Little gap in understanding an argument from Hardy's book.

Question

In the booķ 'Introduction to the Theory of Numbers' by GH Hardy it is argued that $$x^{-n}e^x > \frac {x}{(n+1)!}$$ hence as $x$ increases $$\frac {e^x}{x^n} \rightarrow \infty$$ further since $ln(x)$ is inverse of $e^x$, $\frac {ln(x)}{x^n} \rightarrow 0$ hence $ln(x)=o(x^n)$. I did not understand the last line of the argument where it is mentioned that the reason for $\frac {ln(x)}{x^n} \rightarrow 0$ is that $ln(x)$ is the inverse of $e^x$. Is there an explicit reason why this should be so?

Answer 1

Let $g(x)$ be some increasing and positive function for $x>1$

If we know that $g(x)/x^n \to \infty$ we know that for any $M>0$ there exists some $x_0>1$ st for all $x>x_0$:

$$g(x) > M x^n \tag{1}$$

Now, $g(x)$ has an inverse $h=g^{-1}$ which is also increasing and positive for $x>1$

Then we can apply the inverse to both sides of $(1)$ and get

$$x > h(M x^n)$$

or, setting $u=M x^n$

$$h(u)< \frac{1}{M^{1/n}}u^{1/n}$$

Hence $$\frac{h(x)}{x^\delta}\to 0$$ for any positive $\delta$

Note, BTW, that the book does not use the letter $n$ for exponent in the last formula, and for a good reason: while the hypothesis is that $g(x)/x^n \to \infty$ for an arbitrarily large $n$ (in particular, large integers), the "interesting" conclusion is that $\frac{h(x)}{x^\delta}\to 0$ for arbitrarily small (and real) $\delta$.