Question Number 32376 by rsen3579@gmail.com last updated on 24/Mar/18 Commented by abdo imad last updated on 24/Mar/18 $${we}\:{have}\:{e}^{{x}} \:\sim\mathrm{1}+{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{for}\:{x}\in{V}\left(\mathrm{0}\right)\:{alsowehave} \\ $$$${e}^{{sinx}} \:={e}^{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\mathrm{0}\left({x}^{\mathrm{5}} \right)}…
Question Number 32364 by prof Abdo imad last updated on 23/Mar/18 $${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}}} \\ $$$${find}\:\:{lim}\:{u}_{{n}} \\ $$ Commented by prof Abdo imad…
Question Number 32345 by abdo imad last updated on 23/Mar/18 $${calculate}\:{lim}_{{n}\rightarrow\infty} \:\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\sum_{{j}=\mathrm{1}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{i}+{j}} }{{i}+{j}}\:. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 97840 by bemath last updated on 10/Jun/20 $$\underset{\gamma\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{arc}\:\mathrm{sin}\:\frac{\mathrm{k}\:\mathrm{cos}\:\gamma}{\:\sqrt{\mathrm{1}+\mathrm{k}^{\mathrm{2}} }}\:−\:\mathrm{arc}\:\mathrm{sin}\:\frac{\mathrm{k}}{\:\sqrt{\mathrm{1}+\mathrm{k}^{\mathrm{2}} }}\right)\:=? \\ $$ Commented by john santu last updated on 10/Jun/20 $$=\:\mathrm{0}\: \\…
Question Number 32298 by abdo imad last updated on 22/Mar/18 $${find}\:{tbe}\:{nature}\:{of}\:\:\sum_{{n}\geqslant\mathrm{2}} \:\:\:\frac{\mathrm{1}}{{n}\left({ln}\left({n}\right)\right)^{\mathrm{2}} }\:. \\ $$ Commented by abdo imad last updated on 01/Apr/18 $$\:{let}\:{put}\:\varphi\left({t}\right)\:=\:\frac{\mathrm{1}}{{t}\left({lnt}\right)^{\mathrm{2}} }\:{with}\:{t}\geqslant\mathrm{2}\:\:{we}\:{have}…
Question Number 32297 by abdo imad last updated on 22/Mar/18 $${calculate}\:\sum_{{n}\geqslant\mathrm{0}} \:\frac{{n}+\mathrm{2}^{{n}} }{{n}!}\:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 32296 by abdo imad last updated on 22/Mar/18 $${let}\:\:{u}_{{n}} =\:\frac{\left({n}+\mathrm{1}\right)^{\alpha} \:\:−{n}^{\alpha} }{{n}^{\alpha−\mathrm{1}} }\:\:{with}\:\alpha>\mathrm{1}\:\:{find}\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} \:\:. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 32291 by abdo imad last updated on 22/Mar/18 $${let}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{n}+{k}}\:{prove}\:{that}\:\mathrm{0}\leqslant{u}_{{n}} \leqslant\mathrm{1}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 32286 by abdo imad last updated on 22/Mar/18 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\frac{{e}^{{x}} }{\:\sqrt{\mathrm{1}+{x}}}\:−\mathrm{1}−\frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 32281 by abdo imad last updated on 22/Mar/18 $${calculate}\:{lim}_{{x}\rightarrow\infty} \sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\sqrt{{x}^{\mathrm{2}} \:−{x}+\mathrm{1}}\:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com