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Question Number 93258 by john santu last updated on 12/May/20

Commented by john santu last updated on 12/May/20

$$\underset{x\to 0}{\lim }\frac{3\sin \mathrm{x}-4\sin {}^3\mathrm{x}+4\sin {}^3\mathrm{x}-3\ln (1+\mathrm{x})}{({\mathrm{e}}^{\mathrm{x}}-1)\sin \mathrm{x}}$$$$\underset{x\to 0}{\lim }\frac{3\sin \mathrm{x}-3\ln (1+\mathrm{x})}{({\mathrm{e}}^{\mathrm{x}}-1)\sin \mathrm{x}}=$$$$3\underset{x\to 0}{\lim }\frac{\cos \mathrm{x}-\left(\frac{1}{1+\mathrm{x}}\right)}{{\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}+({\mathrm{e}}^{\mathrm{x}}-1)\cos \mathrm{x}}$$$$3\underset{x\to 0}{\lim }\frac{-\sin \mathrm{x}+\frac{1}{{(1+\mathrm{x})}^2}}{{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}-({\mathrm{e}}^{\mathrm{x}}-1)\sin \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}}=$$$$3\underset{x\to 0}{\lim }\frac{-\sin \mathrm{x}+\frac{1}{{(1+\mathrm{x})}^2}}{{2\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}+\sin \mathrm{x}}=\frac{3}{2}$$

Answered by niroj last updated on 12/May/20

$$\underset{\mathrm{x}\to 0}{\stackrel{\lim }{}}\frac{\sin 3\mathrm{x}+{4\sin }^3\mathrm{x}-3\ln (1+\mathrm{x})}{({\mathrm{e}}^{\mathrm{x}}-1)\sin \mathrm{x}}$$$$\mathrm{Indeterminant}\mathrm{form}\frac{0}{0}$$$$\mathrm{By}{\mathrm{L}}^{{}^{\prime }}\mathrm{hospital}\mathrm{rule},$$$$=\underset{\mathrm{x}\to 0}{\stackrel{\lim }{}}\frac{3\cos 3\mathrm{x}+{12\sin }^2\mathrm{x}.\cos \mathrm{x}-3.\frac{1}{1+\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}-\mathrm{cosx}}\mathrm{form}\frac{0}{0}$$$$=\underset{\mathrm{x}\to 0}{\stackrel{\lim }{}}\frac{-9\sin 3\mathrm{x}-{12\sin }^2\mathrm{x}.\mathrm{sinx}+{24\cos }^2\mathrm{xsinx}.+\frac{3}{{(1+\mathrm{x})}^2}}{-{\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}+\mathrm{sinx}}$$$$=\frac{3}{2}//.$$

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