Question: I want to Show the following:
$\ln(x) = o (e^{\sqrt{\ln x}})$ (Little-o Notation).
So I need to Show that: $\displaystyle{\lim\_{x \rightarrow \infty} \frac{\ln(x)}{e^{\sqrt{\ln x}}}=0}.$

Can you help me, please?

Answer: 
As $r$ and $s$ are rationals, for every rational $t$, $T(t)=r+(s-r)t$ is a rational, right ?

Now if $t$ is irrational, $T(t)=r+(s-r)t$ can't be rational. Otherwise, 
$$t=\frac{T(t)-r}{s-r}$$
would be a rational too, which is excluded.

Hence, $T$ maps rationals to rationals, and irrationals to irrationals.
Let $d \in \mathbb{Q}$ and $d>0$.

We claim that if $r \in \mathbb{Q}$ and $t \in \mathbb{Q}$, then $r+dt \in \mathbb{Q}$, this is due to rational numbers are closed under multiplication and addition.

Also, $r \in \mathbb{Q}$ and $t \notin \mathbb{Q}$, then $r+dt \notin \mathbb{Q}$, suppose on the contrary that $x=r+dt \in \mathbb{Q}$, then $t=\frac{x-r}{d} \in \mathbb{Q}$ since rational numbers are closed under subtraction and remains a rational number upon dividing it by a rational number. However, this contradicts with the assumption that $t \notin \mathbb{Q}$.