Question

How do you find `lim (y^4-y^3+1)/(2y^4-4y^2+5)` as `y->oo`?

Answer 1

`lim_(y->oo) (y^4-y^3+1)/(2y^4-4y^2+5) = 1/2`

Explanation:

When evaluating the limit at `+-oo` of a rational function you can ignore all the terms except the ones of highest order at the numerator and denominator:

`lim_(y->oo) (y^4-y^3+1)/(2y^4-4y^2+5) = lim_(y->oo) y^4/(2y^4) = 1/2`

It's easy to see why this is the case: start noting that around `+-oo` we can assume that `y!=0` so also `y^4 !=0`.

We can then divide both the numerator and denominator by `y^4`:

`lim_(y->oo) (y^4-y^3+1)/(2y^4-4y^2+5) =lim_(y->oo) ((y^4-y^3+1)/y^4)/((2y^4-4y^2+5)/y^4) = lim_(y->oo) (1-1/y+1/y^4)/(2-4/y^2+5/y^4) = (1-0+0)/(2-0+0) = 1/2`