<s>[INST] How to compute $\lim_{ x\to 0} \frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$? How to compute the following limit?
$$\lim_{ x\to 0} \frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
In the context of the book I am reading, I must use some change of variables or multiply the expressions by some "conjugate". I tried to multiply it by $\left(\sqrt{2-x}+\sqrt{x+2}\right)$ to obtain:
$$\frac{2 x}{\left(\sqrt[3]{x+2}-\sqrt[3]{2-x}\right)
   \left(\sqrt{2-x}+\sqrt{x+2}\right)}$$
And I tried to make the following change of variables $2+x =t^6$ and then I obtained:
$$\frac{t^3-\sqrt{4-t^6}}{t^2-\sqrt[3]{4-t^6}}$$
But I couldn't go much farther than this. Can you give me a hint?
 [/INST] Hint
$$\frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
$$=\frac{(\sqrt[3]{x+2} )^3-(\sqrt[3]{2-x})^3  }{  \sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
$$=\frac{(\sqrt[3]{x+2}-\sqrt[3]{2-x}) ((\sqrt[3]{x+2} )^2+(\sqrt[3]{2-x})^2+ \sqrt[3]{x+2}\sqrt[3]{2-x}) }{  \sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
$$  =(x+2)^{2/3} +(2-x)^{2/3}+ \sqrt[3]{4-x^2}   $$
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