# Asymptote Problem

1. Jun 5, 2004

Help!
My problem asks to find the vertical asymptotes of x^3/(x^2+3x-10)

I found –5 & 2 to be vertical asymptotes but what I can’t figure out is
how as x->-5- = -oo and x->-5+ = +oo

I have calculated lim x->-5+ x^3/(x^2+3x-10) = -125/(0)(-5 - -2) = -125/-0 = +oo

But I don’t see –oo lim x->+5+ x^3/(x^2+3x-10) = ???????????????

Can you show me the algebra and how the signs work out to give me –oo

Thanks

2. Jun 5, 2004

Maybe?

I think i have it ...

lim x-> -5+ x^3/(x^2+3x-10) = +oo since x^3/(x+5)(x-2) > 0 for x > -5

and lim x->-5- x^3/(x^2+3x-10) = -oo since x^3/(x+5)(x-2) < 0 for x < -5

3. Jun 5, 2004

### AKG

Didn't look at the actual math, but that's the general idea. You can nomrally if it's going to approach an infinity, then look at the signs to tell whether it would be positive or negative.

4. Jun 5, 2004

### turin

Looking at the graph is always (almost) helpful.