Does the Limit Hold for f'(x)>g'(x) Even if f(x)/g(x) is not a Constant?

[archaic]

Would this $f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0$ hold if $\frac{f(x)}{g(x)}\neq c$?

 

[member 587159]

It is unclear to me what you mean. Can you be a little more careful describe what you mean? You have an $\implies$ and then an 'if' after the implies.

What is $a$ ? What is $c$? Where are $f,g$ defined?

 

[RPinPA]

Would this $f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0$ hold if $\frac{f(x)}{g(x)}\neq c$?

This is a little cryptic. Here's what I think you are saying:

"There is a theorem that if for some $a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)$ and $f,\,g\underset{\infty}{\to}0$ (Does this mean $\lim_{x\to\infty}f, g = 0$?) then $\lim_{x\to\infty}f(x)/g(x)=0$. Does the theorem hold if $\frac{f(x)}{g(x)}\neq c$?"

Or are you asking if there is such a theorem?

 

[WWGD]

If I understood correctly, $1/x ,1/(x+1)$ are counters.

 

[archaic]

It is unclear to me what you mean. Can you be a little more careful describe what you mean? You have an $\implies$ and then an 'if' after the implies.

What is $a$ ? What is $c$? Where are $f,g$ defined?

This is a little cryptic. Here's what I think you are saying:

"There is a theorem that if for some $a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)$ and $f,\,g\underset{\infty}{\to}0$ (Does this mean $\lim_{x\to\infty}f, g = 0$?) then $\lim_{x\to\infty}f(x)/g(x)=0$. Does the theorem hold if $\frac{f(x)}{g(x)}\neq c$?"

Or are you asking if there is such a theorem?

What I want to say is, if there're two functions $f(x)$ and $g(x)$ who are going to $0$ as $x\to\infty$, and such that after a certain $x\,(=a)$ we have $f'(x)>g'(x)$ then $\frac{f(x)}{g(x)} \underset{\infty}{\to} 0$ provided that $f(x)\neq cg(x)$ where $c\in\mathbb{R}$, elsewise the limit would equal $c$.
I realized some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.
However, since what I want to say is if $f$ goes to $0$ faster than $g$ then ..., the correct way to translate this in math is $|f'(x)|<|g'(x)|$ (provided that both $f$ and $g$ are going to $0$).

 

[archaic]

If I understood correctly, $1/x ,1/(x+1)$ are counters.

No, they don't come under this since they have the same rate of change.

 

[WWGD]

No, they don't come under this since they have the same rate of change.

Respective derivatives are $-1/x^2 >-1/(x+1)^2$

 

[archaic]

Respective derivatives are $-1/x^2 >-1/(x+1)^2$

Right sorry. I think, then, that I can say something about the convergence of the limit given the assumptions? And then it follows that it is $0$ if the functions are not equivalent at $\infty$?

 

[PeroK]

What I want to say is, if there're two functions $f(x)$ and $g(x)$ who are going to $0$ as $x\to\infty$, and such that after a certain $x\,(=a)$ we have $f'(x)>g'(x)$ then $\frac{f(x)}{g(x)} \underset{\infty}{\to} 0$ provided that $f(x)\neq cg(x)$ where $c\in\mathbb{R}$, elsewise the limit would equal $c$.
I realized some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.
However, since what I want to say is if $f$ goes to $0$ faster than $g$ then ..., the correct way to translate this in math is $|f'(x)|<|g'(x)|$ (provided that both $f$ and $g$ are going to $0$).

I can't see any reason why this would be true.

 

[archaic]

I can't see any reason why this would be true.

$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$
By $10^{k}$ I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use $O(10^{-n})$? Not sure, I need to study big-O notation.

 

[PeroK]

$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$
By $10^{k}$ I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use $O(10^{-n})$? Not sure, I need to study big-O notation.

I can't see the relationship between that and what you posted before.

 

[archaic]

I can't see the relationship between that and what you posted before.

Well, if $|f'(x)|<|g'(x)|$ and they are both going to $0$, it means that $f$ is going to $0$ faster, and so the limit should be $0$.

 

[PeroK]

Well, if $|f'(x)|<|g'(x)|$ and they are both going to $0$, it means that $f$ is going to $0$ faster, and so the limit should be $0$.

That relationship means that $g(x)$ is changing more quickly than $f(x)$.

You've got that the wrong way round.

 

[PeroK]

Well, if $|f'(x)|<|g'(x)|$ and they are both going to $0$, it means that $f$ is going to $0$ faster, and so the limit should be $0$.

In any case, you already have the counterexample: $f(x) = \frac 1 2 g(x)$.

Even if you exclude that precise counterexample, you only need two functions where $f(x) \approx \frac 1 2 g(x)$. For example:

$f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)$

 

[archaic]

That relationship means that $g(x)$ is changing more quickly than $f(x)$.

You've got that the wrong way round.

If $f$ is going faster to $0$ than $g$, then $|df|$ would be less than $|dg|$, because it is already "at zero" so there isn't much to change.

In any case, you already have the counterexample: $f(x) = \frac 1 2 g(x)$.

Even if you exclude that precise counterexample, you only need two functions where $f(x) \approx \frac 1 2 g(x)$. For example:

$f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)$

Right, see post #8.
through the different counter-examples given, it is clear that this is not true. however, given the premises, it seems that the limit always converges (maybe), and in the special case that $f\underset{\infty}{\not\sim}g$, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.

 

[PeroK]

however, given the premises, it seems that the limit always converges (maybe), and in the special case that $f\underset{\infty}{\not\sim}g$, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.

Not that either. Just amend my last function:

$f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)$

Then the limit $f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}$ is undefined.

 

[archaic]

Not that either. Just amend my last function:

$f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)$

Then the limit $f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}$ is undefined.

wops

 

[archaic]

Not that either. Just amend my last function:

$f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)$

Then the limit $f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}$ is undefined.

It doesn't satisfy $|f'(x)|<|g'(x)|$ so it doesn't come under this.

 

[PeroK]

wops

Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.

The only one we don't have is where $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty$.

Any ideas?

 

[PeroK]

It doesn't satisfy $|f'(x)|<|g'(x)|$ so it doesn't come under this.

With $g(x) = e^{-x}$ it all works out.

 

[archaic]

Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.

The only one we don't have is where $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty$.

Any ideas?

you mean given the premises?

 

[PeroK]

you mean given the premises?

Yes.

 

[PeroK]

you mean given the premises?

Maybe this is the result that could be proved? If $|f'(x)| < |g'(x)|$, then $f(x)$ is changing less than $g(x)$, so $f(x)/g(x)$ cannot become arbitrarily large. So:

$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty$?

 

[archaic]

Maybe this is the result that could be proved? If $|f'(x)| < |g'(x)|$, then $f(x)$ is changing less than $g(x)$, so $f(x)/g(x)$ cannot become arbitrarily large. So:

$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty$?

I thought you meant that there was an example where that limit goes to infinity...
But yes, it seems reasonable that given those premises the limit can't be arbitrarily large.

 

[PeroK]

I thought you meant that there was an example where that limit goes to infinity...

... looking for counterexamples is good way to see what might be true.

 

[archaic]

... looking for counterexamples is good way to see what might be true.

Hello again PeroK. Another way to put $|f'(x)|<|g'(x)|$ would be $|f(x)|'>|g(x)|'$ (for oscillating functions only in intervals where they are both positive, but the last step (the squaring) fixes this). Taking that into account, we have
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\lim_{x\to\infty}\sqrt{\frac{(f'(x))^2}{(g'(x))^2}}=\lim_{x\to\infty}\sqrt{\frac{(|f(x)|')^2}{(|g(x)|')^2}}$$
But since $0>|f(x)|'>|g(x)|'$ we have $0<(|f(x)|')^2<(|g(x)|')^2$, and so the limit cannot go to infinity as the quotient is always less than $1$.

EDIT: I was wrong in using l'Hôpital's rule. I just read the wikipedia page and the limit of the quotient of the derivatives must exist.