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Question Number 93146 by i jagooll last updated on 11/May/20

$$\underset{x\to 0}{\lim }\frac{\underset{0}{\stackrel{\mathrm{x}}{\int }}(\mathrm{a}+\mathrm{bcos}\mathrm{t}+\mathrm{c}\cos \left(2\mathrm{t}\right))\mathrm{dt}}{{\mathrm{x}}^5}=15$$

Commented by i jagooll last updated on 11/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by john santu last updated on 11/May/20

$$\underset{x\to 0}{\lim }\frac{\mathrm{a}+\mathrm{bcos}\mathrm{x}+\mathrm{c}\cos 2\mathrm{x}}{{5\mathrm{x}}^4}=15$$$$\underset{x\to 0}{\lim }\frac{\mathrm{a}+\mathrm{b}(1-\frac{1}{2}{\mathrm{x}}^2+\frac{{\mathrm{x}}^4}{24})+\mathrm{c}(1-\frac{1}{2}{\left(2\mathrm{x}\right)}^2+\frac{1}{24}{\left(2\mathrm{x}\right)}^4)}{{5\mathrm{x}}^4}=15$$$$\underset{x\to 0}{\lim }\frac{(\mathrm{a}+\mathrm{b}+\mathrm{c})+{\mathrm{x}}^2(-\frac{\mathrm{b}}{2}-\frac{4\mathrm{c}}{2})+{\mathrm{x}}^4(\frac{\mathrm{b}}{24}+\frac{16\mathrm{c}}{24})}{{5\mathrm{x}}^4}=15$$$$\left(\mathrm{i}\right)\mathrm{a}+\mathrm{b}+\mathrm{c}=0$$$$\left(\mathrm{ii}\right)\mathrm{b}=-4\mathrm{c}$$$$\left(\mathrm{iii}\right)\frac{\mathrm{b}+16\mathrm{c}}{24}=75\Rightarrow \frac{12\mathrm{c}}{24}=75$$$$\mathrm{c}=150,\mathrm{b}=-600,\mathrm{a}=-(\mathrm{b}+\mathrm{c})$$$$\mathrm{a}=-(-600+150)=450$$$$$$

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