I do not know how to finish this task because I do not see how I could transform the last solution from L-Hospital.
$$\lim _{n \rightarrow \infty} \frac{(\ln n)^{2}}{\sqrt{n}} \underset{\text{L'H}}{=} \lim _{n \rightarrow \infty} \frac{2 \ln (x) \cdot 2 \sqrt{x}}{x} \underset{\text{L'H}}{=} \lim _{n \rightarrow \infty}\frac{2 \ln (x)+4}{\sqrt{x}} = ... $$
\begin{align}\lim_{x\to\infty}\frac{\log^2x}{\sqrt x}&=\lim_{x\to\infty}\frac{\frac2x\log x}{\frac1{2\sqrt x}}\\&=4\lim_{x\to\infty}\frac{\log x}{\sqrt x}\\&=4\lim_{x\to\infty}\frac{\frac1x}{\frac1{2\sqrt x}}\\&=8\lim_{x\to\infty}\frac1{\sqrt x}\\&=0.\end{align}
It's well-known that $x > \ln x$, for $x >0$. So let $x=\sqrt[8]{n}$:
$$\sqrt[8]{n} > \ln \sqrt[8]{n}=\frac{1}{8}\ln n\Rightarrow 64\sqrt[4]{n} > (\ln n)^2$$
Therefore:
$$\frac{(\ln n)^2}{\sqrt{n}} < \frac{64\sqrt[4]{n}}{\sqrt{n}}=\frac{64}{\sqrt[4]{n}} \to 0$$
You were almost there (with a slight typo):
$$\lim _{n \rightarrow \infty} \frac{(\ln n)^{2}}{\sqrt{n}} \underset{\text{L'H}}{=} \lim _{x \rightarrow \infty} \frac{2 \ln (x) \cdot 2 \sqrt{x}}{x} = \lim _{x \rightarrow \infty} \frac{4 \ln (x)}{\sqrt{x}} \underset{\text{L'H}}{=} \lim _{x \rightarrow \infty}\frac{4\cdot 2 \sqrt{x}}{x} = \lim _{x \rightarrow \infty}\frac{8 }{\sqrt{x}} =0.$$
Set $n=e^x$, $x \rightarrow \infty$.
$\dfrac{x^2}{e^{x/2}}<\dfrac{x^2}{(1/3!)(x/2)^3}=$
$(6\cdot 8)\dfrac{x^2}{x^3}=48/x$.
Take the limit.
Used:
$y>0$;
$e^y=1+y+y^2/2!+y^3/3! +\cdots >y^3/3!.$
