# 1=0.

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• 08-19-2008
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 f(x) is x, which -> inf. So we'll get different limits, depending on how f and g vary.
• 08-19-2008
Thantos
Quote:

Originally Posted by tabstop
So if f(x) -> inf and g(x) -> 0, then f(x) * g(x) would represent inf * 0.

That is one way you could arrange it (and the way that directly ties into the previous question)

Quote:

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 f(x) is x, which -> inf.
I think you are using f(x) here to represent two different things. Might be better to say: h(x) = f(x) * g(x).
When f(x) = x and g(x) = 1/x then h(x) = 1
When f(x) = x^2 and g(x) = 1/x then h(x) = x and as gets becomes unbounded so does h(x)

Another example would be f(x) = 2x^2 + x + 1 and g(x) = 1/(x^2 + 10x + 1). As x becomes unbounded then h(x) approaches 2. But if f(x) = x and g(x) = 1/(2x) then the limit would be 1/2.

So as tabstop directly said, different functions will give different results but still have that inf * 0 form.
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