Question Number 159906 by qaz last updated on 22/Nov/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{\mathrm{n}} {\underbrace{\mathrm{tan}\:\mathrm{tan}\:\mathrm{tan}\:…\mathrm{tan}\:\mathrm{x}}}−\underset{\mathrm{n}} {\underbrace{\mathrm{sin}\:\mathrm{sin}\:\mathrm{sin}\:…\mathrm{sin}\:\mathrm{x}}}}{\mathrm{x}^{\mathrm{2}} }=? \\ $$ Answered by kaivan.ahmadi last updated on 24/Nov/21 $$\sim{lim}\frac{\frac{{x}^{\mathrm{3}} }{\mathrm{2}}}{{x}^{\mathrm{2}}…
Question Number 28825 by abdo imad last updated on 30/Jan/18 $${let}\:{give}\:{f}\left({x}\right)=\:{e}^{−{x}} \:{cosx}\:{prove}\:{that} \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}} }{{n}!}\:{cos}\left(\frac{\mathrm{3}{n}\pi}{\mathrm{4}}\right)\:{x}^{{n}} \:\:. \\ $$ Terms of Service Privacy Policy…
Question Number 28818 by abdo imad last updated on 30/Jan/18 $${prove}\:{that}\:\:\frac{\pi}{\mathrm{4}{cos}\left(\frac{\pi\alpha}{\mathrm{2}}\right)}=\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(\mathrm{2}{p}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{p}} }{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} \:−\alpha^{\mathrm{2}} } \\ $$$$\alpha\:\in{R}−{Z}. \\ $$ Terms of Service Privacy Policy…
Question Number 28814 by abdo imad last updated on 30/Jan/18 $${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} \:+\mathrm{2}^{\mathrm{3}} +…+{n}^{\mathrm{3}} }\:. \\ $$ Commented by abdo imad last updated on…
Question Number 159852 by Tawa11 last updated on 21/Nov/21 Answered by tounghoungko last updated on 21/Nov/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}+\mathrm{2}−\mathrm{5}}{{x}+\mathrm{2}}\right)^{{x}} \:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\left(\frac{{x}+\mathrm{2}}{\mathrm{5}}\right)}\right)^{−\frac{{x}+\mathrm{2}}{\mathrm{5}}\:.\frac{−\mathrm{5}{x}}{\left({x}+\mathrm{2}\right)}} \\ $$$$\:\:=\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(−\frac{\mathrm{5}{x}}{{x}+\mathrm{2}}\right)\:} \:=\:{e}^{−\mathrm{5}} \:…
Question Number 28770 by Rasheed.Sindhi last updated on 30/Jan/18 $$\underset{{x}\rightarrow+\infty} {{lim}}\:\frac{\mathrm{5}{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{1}}{−\mathrm{3}{x}^{\mathrm{3}} +\mathrm{10}{x}^{\mathrm{2}} +\mathrm{50}} \\ $$ Commented by Cheyboy last updated on 30/Jan/18 $$\underset{{x}\rightarrow+\infty\:}…
Question Number 159837 by tounghoungko last updated on 21/Nov/21 $$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{2021}{x}^{\mathrm{2}} −\mathrm{6}{x}}\:+\sqrt[{\mathrm{3}}]{\mathrm{1021}{x}^{\mathrm{3}} −\mathrm{5}{x}}\:=? \\ $$ Answered by kaivan.ahmadi last updated on 24/Nov/21 $$\sim{lim}\sqrt{\mathrm{2021}}\mid{x}−\frac{−\mathrm{6}}{\mathrm{2}×\mathrm{2021}}\mid+\sqrt[{\mathrm{3}}]{\mathrm{1021}}\mid{x}\mid= \\ $$$${lim}\sqrt{\mathrm{2021}}\left(−{x}−\frac{\mathrm{3}}{\mathrm{2021}}\right)−\sqrt[{\mathrm{3}}]{\mathrm{1021}}{x}=+\infty…
Question Number 159796 by Ar Brandon last updated on 21/Nov/21 $$\mathrm{Montrer}\:\mathrm{que}\:\mathrm{la}\:\mathrm{suite}\:\mathrm{d}\acute {\mathrm{e}finie}\:\mathrm{par}\: \\ $$$${u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{x}^{{k}} }{{k}}\:\mathrm{est}\:\mathrm{une}\:\mathrm{suite}\:\mathrm{de}\:\mathrm{Cauchy}. \\ $$ Answered by alcohol last updated…
Question Number 94273 by Cheyboy last updated on 17/May/20 Commented by Cheyboy last updated on 17/May/20 $$\boldsymbol{{PLEASE}}\:\boldsymbol{{HELP}}\:\:!! \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 28698 by Rasheed.Sindhi last updated on 29/Jan/18 $$\frac{\infty}{\mathrm{negative}\:\mathrm{number}}=? \\ $$$$\:\infty\:\:\:\:\:\:\:\:\:\:\mathrm{or}\:\:\:\:\:\:\:\:\:−\infty\:\:? \\ $$ Commented by prakash jain last updated on 29/Jan/18 $$−\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−}…