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Question Number 84543 by niroj last updated on 14/Mar/20

$$\mathbf{Determine}\mathbf{the}\mathbf{value}\mathbf{of}\mathbf{a},\mathbf{b},\mathbf{c}\mathbf{so}\mathbf{that}$$$$\underset{\mathrm{x}\to 0}{\stackrel{\lim }{}}\frac{(\mathbf{a}+\mathbf{b}\cos \mathrm{x})\mathrm{x}-\mathbf{c}\sin \mathrm{x}}{{\mathrm{x}}^5}=1$$

Commented by mr W last updated on 14/Mar/20

$$a=120$$$$b=60$$$$c=180$$

Commented by jagoll last updated on 14/Mar/20

$$\underset{x\to 0}{\lim }\frac{(\mathrm{a}+\mathrm{b}(1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{24}+\mathrm{o}\left({\mathrm{x}}^4\right))\mathrm{x}-\mathrm{c}(\mathrm{x}-\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{120}+\mathrm{o}\left({\mathrm{x}}^5\right))}{{\mathrm{x}}^5}=1$$$$\underset{x\to 0}{\lim }\frac{\mathrm{ax}+\mathrm{bx}-\frac{{\mathrm{bx}}^3}{2}+\frac{{\mathrm{bx}}^5}{24}-\mathrm{cx}+\frac{{\mathrm{cx}}^3}{6}-\frac{{\mathrm{cx}}^5}{120}}{{\mathrm{x}}^5}=1$$$$\underset{x\to 0}{\lim }\frac{(\mathrm{a}+\mathrm{b}-\mathrm{c})\mathrm{x}+\left(\frac{\mathrm{c}-3\mathrm{b}}{6}\right){\mathrm{x}}^3+\left(\frac{5\mathrm{b}-\mathrm{c}}{120}\right){\mathrm{x}}^5}{{\mathrm{x}}^5}=1$$$$$$

Commented by jagoll last updated on 14/Mar/20

$$\mathrm{c}=3\mathrm{b}$$$$5\mathrm{b}-\mathrm{c}=120\Rightarrow 2\mathrm{b}=120$$$$\mathrm{b}=60\wedge \mathrm{c}=180$$$$\mathrm{a}+\mathrm{b}-\mathrm{c}=0\Rightarrow \mathrm{a}=\mathrm{c}-\mathrm{b}=120$$

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