Question Number 201680 by cortano12 last updated on 10/Dec/23 $$\:\:\:\Subset \\ $$ Answered by Calculusboy last updated on 11/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{substitute}}\:\boldsymbol{{ditectly}},\:\boldsymbol{{we}}\:\boldsymbol{{get}}\:\frac{\mathrm{0}}{\mathrm{0}}\left(\boldsymbol{{indeterminant}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{p}}=\mathrm{2}\boldsymbol{{sinx}}−\boldsymbol{{sim}}\mathrm{2}\boldsymbol{{x}}\:\:\:\:\:\frac{\boldsymbol{{dp}}}{\boldsymbol{{dx}}}=\mathrm{2}\boldsymbol{{cosx}}−\mathrm{2}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{q}}=\boldsymbol{{sinx}}−\boldsymbol{{xcosx}}\:\:\:\frac{\boldsymbol{{dq}}}{\boldsymbol{{dx}}}=\boldsymbol{{cosx}}−\left(\boldsymbol{{cosx}}−\boldsymbol{{xsinx}}\right) \\…
Question Number 201221 by Calculusboy last updated on 02/Dec/23 Answered by MM42 last updated on 02/Dec/23 $$\bigstar\bigstar\bigstar\:\:{tan}^{−\mathrm{1}} {a}−{tan}^{−\mathrm{1}} {b}={tan}^{−\mathrm{1}} \left(\frac{{a}−{b}}{\mathrm{1}+{ab}}\right) \\ $$$$\Rightarrow{tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{2}}\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}}…
Question Number 201108 by SLVR last updated on 29/Nov/23 Commented by SLVR last updated on 29/Nov/23 $${Now}\:{ok}\:{sir} \\ $$ Commented by mr W last updated…
Question Number 201089 by SLVR last updated on 29/Nov/23 Commented by SLVR last updated on 29/Nov/23 $${kindly}\:{help}\:{me} \\ $$ Commented by SLVR last updated on…
Question Number 200884 by cortano12 last updated on 26/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200821 by Calculusboy last updated on 24/Nov/23 Answered by shunmisaki007 last updated on 24/Nov/23 $$\mathrm{ln}\left(\mathrm{10}!\right) \\ $$$$=\mathrm{ln}\left(\mathrm{10}\centerdot\mathrm{9}\centerdot\mathrm{8}\centerdot\mathrm{7}\centerdot\mathrm{6}\centerdot\mathrm{5}\centerdot\mathrm{4}\centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right) \\ $$$$=\mathrm{ln}\left(\left(\mathrm{2}\centerdot\mathrm{5}\right)\centerdot\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{2}^{\mathrm{3}} \centerdot\mathrm{7}\centerdot\left(\mathrm{2}\centerdot\mathrm{3}\right)\centerdot\mathrm{5}\centerdot\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right) \\…
Question Number 200413 by Mingma last updated on 18/Nov/23 Commented by Mingma last updated on 18/Nov/23 Evaluate! Answered by witcher3 last updated on 18/Nov/23 $$\:\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}}…
Question Number 200304 by Calculusboy last updated on 16/Nov/23 Answered by Sutrisno last updated on 17/Nov/23 $${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty}…
Question Number 200300 by Calculusboy last updated on 16/Nov/23 Answered by MM42 last updated on 17/Nov/23 $${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. \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\underset{{i}=\mathrm{1}} {\overset{{n}}…
Question Number 200298 by Calculusboy last updated on 16/Nov/23 Commented by Frix last updated on 17/Nov/23 $$−\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Answered by Rasheed.Sindhi last updated on…