4. Calculate $\lim _{x \rightarrow 1} \frac{\sqrt{\operatorname{arctg} x}-\sqrt{\arccos \left(\frac{\sqrt{2}}{2} x\right)}}{\sqrt[3]{1+\ln [x(2 x-1)]}-\sqrt[3]{1+\ln \left(x^{3}-4 x+4\right)}}$.

Iulian Danielescu, Brăila
Solution. By amplifying with the conjugate, we obtain

$$
\begin{gathered}
\lim _{x \rightarrow 1}\left[\frac{\operatorname{arctg} x-\arccos \frac{\sqrt{2}}{2} x}{\sqrt{\operatorname{arctg} x}+\sqrt{\arccos \frac{\sqrt{2}}{2} x}} \cdot\right. \\
\left.\frac{\sqrt[3]{1+\ln [x(2 x-1)]^{2}}+\sqrt[3]{1+\ln [x(2 x-1)]} \cdot \sqrt[3]{1+\ln \left(x^{3}-4 x+4\right)}+\sqrt[3]{1+\ln \left(x^{3}-4 x+4\right)}}{1+\ln [x(2 x-1)]-1-\ln \left(x^{3}-4 x+4\right)}\right]= \\
=\frac{3}{2 \sqrt{\frac{\pi}{4}}} \cdot \lim _{x \rightarrow 1}^{2}\left[\frac{\operatorname{arctg} x-\arccos \frac{\sqrt{2}}{2} x}{\sin \left(\operatorname{arctg} x-\arccos \frac{\sqrt{2}}{2} x\right)}\right.
\end{gathered}
$$

$$
\begin{gathered}
\left.\frac{\sin (\operatorname{arctg} x) \cos \left(\arccos \frac{\sqrt{2}}{2} x\right)-\cos (\operatorname{arctg} x) \sin \left(\arccos \frac{\sqrt{2}}{2} x\right)}{\ln \frac{2 x^{2}-x}{x^{3}-4 x+4}}\right]= \\
=\frac{3}{\sqrt{\pi}} \cdot \lim _{x \rightarrow 1} \frac{\frac{x}{\sqrt{1+x^{2}}} \cdot \frac{\sqrt{2}}{2} x-\frac{1}{\sqrt{1+x^{2}}} \cdot \sqrt{1-\frac{2 x^{2}}{4}}}{\frac{\ln \left(1+\frac{2 x^{2}-x-x^{3}+4 x-4}{x^{3}-4 x+4}\right)}{\frac{-x^{3}+2 x^{2}+3 x-4}{x^{3}-4 x+4}}=\frac{-x^{3}+2 x^{2}+3 x-4}{x^{3}-4 x+4}}= \\
=\frac{3}{\sqrt{\pi}} \cdot \lim _{x \rightarrow 1} \frac{1}{\sqrt{1+x^{2}}}\left(\frac{\sqrt{2}}{2} x^{2}-\sqrt{1-\frac{2 x^{2}}{4}}\right) \\
=\frac{3}{2 \sqrt{\pi}} \cdot \lim _{x \rightarrow 1} \frac{\frac{2 x^{2}}{4}-1+\frac{2 x^{2}}{4}}{(x-1)\left(-x^{2}+x+4\right)}=\frac{3}{8 \sqrt{\pi}} \cdot \lim _{x \rightarrow 1} \frac{x^{4}+x^{2}-2}{2(x-1)}= \\
=\frac{3}{16 \sqrt{\pi}} \cdot \lim _{x \rightarrow 1} \frac{(x-1)\left(x^{3}+x^{2}+2 x+2\right)}{x-1}=\frac{3 \cdot 6}{16 \sqrt{\pi}}=\frac{9}{8 \sqrt{\pi}}
\end{gathered}
$$