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Question Number 110760 by bobhans last updated on 30/Aug/20

Answered by john santu last updated on 30/Aug/20

$$\{\begin{array}{l}x+1=m\\ m\to 0\end{array}$$$$L=\underset{m\to 0}{\lim }\frac{\sin 2m-\tan m}{m^3}$$$$L=\underset{m\to 0}{\lim }\frac{2\sin m\cos m-\tan m}{m^3}$$$$L=\underset{m\to 0}{\lim }\frac{\tan m(2\cos {}^{2}m-1)}{m^3}$$$$L=\underset{m\to 0}{\lim }\frac{2\cos {}^{2}m-1}{m^2}=\mathrm{\infty }$$

Answered by mathmax by abdo last updated on 30/Aug/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin (2\mathrm{x}+2)-\tan (\mathrm{x}+1)}{{(\mathrm{x}+1)}^3}\mathrm{changement}\mathrm{x}+1=\mathrm{t}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin \left(2\mathrm{t}\right)-\tan \left(\mathrm{t}\right)}{{\mathrm{t}}^3}=\mathrm{g}\left(\mathrm{t}\right)(\mathrm{x}\to -1\Rightarrow \mathrm{t}\to 0)$$$$\mathrm{we}\mathrm{have}\sin \left(2\mathrm{t}\right)\sim 2\mathrm{t}-\frac{{\left(2\mathrm{t}\right)}^3}{6}+\mathrm{o}\left({\mathrm{t}}^3\right)=2\mathrm{t}-\frac{4}{3}{\mathrm{t}}^3+\mathrm{o}\left({\mathrm{t}}^3\right)$$$$\mathrm{tant}\sim \mathrm{t}+\frac{{\mathrm{t}}^3}{3}+\mathrm{o}\left({\mathrm{t}}^3\right)\Rightarrow \mathrm{g}\left(\mathrm{t}\right)\sim \frac{2\mathrm{t}-\frac{4}{3}{\mathrm{t}}^3-\mathrm{t}-\frac{{\mathrm{t}}^3}{3}}{{\mathrm{t}}^3}=\frac{1}{{\mathrm{t}}^2}-\frac{5}{3}$$$$\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=+\mathrm{\infty }={\lim }_{\mathrm{x}\to -1}\mathrm{f}\left(\mathrm{x}\right)$$

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