# Math Help - 1/((x^2)(x + 7)) = INFINITY

1. ## 1/((x^2)(x + 7)) = INFINITY

lim x-->0 1/((x^2)(x + 7)) = INFINITY

but my question is:
if it says x-->0 instead of x-->0+ or x-->0-, so how do you know that both x-->0+ AND x-->0- go to a infinite y value?

Any help would be greatly appreciated!
In either case $x^2$ is positive and near $0$ while $x+7$ is near $7$.