1. ## Orders of iinfinity

This is correct isn't it?

$\forall{n}\in\mathbb{Z^{+}}\text{ }\lim_{x\to\infty}\frac{\ln(x)}{x^n}=0\Rightarrow{ \ln(x)\prec{x^n}}$

If not could someone point me in the right drection?

Also, does it suffice to say $\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$ the justification being $f(x)\prec{g(x)}$

Or would that be using circular logic?

2. Originally Posted by Mathstud28
This is correct isn't it?

$\forall{n}\in\mathbb{Z^{+}}\text{ }\lim_{x\to\infty}\frac{\ln(x)}{x^n}=0\Rightarrow{ \ln(x)\prec{x^n}}$

If not could someone point me in the right drection?

Also, does it suffice to say $\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$ the justification being $f(x)\prec{g(x)}$

Or would that be using circular logic?
You appear to be using the notation from Hardy's "Orders of Infinity". Not the most common from of representing asymtotics these days.

But both are correct since this is the definition of the meaning of the relation

$f\prec{g}$,

this is (ii) on page 1 of "Orders of Infinity".

The more common notation for these things is the Landau-like notation shown in the attachment which comes from the Wikipedia article on Big-O notation

RonL

3. Originally Posted by CaptainBlack
You appear to be using the notation from Hardy's "Orders of Infinity". Not the most common from of representing asymtotics these days.

But both are correct since this is the definition of the meaning of the relation

$f\prec{g}$,

this is (ii) on page 1 of "Orders of Infinity"

RonL
Thanks Captain Black! I was just wondering though, I do not know this Hardy book you speak of, like Hardy as in relation with Ramajaun Hardy? I am using a Dover book called Advanced Calculus by David Widder

But I see you say both are correct, but it seems strange that you can justify a limit by order of infinity

For example

$lim_{x\to\infty}\frac{\ln(\ln(x))}{\ln(x)}=0$

How can we say this is true by the fact that $\ln(\ln(x))\prec\ln(x)$ when this is the defintion of orders of infinity?

It would be like showing

$\frac{d}{dx}\bigg[x^n\bigg]=nx^{n-1}$

But showing it by the difference quotient, and then using L'hopital's, but we cannot do this for we would use the conclusion of our problem in the problem before proving it

Any clarification would be greatly appreciated

4. Originally Posted by Mathstud28
Thanks Captain Black! I was just wondering though, I do not know this Hardy book you speak of, like Hardy as in relation with Ramajaun Hardy? I am using a Dover book called Advanced Calculus by David Widder
Yes that Hardy. The G H Hardy book is almost unobtainable at a sensible price, depending on the edition it goes fro between US$100 and US$500. But I do have a copy as I am trying to collect Hardy's works.

But I see you say both are correct, but it seems strange that you can justify a limit by order of infinity

For example

$lim_{x\to\infty}\frac{\ln(\ln(x))}{\ln(x)}=0$

How can we say this is true by the fact that $\ln(\ln(x))\prec\ln(x)$ when this is the defintion of orders of infinity?

It would be like showing

$\frac{d}{dx}\bigg[x^n\bigg]=nx^{n-1}$

But showing it by the difference quotient, and then using L'hopital's, but we cannot do this for we would use the conclusion of our problem in the problem before proving it

Any clarification would be greatly appreciated
If you knew that $f\prec g$ you know the limit form is true, and vice versa, which is why they are equivalent statements. You could arrive at $f\prec g$ by a derivation by other means than the limit, so could deduce the limit from this. It only looks peculiar because you have not seen examples of the argument going both ways.

For instance as we know $\ln(x)\prec x$ we can deduce that $\ln(\ln(x)\prec \ln(x)$ (we actualy need a bit more knowlege about the properties of $\prec$ and monotonic increasing functions, but lets just suppose we have that knowlege for now), and so:

$\lim_{x\to\infty}\frac{\ln(\ln(x))}{\ln(x)}=0$

RonL

5. Originally Posted by CaptainBlack
Yes that Hardy. The G H Hardy book is almost unobtainable at a sensible price, depending on the edition it goes fro between US$100 and US$500. But I do have a copy as I am trying to collect Hardy's works.

If you knew that $f\prec g$ you know the limit form is true, and vice versa, which is why they are equivalent statements. You could arrive at $f\prec g$ by a derivation by other means than the limit, so could deduce the limit from this. It only looks peculiar because you have not seen examples of the argument going both ways.

For instance as we know $\ln(x)\prec x$ we can deduce that $\ln(\ln(x)\prec \ln(x)$ (we actualy need a bit more knowlege about the properties of $\prec$ and monotonic increasing functions, but lets just suppose we have that knowlege for now), and so:

$\lim_{x\to\infty}\frac{\ln(\ln(x))}{\ln(x)}=0$

RonL
So from your workings I can imply that

if $f(x)\prec{g(x)}$

Then obvsiously

$u(f(x))\prec{u(g(x))}$

So it has an equality like tendency?

Everything is telling me this is correct, but I should probably verify

6. Originally Posted by Mathstud28
So from your workings I can imply that

if $f(x)\prec{g(x)}$

Then obvsiously

$u(f(x))\prec{u(g(x))}$

So it has an equality like tendency?

Everything is telling me this is correct, but I should probably verify

What I was using was:

if $h(x)\prec g(x)$, and $u(x)$ is monotonic increasing and $\lim_{x \to \infty}u(x)=\infty$ (and we probably only need the latter condition on $u(x)$ ), then:

$h(u(x))\prec g(u(x))$

Then if $h(x)=\ln(x),\ g(x)=x$ and $u(x)=\ln(x)$ we have:

$\ln(x)\prec x$ implies $\ln(\ln(x))\prec \ln(x)$.

(what you label as obviously true, may be true, but it is not obviously true to my eye, and I have not checked it)

RonL