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Question Number 195619 by cortano12 last updated on 06/Aug/23

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

Answered by MM42 last updated on 06/Aug/23

$$\mathrm{\infty }$$

Answered by dimentri last updated on 06/Aug/23

$$\underset{x\to 0^{-}}{\lim }\left(\frac{x+2-2\sqrt[3]{1+\sin x}\cos x}{x^2}\right)=-\mathrm{\infty }$$$$\underset{x\to 0^{+}}{\lim }\left(\frac{x+2-2\sqrt[3]{1+\sin x}\cos x}{x^2}\right)=\mathrm{\infty }$$$$\mathrm{so}\underset{x\to 0}{\lim }\left(\frac{\mathrm{x}+2-2\sqrt[3]{1+\sin \mathrm{x}}\cos \mathrm{x}}{{\mathrm{x}}^2}\right)$$$$\mathrm{doesnot}\mathrm{exist}$$

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