5.14. Calculate the limit

$$
\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\operatorname{arctan} 4 x}-\sqrt[3]{1-\operatorname{arctan} 4 x}}{\sqrt{1-\arcsin 3 x}-\sqrt{1+\operatorname{arctan} 3 x}}
$$
Solution. Here we have an indeterminate form of type "0/0". To resolve it, we replace the functions under the limit sign with their equivalent functions.

$$
A=\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\operatorname{arctg} 4 x}-\sqrt[3]{1-\operatorname{arctg} 4 x}}{\sqrt{1-\arcsin 3 x}-\sqrt{1+\operatorname{arctg} 3 x}}=\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+4 x}-\sqrt[3]{1-4 x}}{\sqrt{1-3 x}-\sqrt{1+3 x}}
$$

We eliminate the indeterminacy by multiplying and dividing by the corresponding conjugate radicals. We now have

$$
\begin{gathered}
\lim _{x \rightarrow 0}\left(\frac{(\sqrt[3]{1+4 x}-\sqrt[3]{1-4 x})\left(\sqrt[3]{(1+4 x)^{2}}+\sqrt[3]{1+4 x} \sqrt[3]{1-4 x}+\sqrt[3]{(1-4 x)^{2}}\right)}{\left(\sqrt[3]{(1+4 x)^{2}}+\sqrt[3]{1+4 x} \sqrt[3]{1-4 x}+\sqrt[3]{(1-4 x)^{2}}\right)(\sqrt{1-3 x}+\sqrt{1+3 x})} \times\right. \\
\left.\times \frac{(\sqrt{1-3 x}+\sqrt{1+3 x})}{(\sqrt{1-3 x}-\sqrt{1+3 x}}\right)= \\
=\lim _{x \rightarrow 0} \frac{(1+4 x-1+4 x)(\sqrt{1-3 x}+\sqrt{1+3 x})}{\left(\sqrt[3]{(1+4 x)^{2}}+\sqrt[3]{1+4 x} \sqrt[3]{1-4 x}+\sqrt[3]{(1-4 x)^{2}}\right)(1-3 x-1-3 x)}= \\
=\lim _{x \rightarrow 0} \frac{8 x(\sqrt{1-3 x}+\sqrt{1+3 x})}{\left.-\sqrt[3]{(1+4 x)^{2}}+\sqrt[3]{1+4 x} \sqrt[3]{1-4 x}+\sqrt[3]{(1-4 x)^{2}}\right)}= \\
=\frac{-4}{3} \cdot \frac{1+1}{1+1+1}=-\frac{8}{9}
\end{gathered}
$$