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Question Number 106217 by bobhans last updated on 03/Aug/20

$$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin \mathrm{x}-\mathrm{x}\cos \mathrm{x}}{{\mathrm{x}}^2}\mathrm{and}\mathrm{f}\left(0\right)=0\mathrm{when}\mathrm{x}$$$$=0,\mathrm{what}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{value}\mathrm{of}\underset{\mathrm{t}\to 0}{\lim }\frac{\underset{0}{\stackrel{\mathrm{t}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}}{{\mathrm{t}}^2}$$$$$$

Answered by john santu last updated on 03/Aug/20

$${\mathrm{L}}^{\prime }\mathrm{Hopital}:\underset{\mathrm{t}\to 0}{\lim }\frac{\underset{0}{\stackrel{\mathrm{t}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}}{{\mathrm{t}}^2}=$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{\mathrm{f}\left(\mathrm{t}\right)}{2\mathrm{t}}=\underset{\mathrm{t}\to 0}{\lim }\frac{\sin \mathrm{t}-\mathrm{t}.\cos \mathrm{t}}{2\mathrm{t}.{\mathrm{t}}^2}$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{(\mathrm{t}-\frac{{\mathrm{t}}^3}{6})-\mathrm{t}(1-\frac{{\mathrm{t}}^2}{2})}{{2\mathrm{t}}^3}$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{\frac{{\mathrm{t}}^3}{2}-\frac{{\mathrm{t}}^3}{6}}{{2\mathrm{t}}^3}=\frac{2}{12}=\frac{1}{6}.\left(\mathrm{JS}\right)$$

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