Question Number 30282 by Tinkutara last updated on 19/Feb/18 $${Find}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}cos}^{{n}} \:\left(\frac{\mathrm{2}\pi}{{n}}\right) \\ $$ Commented by prof Abdo imad last updated on 20/Feb/18 $${we}\:{have}\:\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{for}\:{x}\in{v}\left(\mathrm{0}\right)\Rightarrow…
Question Number 161337 by bobhans last updated on 16/Dec/21 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:\left(\mathrm{p}+\mathrm{x}\right)−\mathrm{cos}\:\left(\mathrm{p}+\mathrm{2x}\right)−\mathrm{cos}\:\mathrm{p}}{\mathrm{x}^{\mathrm{2}} }\:? \\ $$$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2x}+\mathrm{q}\right)−\mathrm{2tan}\:\left(\mathrm{x}+\mathrm{q}\right)+\mathrm{tan}\:\mathrm{q}}{\mathrm{x}^{\mathrm{2}} }\:? \\ $$ Commented by cortano last updated on 16/Dec/21…
Question Number 161319 by cortano last updated on 16/Dec/21 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}−\mathrm{sin}\:{x}}=? \\ $$ Answered by som(math1967) last updated on 16/Dec/21 $$\underset{\boldsymbol{{x}}\rightarrow\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{cosx}}\left\{\left(\boldsymbol{{sinx}}+\boldsymbol{{cosx}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\boldsymbol{{sinx}}+\boldsymbol{{cosx}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{sinx}}+\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}\right\}}{\boldsymbol{{sinx}}+\boldsymbol{{cosx}}−\boldsymbol{{sin}}^{\mathrm{3}}…
Question Number 30192 by abdo imad last updated on 17/Feb/18 $${let}\:\:{u}_{{n}} =\:\frac{\mathrm{1}}{{n}!}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}!\:\:\:\:{find}\:{lim}_{{n}\rightarrow\infty} \:{u}_{{n}} \:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 30189 by abdo imad last updated on 17/Feb/18 $${find}\:{lim}_{{n}\rightarrow\infty} \:\:{n}^{−\mathrm{4}} \:\prod_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\left({n}^{\mathrm{2}} \:+{k}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{{n}}} \:?. \\ $$ Terms of Service Privacy Policy…
Question Number 161248 by cortano last updated on 15/Dec/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{3}} {x}\right)^{\mathrm{4}} −\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{3}} {x}\right)^{\mathrm{4}} }{{x}^{\mathrm{5}} }\:=?\: \\ $$ Answered by bobhans last updated on 15/Dec/21…
Question Number 30176 by abdo imad last updated on 17/Feb/18 $${prove}\:{that}\:\:{v}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}{k}\:+\mathrm{1}}\:{is}\:{convergente}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 30175 by abdo imad last updated on 17/Feb/18 $${prove}\:{that}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{n}+{k}}\:{is}\:{convergente}\:. \\ $$ Commented by abdo imad last updated on 21/Feb/18 $${we}\:{have}\:{u}_{{n}}…
Question Number 161202 by qaz last updated on 14/Dec/21 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{n}} =? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161186 by mathlove last updated on 13/Dec/21 Answered by Rasheed.Sindhi last updated on 13/Dec/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\sqrt{\mathrm{2}}\:{x}+\mathrm{10}}{\:\sqrt{\mathrm{2}}\:{x}−\mathrm{11}}\right)^{\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{1}}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\cancel{{x}}\left(\sqrt{\mathrm{2}}\:+\frac{\mathrm{10}}{{x}}\right)}{\:\cancel{{x}}\left(\sqrt{\mathrm{2}}\:−\frac{\mathrm{11}}{{x}}\right)}\right)^{\frac{\cancel{{x}}\left(\mathrm{2}−\frac{\mathrm{1}}{{x}}\right)}{\cancel{{x}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\sqrt{\mathrm{2}}\:+\frac{\mathrm{10}}{{x}}}{\:\sqrt{\mathrm{2}}\:−\frac{\mathrm{11}}{{x}}}\right)^{\frac{\mathrm{2}−\frac{\mathrm{1}}{{x}}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}} \\…