Question

How do you find `lim (10x^2+x+2)/(x^3-4x^2-1)` as `x->oo`?

Answer 1

`lim_(x->oo) (10x^2+x+2)/(x^3-4x^2-1) = 0`

Explanation:

When calculating the limit of a rational function for `x->+-oo` you can ignore all the monomials above and below the line except the ones with the highest order:

`lim_(x->oo) (10x^2+x+2)/(x^3-4x^2-1) = lim_(x->oo) (10x^2)/x^3 = lim_(x->oo) 10/x = 0`

If you want to see why this is the case, split the rational function as a sum and then bring each monomial below the line:

`(10x^2+x+2)/(x^3-4x^2-1) = (10x^2)/(x^3-4x^2-1)+x/(x^3-4x^2-1)+2/(x^3-4x^2-1)=10/((x^3-4x^2-1)/x^2)+1/((x^3-4x^2-1)/x)+2/(x^3-4x^2-1)= 10/(x-4-1/x^2)+1/(x^2-4x-1/x)+2/(x^3-4x^2-1)`

Evidently all three addenda are infinitesimal for `x->oo`