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Question Number 201724 by LimPorly last updated on 11/Dec/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} \left({x}−\mathrm{2}\right)+{x}+\mathrm{2}}{{x}^{\mathrm{3}} }\:{solve}\:{it}\:{by}\:{not}\:{using} \\ $$$${taylor}\:{series}\:{or}\:{l}'{hopital}\:{rule}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201657 by LimPorly last updated on 10/Dec/23 $${if}\:\:{f}\left({x}\right)=\begin{cases}{\frac{\mathrm{sin}\:\left(\mathrm{1}+\left[{x}\right]\right)}{\left[{x}\right]}\:\:{for}\:\left[{x}\right]\neq\mathrm{0}}\\{\mathrm{0}\:\:{for}\:\left[{x}\right]=\mathrm{0}}\end{cases} \\ $$$${where}\:\left[{x}\right]\:{represents}\:{an}\:{integer}\:\boldsymbol{{x}}\:{greatest}\:\leqslant\:\boldsymbol{{x}} \\ $$$${Find}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right). \\ $$ Answered by aleks041103 last updated on 10/Dec/23…
Question Number 201708 by Calculusboy last updated on 10/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
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) \\…