Question Number 160609 by cortano last updated on 03/Dec/21 $$\:\:\:\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right)=? \\ $$ Answered by Ar Brandon last updated on 03/Dec/21 $$\mathscr{L}=\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left(\frac{\mathrm{tan}{x}}{\mathrm{1}+\mathrm{cos}{x}}\right),\:{u}={x}−\pi \\ $$$$\:\:\:\:\:=\underset{{u}\rightarrow\mathrm{0}}…
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Question Number 160604 by cortano last updated on 03/Dec/21 $$\:\:\mathrm{S}_{\mathrm{n}} =\:\frac{\mathrm{12}}{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)}+\frac{\mathrm{12}^{\mathrm{2}} }{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{3}} −\mathrm{3}^{\mathrm{3}} \right)}+\frac{\mathrm{12}^{\mathrm{3}} }{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{4}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \right)}+…+\frac{\mathrm{12}^{\mathrm{n}}…
Question Number 160589 by Ar Brandon last updated on 02/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29511 by abdo imad last updated on 09/Feb/18 $${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\:\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:. \\ $$ Commented by prof Abdo imad last updated on 12/Feb/18 $${let}\:{use}\:{stirling}\:{formula}\:{we}\:{have}…
Question Number 29510 by abdo imad last updated on 09/Feb/18 $${let}\:\:\:{w}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\:\mathrm{1}\:+{e}^{\frac{\mathrm{1}}{{n}}} \:+{e}^{\frac{\mathrm{2}}{{n}}} \:+\:…\:{e}^{\frac{{n}−\mathrm{1}}{{n}}} \right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {w}_{{n}} . \\ $$ Commented by abdo imad last updated…
Question Number 29509 by abdo imad last updated on 09/Feb/18 $${find}\:{lim}_{{n}\rightarrow+\:\infty} \:\:\:\:\frac{{n}!}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$ Commented by abdo imad last updated on 11/Feb/18 $${let}\:{use}\:{stirling}\:{formula}\:\:{n}!\:\sim\:{n}^{{n}} \:{e}^{−{n}}…
Question Number 160576 by mnjuly1970 last updated on 02/Dec/21 Commented by aleks041103 last updated on 02/Dec/21 $${What}\:{do}\:{the}\:{square}\:{brakets}\:{mean}? \\ $$ Commented by mnjuly1970 last updated on…
Question Number 29505 by abdo imad last updated on 09/Feb/18 $${let}\:{give}\:{u}_{{n}} =\:\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{2}^{{k}} }\:\:{find}\:\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} \:\:. \\ $$ Commented by abdo imad last…
Question Number 29500 by abdo imad last updated on 09/Feb/18 $${nature}\:{of}\:{the}\:{sequence}\:\:{u}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{kln}\left({k}\right)}\:. \\ $$ Commented by abdo imad last updated on 11/Feb/18…
Question Number 29499 by abdo imad last updated on 09/Feb/18 $${find}\:\:{lim}_{{n}\rightarrow+\infty} \left(\:\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{n}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}{n}}}\:+….+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} }}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com