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Question Number 196806 by cortano12 last updated on 01/Sep/23

$$\cancel{\underset{―}{\underbrace{}}}$$

Answered by dimentri last updated on 01/Sep/23

$$=\underset{x\to 0}{\lim }\frac{x^{1/6}{(2-2x)}^{1/6}-x}{2x-x\sqrt[3]{\frac{1}{1+\sqrt[3]{x^3+1}+\sqrt[3]{{(x^3+1)}^2}}}}$$$$=\underset{x\to 0}{\lim }\frac{{(2-2x)}^{1/6}-x^{5/6}}{x^{5/6}(2-\sqrt[3]{\frac{1}{1+\sqrt[3]{x^3+}+\sqrt[3]{{(x^3+1)}^2}}}}$$$$=\mathrm{\infty }$$

Commented by jabarsing last updated on 02/Sep/23

$$L^{+}\left(0\right):+\mathrm{\infty }$$$$L^{-}\left(0\right):NotExist$$

Answered by jabarsing last updated on 01/Sep/23

$$NotExist$$$$$$$$becausex\to 0^{-}\Rightarrow 2x-2x^2<0\Rightarrow \sqrt{⊝}$$

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