<s>[INST] 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:


*

*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 $\infty\cdot 0$?

*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 $f_1>f_2, \ x>4.$
 [/INST] For any $a > c > 0, x > 1$,
we have
$\begin{array}\\
\ln(x)
&=\int_1^x \frac{dt}{t}\\
&\lt\int_1^x \frac{dt}{t^{1-c}}\\
&=\int_1^x t^{c-1}dt\\
&=\dfrac{t^c}{c}\big|_1^x\\
&=\dfrac{x^c-1}{c}\\
&<\dfrac{x^c}{c}\\
\text{so}\\
\dfrac{\ln(x)}{x^a}
&<\dfrac1{x^a}\dfrac{x^c}{c}\\
&=\dfrac{x^{c-a}}{c}\\
&\to 0
\qquad\text{as } x \to \infty
\text{ since } a > c\\
\end{array}
$
For your case,
where $a = \frac12$,
we can choose
$c = \frac14$
to get
$\dfrac{\ln(x)}{x^{1/2}}
\lt\dfrac{x^{\frac14-\frac12}}{\frac14}
=4x^{-\frac14}
\to 0
$.
In general
we can choose
$c = a/2$
to get
$\dfrac{\ln(x)}{x^{a}}
\lt\dfrac{x^{a/2-a}}{a/2}
=\dfrac{2}{ax^{a/2}}
\to 0
$.
</s>