Infinitesimals' rates of approaching 0

[feynman1]

When comparing 2 infinitesimals, does the higher order one approach 0 faster or slower?

 

[anuttarasammyak]

Why don't you take the ratio of the two to see it ?

 

[feynman1]

Why don't you take the ratio of the two to see it ?

I know how to do the maths, here I'm asking just about the statement.

 

[anuttarasammyak]

$x^3$ goes to zero faster than $x^2$, $|x^3|<|x^2|$, when x of -1< x < 1 approaches to zero.

 

[feynman1]

$x^3$ goes to zero faster than $x^2$, $|x^3|<|x^2|$, when x of -1< x < 1 approaches to zero.

how did you define speed/fast

 

[anuttarasammyak]

I am sorry to say I do not have confidence on my rigorous use of mathematics and English how to express $|x^3|<|x^2|$ for $|x|<1$ which approaches zero or is numerical sequence $x_n$ such that
$$\forall \epsilon>0\ \exists N \ \ N<n\ \ |x_n|<\epsilon$$

 

[Office_Shredder]

how did you define speed/fast

Usually speed is defined by looking at a ratio or just comparing magnitudes If $f(0)=g(0)=0$ and f and g are continuous at 0, we could say $g(x)$ goes to 0 faster than $f(x)$ if $|g(\epsilon)| < |f(\epsilon)|$ for all sufficiently small $\epsilon$. Sometimes you want to break ties a bit blunter - e.g. you might want to say that $\sin(x)$ and $x$ go to 0 at the same speed, so you would want $|g(\epsilon)/f(\epsilon)| < 1$ to say $g$ goes to zero faster. Sometimes you really want things like $x$ and $2x$ to count as going to 0 at the same speed, in which case you might require the ratio to go to 0 as $\epsilon$ goes to 0. It really depends on the context and why you are trying to pick one thing going to zero faster than the other.

I would say the most common context is you have several terms you're adding together and you want to ignore one entirely, in which case you would probably say $g$ goes to 0 faster than $f$ if $\lim_{x\to 0} g(x)/f(x)=0$