Question Number 43000 by abdo.msup.com last updated on 06/Sep/18 $${prove}\:{that}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+….+\sqrt{\mathrm{2}}}}\:\:=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18 $${cos}\mathrm{2}\alpha=\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$${cos}^{\mathrm{2}} \alpha=\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}}…
Question Number 42999 by abdo.msup.com last updated on 06/Sep/18 $${let}\:{f}\left({x}\right)=\sqrt{{x}}+\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{2}\right) \\ $$$$\left.\mathrm{2}\right)\:{if}\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} \left({x}−\mathrm{2}\right)^{{n}} \:{find}\:{the} \\ $$$${sequence}\:{a}_{{n}} \\ $$ Commented by…
Question Number 42805 by maxmathsup by imad last updated on 02/Sep/18 $${calculate}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\:\:{with} \\ $$$${S}_{{n}} =\:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{k}^{\mathrm{3}} }{\:\sqrt{\left(\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} \right)^{\mathrm{3}} }} \\ $$…
Question Number 42789 by maxmathsup by imad last updated on 02/Sep/18 $${let}\:{u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}+\mathrm{1}} \:={u}_{{n}} \:+\:\frac{\mathrm{2}}{{u}_{{n}} } \\ $$$${study}\:{the}\:{convervence}\:{of}\:\left({u}_{{n}} \right) \\ $$ Commented by maxmathsup by…
Question Number 42788 by maxmathsup by imad last updated on 02/Sep/18 $${calculate}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{2}{x}}{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}\:−{cosx} \\ $$ Commented by maxmathsup by imad last updated on 30/Sep/18 $${we}\:{have}\:\frac{\mathrm{2}{x}}{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}\:=\frac{\mathrm{2}{x}}{{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)}\:{but}…
Question Number 42786 by maxmathsup by imad last updated on 02/Sep/18 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}−\frac{{x}}{{sinx}}}{{x}^{\mathrm{2}} } \\ $$ Commented by maxmathsup by imad last updated on 30/Sep/18…
Question Number 42787 by maxmathsup by imad last updated on 02/Sep/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{1}}{\left({sinx}\right)^{\mathrm{4}} }\left\{\:{sin}\left(\frac{{x}}{\mathrm{1}−{x}}\right)−\frac{{sinx}}{\mathrm{1}−{sinx}}\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42785 by maxmathsup by imad last updated on 02/Sep/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}+{x}\:−{e}^{{arcsinx}} }{{x}^{\mathrm{2}} } \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 42783 by maxmathsup by imad last updated on 02/Sep/18 $${calculate}\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\:\:\:\mid{tan}\left(\mathrm{2}{x}\right)\mid^{{sin}\left(\mathrm{4}{x}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42784 by maxmathsup by imad last updated on 02/Sep/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:{ln}\left(\frac{{e}^{{x}^{\mathrm{2}} −{x}} \:−\mathrm{1}}{{x}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com