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Category: Limits

Question-201221

Question Number 201221 by Calculusboy last updated on 02/Dec/23 Answered by MM42 last updated on 02/Dec/23 $$\bigstar\bigstar\bigstar\:\:{tan}^{−\mathrm{1}} {a}−{tan}^{−\mathrm{1}} {b}={tan}^{−\mathrm{1}} \left(\frac{{a}−{b}}{\mathrm{1}+{ab}}\right) \\ $$$$\Rightarrow{tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{2}}\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}}…

Question-200821

Question Number 200821 by Calculusboy last updated on 24/Nov/23 Answered by shunmisaki007 last updated on 24/Nov/23 $$\mathrm{ln}\left(\mathrm{10}!\right) \\ $$$$=\mathrm{ln}\left(\mathrm{10}\centerdot\mathrm{9}\centerdot\mathrm{8}\centerdot\mathrm{7}\centerdot\mathrm{6}\centerdot\mathrm{5}\centerdot\mathrm{4}\centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right) \\ $$$$=\mathrm{ln}\left(\left(\mathrm{2}\centerdot\mathrm{5}\right)\centerdot\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{2}^{\mathrm{3}} \centerdot\mathrm{7}\centerdot\left(\mathrm{2}\centerdot\mathrm{3}\right)\centerdot\mathrm{5}\centerdot\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right) \\…

Question-200304

Question Number 200304 by Calculusboy last updated on 16/Nov/23 Answered by Sutrisno last updated on 17/Nov/23 $${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty}…

Question-200300

Question Number 200300 by Calculusboy last updated on 16/Nov/23 Answered by MM42 last updated on 17/Nov/23 $${lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\left[{ln}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{1}} \right)+{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} +…+{ln}\left(\mathrm{1}+\left(\frac{{n}}{{n}}\right)^{\mathrm{2}} \right)\right.\right. \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\underset{{i}=\mathrm{1}} {\overset{{n}}…