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Question Number 84873 by john santu last updated on 17/Mar/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{x}}^2\sin \left(\frac{\mathrm{x}!}{\mathrm{x}}\right)}{{\mathrm{x}}^2+1}$$

Commented by john santu last updated on 17/Mar/20

$$\mathrm{yes}.\mathrm{i}\mathrm{agree}\mathrm{sir}$$

Answered by bshahid010@gmail.com last updated on 17/Mar/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\left({\mathrm{x}}^2\sin \left(\frac{\mathrm{x}!}{\mathrm{x}}\right)\right)}{{\mathrm{x}}^2(1+\frac{1}{{\mathrm{x}}^2})}=\underset{x\to \mathrm{\infty }}{\lim }\frac{\sin ((\mathrm{x}-1)!)}{1}$$$$\mathrm{this}\mathrm{value}\mathrm{is}\mathrm{ossilatinig}\mathrm{between}1\mathrm{and}-1$$$$\mathrm{so}\mathrm{limit}\mathrm{does}\mathrm{not}\mathrm{exist}$$

Answered by TANMAY PANACEA last updated on 17/Mar/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{sin\left(\frac{x!}{x}\right)}{1+\frac{1}{x^2}}\to$$$$1⩾sin\left(\frac{x!}{x}\right)⩾-1foranyvalueofx$$$$sothevalueoftheaboveexpression$$$$liesbetween\pm 1$$$$hencelimitdoesnotexist$$

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