$\underset{x\to 0}{\lim }\frac{x}{\sqrt[3]{x^2+3x+8}-\sqrt[3]{x^2-5x+8}}=$

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\mathrm{lt}}\frac{\mathrm{x}}{\sqrt[3]{{\mathrm{x}}^2+3\mathrm{x}+8} - \sqrt[3]{{\mathrm{x}}^2-5\mathrm{x}+8}} \\ =\underset{\mathrm{x}\to 0}{\mathrm{lt}}\frac{\mathrm{x}}{{({\mathrm{x}}^2+3\mathrm{x}+8)}^{1/3} - {({\mathrm{x}}^2-5\mathrm{x}+8)}^{1/3}} \left(\frac{0}{0}\mathrm{form}\right) \\ =\underset{\mathrm{x}\to 0}{\mathrm{lt}}\frac{ 1 }{{\displaystyle \frac{1}{3}} {({\mathrm{x}}^2+3\mathrm{x}+8)}^{-2/3} (2\mathrm{x}+3)-{\displaystyle \frac{1}{3}}{({\mathrm{x}}^2-5\mathrm{x}+8)}^{-2/3}.(2\mathrm{x}-5)} (\mathrm{By} \mathrm{L}\text{'}\mathrm{Hospital}) \\ =\frac{1}{{\displaystyle \frac{1}{3}}{\left(8\right)}^{-2/3}\left(3\right)-{\displaystyle \frac{1}{3}}{\left(8\right)}^{-2/3}(-5)} \\ =\frac{1}{{\displaystyle \frac{1}{3}} {\displaystyle \frac{1}{4}}(3+5)} \\ =\frac{12}{8}=\frac{3}{2} \end{gathered}$$