Question Number 97398 by RAMANA last updated on 07/Jun/20 $$\mathrm{The}\:\mathrm{discontinuty}\:\mathrm{of}\:\left[\mathrm{x}\right]^{\mathrm{2}} −\left[\mathrm{x}^{\mathrm{2}} \right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162925 by qaz last updated on 02/Jan/22 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\left[\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}}} −\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}}} \right]=? \\…
Question Number 162926 by qaz last updated on 02/Jan/22 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }\left[\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}}} −\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2}}+…+\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}}} \right]=? \\ $$ Terms of Service Privacy Policy…
Question Number 97355 by bagjamath last updated on 07/Jun/20 Commented by PRITHWISH SEN 2 last updated on 07/Jun/20 $$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\left(\frac{\pi}{\mathrm{n}+\mathrm{1}}+\frac{\pi}{\mathrm{n}+\mathrm{1}}+……+\frac{\pi}{\mathrm{n}+\mathrm{1}}\:\mathrm{ntimes}\:\right) \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\pi}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\:=\:\pi \\ $$…
Question Number 162827 by saboorhalimi last updated on 01/Jan/22 Answered by Ar Brandon last updated on 01/Jan/22 $$\mathrm{g}\left({x}\right)=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\left({x}+\mathrm{1}\right)^{{r}+\mathrm{1}} −{x}^{{r}+\mathrm{1}} \right)^{\frac{\mathrm{1}}{{r}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{g}\left({x}\right)}{{x}}=\underset{{r}\rightarrow\mathrm{0},\:{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{1}+\frac{\mathrm{1}}{{r}}−\mathrm{1}}…
Question Number 31709 by gunawan last updated on 13/Mar/18 $$\mathrm{Given}\:\mathrm{sequence}\:\left({y}_{{n}} \right)\:\mathrm{with}\:{y}_{\mathrm{1}} =\mathrm{1}\:, \\ $$$${y}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}\left({y}_{{n}} ^{\mathrm{3}} +{y}_{{n}} ^{\mathrm{2}} \right)−\mathrm{1}\: \\ $$$$\mathrm{find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{y}_{{n}} \\ $$ Terms…
Question Number 31708 by gunawan last updated on 13/Mar/18 $$\mathrm{Given}\:\mathrm{sequence}\:\mathrm{real}\:\mathrm{function}\:\left({f}_{{n}} \right)\:, \\ $$$${f}_{{n}} :\:\left[\mathrm{0},\:\mathrm{2}\right]\:\rightarrow\:\mathbb{R}\:,\mathrm{with}\: \\ $$$${f}_{{n}} \left({x}\right)=\frac{{x}^{{n}} }{\mathrm{1}+{x}^{{n}} }\:\:.\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:… \\ $$$$\mathrm{a}.\mathrm{Prove}\:\left({f}_{{n}} \right)\:\mathrm{not}\:\mathrm{uniformly}\:\mathrm{convergent}\:\mathrm{on}\:\left[\mathrm{0},\:\mathrm{2}\right] \\ $$$$\mathrm{b}.\:\mathrm{Find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{f}_{{n}}…
Question Number 31707 by gunawan last updated on 13/Mar/18 $$\mathrm{Find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}\frac{\mathrm{1}}{{n}}\mathrm{sin}\frac{\mathrm{2}\pi}{{n}}\: \\ $$ Commented by abdo imad last updated on 18/Mar/18 $${i}\:{think}\:{the}\:{Q}.{is}\:{find}\:{lim}_{{n}\rightarrow\infty} \:\sum_{{k}=\mathrm{1}}…
Question Number 97238 by bobhans last updated on 07/Jun/20 Commented by john santu last updated on 07/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}\:\left(\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{1}\right)}{\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{2}}−\mathrm{1}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 31705 by gunawan last updated on 12/Mar/18 $$\mathrm{Given}\:\mathrm{sequence}\left({x}_{{n}} \right)\:\mathrm{with}\:\mathrm{0}<{a}={x}_{\mathrm{1}} <{x}_{\mathrm{2}} ={b} \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}+\mathrm{1}} +{x}_{{n}} \:,\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},… \\ $$$$\mathrm{review}\:\mathrm{sequence}\:\left({r}_{{n}} \right)\:\mathrm{with}\:{r}_{{n}} =\frac{{x}_{{n}+\mathrm{1}} }{{x}_{{n}} }\:,\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},… \\…