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Question Number 126389 by shaker last updated on 20/Dec/20

Answered by liberty last updated on 20/Dec/20

$$\underset{x\to 0}{\lim }\frac{\left[\frac{3\cos 3x}{\sin 3x}\right]}{\left[\frac{2\cos 2x}{\sin 2x}\right]}=\underset{x\to 0}{\lim }\frac{3\cos 3x}{\sin 3x}\times \frac{\sin 2x}{2\cos 2x}$$$$=\frac{3}{2}\times \underset{x\to 0}{\lim }\frac{\sin 2x}{\sin 3x}=1$$

Answered by mathmax by abdo last updated on 20/Dec/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln \left(\sin \left(3\mathrm{x}\right)\right)}{\ln \left(\sin \left(2\mathrm{x}\right)\right)}\mathrm{for}\mathrm{x}\sim 0\mathrm{we}\mathrm{have}\sin \left(3\mathrm{x}\right)\sim 3\mathrm{x}\mathrm{and}\sin \left(2\mathrm{x}\right)\sim 2\mathrm{x}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{\ln \left(3\mathrm{x}\right)}{\ln \left(2\mathrm{x}\right)}\mathrm{by}\mathrm{hospital}\mathrm{we}\mathrm{get}{\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to 0}\frac{\frac{3}{3\mathrm{x}}}{\frac{2}{2\mathrm{x}}}=1$$

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