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Lebombo
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Learning about the Limit Comparison Test for Improper Integrals. I haven't gotten to any applications or actual problems yet. Just learning the theory so far, and have a question on the very beginning of it.
f(x) ~ g(x) as x→a, then [itex]\frac{f(x)}{g(x)} = 1[/itex]
(that is, f(x) is asymptotic to g(x) as x→a, then [itex]\frac{f(x)}{g(x)} = 1[/itex])
so,
sin(x) ~ x as x→0, because of the fact [itex]\frac{sin(x)}{x}=1[/itex]
My question here is what order are the assumptions or logic being made?
Is the idea to say (A):
"Well, since we know the limit [itex]\frac{sin(x)}{x}=1[/itex], then we can conclude sin(x)~x"
or is the idea to say (B):
"Well, since we know sin(x)~x as x→0, then we can conclude the limit [itex]\frac{sin(x)}{x}=1[/itex]"
The reason it doesn't make sense is because if x→0, then sin(x)=1 and x=0. So it is not possible to say [itex]\frac{1}{0}=1[/itex]
Yet, if we first prove that [itex]\frac{sin(x)}{x}=1[/itex], then we can say that sin(x)~x.
I am having trouble reconciling the fact that sin(x) is said to "behave" like x as x→0 while f(0) = sin(0) is a completely different number than g(0)= x.
When you look at f(x) and g(x) separately first, it seems it can't be concluded that f(x)~g(x), but when we look first at [itex]\frac{f(x)}{g(x)}[/itex], then since their ratio is 1, we can conclude that f(x)~g(x), even though they are completely different functions when the numerator and denominator are evaluated separately.
Homework Statement
f(x) ~ g(x) as x→a, then [itex]\frac{f(x)}{g(x)} = 1[/itex]
(that is, f(x) is asymptotic to g(x) as x→a, then [itex]\frac{f(x)}{g(x)} = 1[/itex])
so,
sin(x) ~ x as x→0, because of the fact [itex]\frac{sin(x)}{x}=1[/itex]
Homework Equations
My question here is what order are the assumptions or logic being made?
Is the idea to say (A):
"Well, since we know the limit [itex]\frac{sin(x)}{x}=1[/itex], then we can conclude sin(x)~x"
or is the idea to say (B):
"Well, since we know sin(x)~x as x→0, then we can conclude the limit [itex]\frac{sin(x)}{x}=1[/itex]"
The Attempt at a Solution
The reason it doesn't make sense is because if x→0, then sin(x)=1 and x=0. So it is not possible to say [itex]\frac{1}{0}=1[/itex]
Yet, if we first prove that [itex]\frac{sin(x)}{x}=1[/itex], then we can say that sin(x)~x.
I am having trouble reconciling the fact that sin(x) is said to "behave" like x as x→0 while f(0) = sin(0) is a completely different number than g(0)= x.
When you look at f(x) and g(x) separately first, it seems it can't be concluded that f(x)~g(x), but when we look first at [itex]\frac{f(x)}{g(x)}[/itex], then since their ratio is 1, we can conclude that f(x)~g(x), even though they are completely different functions when the numerator and denominator are evaluated separately.