Question

How do you evaluate the limit `(x^4-10)/(4x^3+x)` as x approaches `oo`?

Answer 1

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

Explanation:

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

So:

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

You can see that by separating the sum:

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

Evidently:

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

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

graph{(x^4-10)/(4x^3+x) [-10, 10, -5, 5]}