + mathcal O (1/x)^0 = oo`

Answer 2

`lim_(x->oo) (x^3/(x-1)-x^2) = oo`

Explanation:

When assessing:

`lim_(x->oo) (x^3/(x-1)-x^2)`

you cannot simply substitute `x=oo` since this results in the indeterminate form:

`oo - oo`

Instead, we can simplify the expression first, like this:

`lim_(x->oo) (x^3/(x-1)-x^2) = lim_(x->oo) ((x^3-x^2+x^2-x+x-1+1)/(x-1)-x^2)`

`color(white)(lim_(x->oo) (x^3/(x-1)-x^2)) = lim_(x->oo) (((x^2+x+1)(x-1)+1)/(x-1)-x^2)`

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

`color(white)(lim_(x->oo) (x^3/(x-1)-x^2)) = lim_(x->oo) (x+1+1/(x-1))`

`color(white)(lim_(x->oo) (x^3/(x-1)-x^2)) >= lim_(x->oo) x = oo`