1. Ok

I will start a new thread here to ask this question...suppose that$\displaystyle f(x)<g(x),\forall{x}\in\mathbb{R}$ and also suppose that $\displaystyle f(x)\sim{g(x)}$ could you argue that the original inequality could be untrue?

2. Hello,

It depends on where f(x) ~ g(x)

And what does f(x) ~ g(x) mean ? It means that there exists $\displaystyle \epsilon (x)$ such as $\displaystyle f(x)=g(x)+\epsilon (x)$, where $\displaystyle \epsilon (x)$ tends to be very small as approaching the domain where f(x) ~ g(x).

This doesn't impeach f(x) to be inferior to g(x)...

3. That is not what I meant

I have always known $\displaystyle f(x)\sim{g(x)}$ to mean $\displaystyle \lim_{x \to {\infty}}\frac{f(x)}{g(x)}=1$ or $\displaystyle \lim_{x \to {\infty}}\frac{g(x)}{f(x)}=1$...or in other words they are asymptotically equivalent

4. Originally Posted by Mathstud28
I have always known $\displaystyle f(x)\sim{g(x)}$ to mean $\displaystyle \lim_{x \to {\infty}}\frac{f(x)}{g(x)}=1$ or $\displaystyle \lim_{x \to {\infty}}\frac{g(x)}{f(x)}=1$...or in other words they are asymptotically equivalent
well, $\displaystyle f(x) = g(x)$ certainly satisfies that definition, and $\displaystyle f(x) > g(x)$ for all $\displaystyle x$ could satisfy it as well. so yes, we could argue that the original inequality was false. but of course, to do that, we would need more information that you have provided.

5. I dont know if this is a good example

but what relating to the classic $\displaystyle \pi(x)$....I am pretty sure that $\displaystyle Li(x)>\frac{x}{\ln(x)} \forall{x}\in\mathbb{R}$...but assuming the limit exists that then $\displaystyle Li(x)\sim\frac{x}{\ln(x)}$...I am not sure that is the best example but Its the only one I can think of right now...so would you still agree with the original inequality?

6. Another example is Stirling approximation.