calculus - How to show that $sqrt{x}$ grows faster than $ln{x}$.

So I have the limit $$\lim_{x\rightarrow \infty}\left(\frac{1}{2-\frac{3\ln{x}}{\sqrt{x}}}\right)=\frac{1}2,$$ I now want to motivate why $(3\ln{x}/\sqrt{x})\rightarrow0$ as $x\rightarrow\infty.$ I cam up with two possibilites:

1. Algebraically it follows that $$\frac{3\ln{x}}{\sqrt{x}}=\frac{3\ln{x}}{\frac{x}{\sqrt{x}}}=\frac{3\sqrt{x}\ln{x}}{x}=3\sqrt{x}\cdot\frac{\ln{x}}{x},$$
   
   Now since the last factor is a standard limit equal to zero as $x$ approaches infinity, the limit of the entire thing should be $0$. However, isn't it a problem because $\sqrt{x}\rightarrow\infty$ as $x\rightarrow \infty$ gives us the indeterminate value $\infty\cdot 0$?




2. One can, without having to do the arithmeticabove, directly motivate that the function $f_1:x\rightarrow \sqrt{x}$ increases faster than the function $f_2:x\rightarrow\ln{x}.$ Is this motivation sufficient? And, is the proof below correct?

We have that $D(f_1)=\frac{1}{2\sqrt{x}}$ and $D(f_2)=\frac{1}{x}$. In order to compare these two derivateives, we have to look at the interval $(0,\infty).$ Since $D(f_1)\geq D(f_2)$ for $x\geq4$, it follows that $f_1>f_2, \ x>4.$

Answer

1. This is a standard result from high school


2. If you nevertheless want to deduce it from the limit of $\dfrac{\ln x}x$, use the properties of logarithm:
   $$\frac{\ln x}{\sqrt x}=\frac{2\ln(\sqrt x)}{\sqrt x}\xrightarrow[\sqrt x\to\infty]{}2\cdot 0=0$$