Question Number 116039 by bobhans last updated on 30/Sep/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:^{\mathrm{4}} \left({x}\right)}}{\left(\mathrm{1}−\mathrm{sin}\:\left({x}\right)\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }\:? \\ $$ Answered by Dwaipayan Shikari last updated on 30/Sep/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{\left(\mathrm{cosx}\right)^{\frac{\mathrm{4}}{\mathrm{3}}}…
Question Number 116029 by bemath last updated on 30/Sep/20 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}}{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }\:=?\: \\ $$ Answered by bobhans last updated on 30/Sep/20 $${set}\:{x}\:=\:\frac{\pi}{\mathrm{2}}\:+\:{z} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:{z}}{\left(\mathrm{1}−\mathrm{cos}\:{z}\right)^{\frac{\mathrm{2}}{\mathrm{3}}}…
Question Number 181553 by Frix last updated on 26/Nov/22 $$\mathrm{Reposting}\:\mathrm{question}\:\mathrm{181462} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{{n}}]{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}}{\:\sqrt[{{n}+\mathrm{1}}]{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}\overset{?} {=}\frac{\mathrm{1}}{\mathrm{2e}} \\ $$ Answered by aleks041103 last updated on 26/Nov/22 $${b}_{{n}} =\frac{\sqrt[{{n}}]{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}}{\:\sqrt[{{n}+\mathrm{1}}]{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}=\sqrt[{{n}}]{\frac{\sqrt{\mathrm{2}!}×\sqrt[{\mathrm{3}}]{\mathrm{3}!}×\sqrt[{\mathrm{4}}]{\mathrm{4}!}×…×\sqrt[{{n}}]{{n}!}}{\left(\left(\mathrm{2}{n}+\mathrm{1}\right)!!\right)^{\frac{{n}}{{n}+\mathrm{1}}}…
Question Number 116019 by bemath last updated on 30/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\:\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\mathrm{sin}\:\left({t}^{\mathrm{3}} \right)\:{dt}}{\mathrm{2}{x}^{\mathrm{4}} }\:? \\ $$ Commented by bemath last updated on 30/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 116005 by mnjuly1970 last updated on 30/Sep/20 $$\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{mathematics}… \\ $$$$ \\ $$$$\:\:\:\:\:{prove}\:\:{that}::: \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{lim}_{{x}\rightarrow\mathrm{1}^{+} } \left(\:\zeta\left(\:{x}\:\right)\:−\frac{\mathrm{1}}{{x}\:−\:\mathrm{1}}\right)\:\overset{???} {=}\gamma\:\:\: \\ $$$$\:\:\gamma::\:\mathscr{E}{uler}\:−\:{mascheroni}\:{constant}. \\ $$$$…
Question Number 115999 by bemath last updated on 30/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{3}{x}}\:\sqrt[{\mathrm{4}\:}]{\mathrm{cos}\:\mathrm{4}{x}}}{{x}^{\mathrm{2}} } \\ $$ Answered by bobhans last updated on 30/Sep/20 $${short}\:{cut}\:'{mr}\:{john}\:{santu}\:' \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{3}{x}}\:\sqrt[{\mathrm{4}\:}]{\mathrm{cos}\:\mathrm{4}{x}}}{{x}^{\mathrm{2}}…
Question Number 115961 by Ar Brandon last updated on 29/Sep/20 Commented by Dwaipayan Shikari last updated on 29/Sep/20 $$\mathrm{Indetermined} \\ $$ Commented by Ar Brandon…
Question Number 181464 by cortano1 last updated on 25/Nov/22 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{tan}\:\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{x}\right)}{\mathrm{x}−\mathrm{tan}\:\left(\mathrm{x}\right)}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50395 by Abdo msup. last updated on 16/Dec/18 $${let}\:{U}_{{n}} =\left({e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \right)^{\sqrt{{n}^{\mathrm{2}} \:+\mathrm{2}}−\sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:{U}_{{n}} \\ $$ Terms of Service Privacy Policy…
Question Number 181462 by mathlove last updated on 25/Nov/22 Commented by Frix last updated on 26/Nov/22 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\frac{\mathrm{1}}{\mathrm{2e}}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{100\%}\:\mathrm{sure} \\ $$$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{confirm}\:\mathrm{this}? \\ $$ Answered by SEKRET last…