How to compute $\lim_{ x\to 0} \frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$?

Question

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?

Answer 1

Hint: Multiply by $\sqrt[3]{(x+2)^2}+\sqrt[3]{(2-x)^2}+\sqrt[3]{(x+2)(2-x)}$ and then you are basically done.

The idea is that $A^3-B^3=(A-B)(A^2+AB+B^2)$

More in general, https://brilliant.org/wiki/factorization-of-polynomials/ (sorry if it is brilliant, it's the first I found). Very helpful trick though.

Answer 2

Hint

$$\frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$$ $$=\frac{(\sqrt[3]{x+2} )^3-(\sqrt[3]{2-x})^3 }{ \sqrt[3]{x+2}-\sqrt[3]{2-x}}$$ $$=\frac{(\sqrt[3]{x+2}-\sqrt[3]{2-x}) ((\sqrt[3]{x+2} )^2+(\sqrt[3]{2-x})^2+ \sqrt[3]{x+2}\sqrt[3]{2-x}) }{ \sqrt[3]{x+2}-\sqrt[3]{2-x}}$$ $$ =(x+2)^{2/3} +(2-x)^{2/3}+ \sqrt[3]{4-x^2} $$

Answer 3

Use L'Hôpital's rule.
With

$$f(x){\mapsto }\sqrt{x+2}-\sqrt{2-x}$$

$$g(x){\mapsto }\left(x+2\right)^{\frac{1}{3}}-\left(2-x\right)^{\frac{1}{3}}$$

we compute the limit of the quotient of the first derivatives

$$\underset{x\to 0}{\text{lim}}\frac{f'(x)}{g'(x)}=\underset{x\to 0}{\lim }\, \frac{\frac{1}{2 \sqrt{x+2}}+\frac{1}{2 \sqrt{2-x}}}{\frac{1}{3 (x+2)^{2/3}}+\frac{1}{3 (2-x)^{2/3}}}=\frac{3 \sqrt[6]{2}}{2}$$