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Question Number 131836 by liberty last updated on 09/Feb/21

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}(2+\cos \mathrm{x})-3\sin \mathrm{x}}{{\mathrm{x}}^5}$$

Commented by EDWIN88 last updated on 10/Feb/21

$$\mathrm{another}\mathrm{way}{\mathrm{L}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}$$$$\underset{x\to 0}{\lim }\frac{2\mathrm{x}+\mathrm{xcos}\mathrm{x}-3\sin \mathrm{x}}{{\mathrm{x}}^5}=$$$$\underset{x\to 0}{\lim }\frac{2+\cos \mathrm{x}-\mathrm{xsin}\mathrm{x}-3\cos \mathrm{x}}{{5\mathrm{x}}^4}=\underset{x\to 0}{\lim }\frac{2-2\cos \mathrm{x}-\mathrm{xsin}\mathrm{x}}{{5\mathrm{x}}^4}$$$$\underset{x\to 0}{\lim }\frac{2\sin \mathrm{x}-(\sin \mathrm{x}+\mathrm{xcos}\mathrm{x})}{{20\mathrm{x}}^3}=\underset{x\to 0}{\lim }\frac{\sin \mathrm{x}-\mathrm{xcos}\mathrm{x}}{{20\mathrm{x}}^3}$$$$=\underset{x\to 0}{\lim }\frac{\cos \mathrm{x}-(\cos \mathrm{x}-\mathrm{xsin}\mathrm{x})}{{60\mathrm{x}}^2}$$$$=\underset{x\to 0}{\lim }\frac{\mathrm{xsin}\mathrm{x}}{{60\mathrm{x}}^2}=\frac{1}{60}$$

Answered by EDWIN88 last updated on 10/Feb/21

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}(2+1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{24})-3(\mathrm{x}-\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{120})}{{\mathrm{x}}^5}=$$$$\underset{x\to 0}{\lim }\frac{3\mathrm{x}-\frac{{\mathrm{x}}^3}{2}+\frac{{\mathrm{x}}^5}{24}-3\mathrm{x}+\frac{{\mathrm{x}}^3}{2}-\frac{{\mathrm{x}}^5}{40}}{{\mathrm{x}}^5}=$$$$\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^5(\frac{1}{8.3}-\frac{1}{8.5})}{{\mathrm{x}}^5}=\frac{5-3}{8.5.3}=\frac{1}{4.5.3}=\frac{1}{60}$$

Commented by malwan last updated on 10/Feb/21

$$thankyousir$$

Answered by malwan last updated on 09/Feb/21

$$\underset{x\to 0}{lim}\frac{x(2+1-\frac{x^2}{2!}+\frac{x^4}{4!}\mp ..)-3(x-\frac{x^3}{3!}+\frac{x^5}{5!}\mp ..)}{x^5}$$$$=\underset{x\to 0}{lim}\frac{(3x-\frac{x^3}{2}+\frac{x^5}{24}\mp ..)-(3x-\frac{x^3}{2}+\frac{x^5}{40}\mp ..)}{x^5}$$$$=\underset{x\to 0}{lim}\frac{\left(\frac{5-3}{120}\right)x^5+..}{x^5}=\frac{2}{120}=\frac{1}{60}$$

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