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Question Number 111080 by bemath last updated on 02/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\left(1\right)\underset{\mathrm{k}=50}{\sum ^{100}}\frac{1}{\mathrm{k}(151-\mathrm{k})}?$$$$\left(2\right)\mathrm{without}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{and}\mathrm{series}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\underset{x\to 0}{\lim }\frac{\mathrm{xcos}\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^2\sin \mathrm{x}}$$

Answered by john santu last updated on 02/Sep/20

$$\left(2\right)\underset{x\to 0}{\lim }\frac{x\cos x-\sin x+x-x}{x^2\sin x}=$$$$\underset{x\to 0}{\lim }\frac{x(\cos x-1)}{x^2\sin x}+\underset{x\to 0}{\lim }\frac{x-\sin x}{x^2\sin x}=$$$$thefirstterm$$$$L_1=\underset{x\to 0}{\lim }\frac{x(\cos x-1)}{x^2\sin x}=\underset{x\to 0}{\lim }\frac{\cos x-1}{x\sin x}$$$$L_1=\underset{x\to 0}{\lim }\frac{-2\sin {}^2\left(\frac{x}{2}\right)}{x\sin x}=-2\times \frac{1}{4}=-\frac{1}{2}$$$$thesecondterm$$$$L_2=\underset{x\to 0}{\lim }\frac{x-\sin x}{x^2\sin x}=\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}.\frac{x}{\sin x}$$$$[notethat\underset{x\to 0}{\lim }\frac{x}{\sin x}=1]$$$$L_2=\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}=\frac{1}{6}$$$$[wecanfinditbylettingx=2t]$$$$Hence,weconcludethat$$$$\underset{x\to 0}{\lim }\frac{x\cos x-\sin x}{x^2\sin x}=-\frac{1}{2}+\frac{1}{6}=-\frac{1}{3}$$

Commented by bemath last updated on 02/Sep/20

$$\mathrm{greatt}....$$

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