Question Number 98952 by Ar Brandon last updated on 17/Jun/20
Question Number 98955 by Ar Brandon last updated on 17/Jun/20
Question Number 98938 by john santu last updated on 17/Jun/20 Answered by mathmax by abdo last updated on 17/Jun/20 $$=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{n}^{\mathrm{2}} \left(\mathrm{3}+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\right)\left(\mathrm{2}+\frac{\mathrm{5}}{\mathrm{n}}\right)}{\mathrm{n}^{\mathrm{2}} \left(\mathrm{7}+\frac{\mathrm{2}}{\mathrm{n}}\right)\left(\mathrm{5}−\frac{\mathrm{cosn}}{\mathrm{n}}\right)}\:=\frac{\mathrm{3}×\mathrm{2}}{\mathrm{7}×\mathrm{5}}\:=\frac{\mathrm{6}}{\mathrm{35}} \…
Question Number 98880 by Ar Brandon last updated on 16/Jun/20
Question Number 98858 by M±th+et+s last updated on 16/Jun/20
Question Number 33313 by abdo imad last updated on 14/Apr/18
Question Number 33312 by abdo imad last updated on 14/Apr/18
Question Number 98833 by john santu last updated on 16/Jun/20 Commented by bobhans last updated on 16/Jun/20 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}^{\mathrm{n}} \:\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \right)\:=\:\mathrm{3}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \…
Question Number 164364 by Zaynal last updated on 16/Jan/22
Question Number 98823 by john santu last updated on 16/Jun/20 Commented by bramlex last updated on 16/Jun/20