Question

How do you find `\lim _ { n \rightarrow \infty } \frac { 2x ^ { 5} - 3x ^ { 2} } { 3x ^ { 5} - x ^ { 3} }`?

Answer 1

I think you meant `x->oo`

Explanation:

If it is how I think it is you have:
`lim_(x->oo)((2x^5-3x^2)/(3x^5-x^3))=`
collect `x^5`:
`lim_(x->oo)(cancel(x^5)(2-3/x^3))/(cancel(x^5)(3-1/x^2))=`
as `x->oo` then `1/x^n->0` and so:

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