Answer :
[tex]\lim\limits_{x\to0}\frac{\sqrt{cos2x}-\sqrt[3]{cosx}}{arctanx^2}=\left[\frac{0}{0}\right]\\\\De\ L'Hospital's\ rule:\\\\\lim\limits_{x\to0}\frac{\sqrt{cos2x}-\sqrt[3]{cosx}}{arctanx^2}=\lim\limits_{x\to0}\frac{(\sqrt{cos2x}-\sqrt[3]{cosx})'}{(arctanx^2)'}=(*)\\-------------------------------\\(\sqrt{cos2x}-\sqrt[3]{cosx})'=\frac{-sin2x}{\sqrt{cos2x}}-\frac{-sinx}{3\sqrt[3]{cos^2x}}=\frac{-2sinxcosx\cdot3\sqrt[3]{cos^2x}+sinx\sqrt{cos2x}}{3\sqrt{cos2x}\cdot\sqrt[3]{cos^2x}}\\\\(arctanx^2)'=\frac{2x}{x^2+1}[/tex]
[tex]-------------------------------\\(*)=\lim\limits_{x\to0}\frac{-6sinxcosx\sqrt[3]{cos^2x}+sinx\sqrt{cos2x}}{3\sqrt[3]{cos^2x}\cdot\sqrt{cos2x}}:\frac{2x}{x^4+1}\\\\=\lim\limits_{x\to0}\frac{-sinx(6cosx\sqrt[3]{cos^2x}-\sqrt{cos2x})}{3\sqrt[3]{cos^2x}\cdot\sqrt{cos2x}}\cdot\frac{x^4+1}{2x}\\\\=\lim\limits_{x\to0}\frac{-sinx}{x}\cdot\frac{(6cosx\sqrt[3]{cos^2x}-\sqrt{cos2x})(x^4+1)}{6}=-1\cdot\frac{(6\cdot1\sqrt[3]{1^2}-\sqrt1)(0+1)}{6}\\\\=-1\cdot\frac{5}{6}=-\frac{5}{6}[/tex]
[tex]-------------------------------\\(*)=\lim\limits_{x\to0}\frac{-6sinxcosx\sqrt[3]{cos^2x}+sinx\sqrt{cos2x}}{3\sqrt[3]{cos^2x}\cdot\sqrt{cos2x}}:\frac{2x}{x^4+1}\\\\=\lim\limits_{x\to0}\frac{-sinx(6cosx\sqrt[3]{cos^2x}-\sqrt{cos2x})}{3\sqrt[3]{cos^2x}\cdot\sqrt{cos2x}}\cdot\frac{x^4+1}{2x}\\\\=\lim\limits_{x\to0}\frac{-sinx}{x}\cdot\frac{(6cosx\sqrt[3]{cos^2x}-\sqrt{cos2x})(x^4+1)}{6}=-1\cdot\frac{(6\cdot1\sqrt[3]{1^2}-\sqrt1)(0+1)}{6}\\\\=-1\cdot\frac{5}{6}=-\frac{5}{6}[/tex]
