LCT Limit Comparison Test for Improper Integrals

[Lebombo]

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.

Homework Statement

f(x) ~ g(x) as x→a, then $\frac{f(x)}{g(x)} = 1$
(that is, f(x) is asymptotic to g(x) as x→a, then $\frac{f(x)}{g(x)} = 1$)

so,

sin(x) ~ x as x→0, because of the fact $\frac{sin(x)}{x}=1$

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 $\frac{sin(x)}{x}=1$, 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 $\frac{sin(x)}{x}=1$"

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 $\frac{1}{0}=1$

Yet, if we first prove that $\frac{sin(x)}{x}=1$, 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 $\frac{f(x)}{g(x)}$, 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.

 

[Mark44]

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.

Homework Statement

f(x) ~ g(x) as x→a, then $\frac{f(x)}{g(x)} = 1$
(that is, f(x) is asymptotic to g(x) as x→a, then $\frac{f(x)}{g(x)} = 1$)

No, $\lim_{x \to a} \frac{f(x)}{g(x)} = 1$
The operative concept there is that the limit is 1. It does not say that f(x)/g(x) = 1.

so,

sin(x) ~ x as x→0, because of the fact $\frac{sin(x)}{x}=1$

sin(x)/x ≠ 1, but $\lim_{x \to 0} \frac{sin(x)}{x} = 1$

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 $\frac{sin(x)}{x}=1$, 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 $\frac{sin(x)}{x}=1$"

I would say the latter.

The Attempt at a Solution

The reason it doesn't make sense is because if x→0, then sin(x)=1 and x=0.

No. sin(0) = 0.

So it is not possible to say $\frac{1}{0}=1$

Yet, if we first prove that $\frac{sin(x)}{x}=1$, 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.

sin(0) = 0

If you look at the graphs of y = sin(x) and y = x close to x = 0, both graphs nearly coincide.

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 $\frac{f(x)}{g(x)}$, 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.

 

[Lebombo]

Oh wow, its been so long since looking at trig or calculus that I was treating the x variable in sin(x) as an x-axis point on the Cartesian instead of the angle. When the "x" or adjacent side of the triangle is 0 on the unit circle, then the ratio of opposite/hypotenuse is of course 1. That was the mistake I made in the latter part of the post.

As for the beginning of the post, thank you for also correcting the theorem. I'm trying to learn it from a video of a university lecture and for many parts of the video, the camera man seems to scan past what the professor is writing on the board, so it takes a lot of rewind and replay to see and absorb it all.

Best Regards,
Lebombo