Upper and lower bounds for $x/\ln x$

Question

It is well known that $$ (x-1)-\frac{3}{2}\left(x-1\right)^{2}<\frac{\ln x}{x}<x-1 $$ for every $x>1$. This inequalities are good around $x=1$. I found that $$ \frac{2x}{x^2+1}<\frac{\ln x}{x} $$ for $x>8$. Are there other rational functions $f,g$ such that we get a better inequality $f(x)<\frac{x}{\ln x}<g(x)$ for large values ​​of $x$?

Answer 1

Let $\phi(x)=x/\log x$. Note the following:

1. $\phi(x)\to\infty$ as $x\to\infty$.
2. $\phi'(x)=\frac{\log x-1}{{\log(x)}^2}\to 0$ as $x\to\infty$.

For a function $f$ to be a "good" approximation of $\phi$ on all of $\mathbb R_+$, it must also obey these two criteria.

Now let $f=p/q$ be a rational approximation of $\phi$. Letting $a=\operatorname{deg}p$ and $b=\operatorname{deg}q$, there are three options:

A) $a>b$. In this case, then criterion 1 is satisfied, but not criterion 2. $f$ will grow asymptotically at a linear rate whereas $\phi$ grows at a less than linear rate. The function $|f-\phi|$ will grow arbitrarily large for large inputs.

B) $a=b$. In this case criterion 2 is satisfied, but not criterion 1. $f$ will reach a finite limit $c$, and once again the function $|f-\phi|$ will grow arbitrarily large for large inputs.

C) $a<b$. Again in this case criterion 2 is satisfied, but not criterion 1. This is essentially identical to case B, in the special case of the limit $c$ being zero.

Answer 2

No rational function will be a good approximation for $\log x$:

If $f(x)=\frac{p(x)}{g(x)}$ has degree $0$ then $\lim\limits_{x\to\infty} f(x) = c$ for some constant, while $\log x \to \infty$.

If $f$ has degree $1$, then $f(x)\sim x$ which is a very loose upper bound for $\log x$.

For others degrees the approximation will be even worse.

You will get much better bounds if you're willing to use functions that involve radicals, instead of only rational functions.