So if f(x) -> inf and g(x) -> 0, then f(x) * g(x) would represent inf * 0. So if f(x) = x, and g(x) = 1/x as x gets large, then f(x) is always 1. If f(x) = x^2 and g(x) = 1/x as x gets large, then...
Type: Posts; User: tabstop
So if f(x) -> inf and g(x) -> 0, then f(x) * g(x) would represent inf * 0. So if f(x) = x, and g(x) = 1/x as x gets large, then f(x) is always 1. If f(x) = x^2 and g(x) = 1/x as x gets large, then...
And since when did canceling in multiplication give zero?
Because infinity * 0 = infinity / (1/0) = infinity / infinity, which I think we already did.