Question Number 170824 by 0731619 last updated on 31/May/22
Answered by floor(10²Eta[1]) last updated on 01/Jun/22
$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{a}}\sqrt[{\mathrm{3}}]{\mathrm{a}−\mathrm{x}}−\sqrt{\mathrm{a}−\mathrm{x}}\sqrt{\mathrm{a}+\mathrm{x}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}−\mathrm{x}}\sqrt[{\mathrm{3}}]{\mathrm{a}+\mathrm{x}}+\mid\mathrm{a}\mid\sqrt{\mathrm{a}−\mathrm{x}}}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}+\mathrm{x}}+\mid\mathrm{a}\mid\sqrt[{\mathrm{6}}]{\mathrm{a}−\mathrm{x}}}−\frac{\sqrt{\mathrm{a}+\mathrm{x}}}{\:\sqrt[{\mathrm{6}}]{\mathrm{a}−\mathrm{x}}\sqrt[{\mathrm{3}}]{\mathrm{a}+\mathrm{x}}+\mid\mathrm{a}\mid} \\ $$$$\Rightarrow\mathrm{lim}\left(…\right)=\frac{\sqrt[{\mathrm{3}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2a}}}−\frac{\sqrt{\mathrm{2a}}}{\mid\mathrm{a}\mid}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{a}}} \\ $$