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Question Number 117673 by huotpat last updated on 13/Oct/20

Answered by bemath last updated on 13/Oct/20

$$\underset{x\to 0}{\lim }\frac{3\mathrm{x}-(3\mathrm{x}-\frac{{27\mathrm{x}}^3}{6})}{2\mathrm{x}-(2\mathrm{x}-\frac{{8\mathrm{x}}^3}{6})}=$$$$\underset{x\to 0}{\lim }\frac{{27\mathrm{x}}^3}{{8\mathrm{x}}^3}=\frac{27}{8}$$

Commented by MJS_new last updated on 13/Oct/20

$$\mathrm{typo}:{\mathrm{it}}^{\prime }\mathrm{s}\frac{27x^3}{8x^3}$$

Commented by bemath last updated on 13/Oct/20

$$\mathrm{yes}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

Answered by Olaf last updated on 13/Oct/20

$$\frac{3x-\sin 3x}{2x-\sin 2x}\underset{0}{\sim }\frac{3x-(3x-\frac{{\left(3x\right)}^3}{3!})}{2x-(2x-\frac{{\left(2x\right)}^3}{3!})}=\frac{27}{8}$$$$$$$$\mathrm{With}{\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{rule}:$$$$\underset{x\to 0}{\lim }\frac{3x-\sin 3x}{2x-\sin 2x}=\underset{x\to 0}{\lim }\frac{3-3\cos 3x}{2-2\cos 2x}$$$$=\underset{x\to 0}{\lim }\frac{9\sin 3x}{4\sin 2x}=\underset{x\to 0}{\lim }\frac{27\cos 3x}{8\cos 2x}=\frac{27}{8}$$

Answered by Dwaipayan Shikari last updated on 13/Oct/20

$$\underset{x\to 0}{\lim }\frac{3x-3x+\frac{27}{6}{x}^3}{2x-2x+\frac{8}{6}{x}^3}=\frac{27}{8}$$

Answered by 1549442205PVT last updated on 13/Oct/20

$$\mathrm{This}\mathrm{is}\mathrm{the}\mathrm{form}\frac{0}{0}\Rightarrow \mathrm{using}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\mathrm{rule}\mathrm{we}\mathrm{get}\mathrm{I}=\underset{\mathrm{x}\to 0}{\lim }\frac{3\mathrm{x}-\sin 3\mathrm{x}}{2\mathrm{x}-\sin 2\mathrm{x}}=$$$$\underset{\mathrm{x}\to 0}{\lim }\frac{3-3\cos 3\mathrm{x}}{2-2\cos 2\mathrm{x}}\underset{{\mathrm{L}}^{\prime }\mathrm{Hopital}}{={}^{\frac{0}{0}}}\underset{\mathrm{x}\to 0}{\lim }\frac{9\sin 3\mathrm{x}}{4\sin 2\mathrm{x}}$$$$\underset{\mathrm{x}\to 0}{\lim }\frac{27\cos 3\mathrm{x}}{8\cos 2\mathrm{x}}=\frac{27}{8}$$

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