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

1 Answer
Jan 9, 2017

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]}