Question Number 98938 by john santu last updated on 17/Jun/20
Answered by mathmax by abdo last updated on 17/Jun/20
$$=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{n}^{\mathrm{2}} \left(\mathrm{3}+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\right)\left(\mathrm{2}+\frac{\mathrm{5}}{\mathrm{n}}\right)}{\mathrm{n}^{\mathrm{2}} \left(\mathrm{7}+\frac{\mathrm{2}}{\mathrm{n}}\right)\left(\mathrm{5}−\frac{\mathrm{cosn}}{\mathrm{n}}\right)}\:=\frac{\mathrm{3}×\mathrm{2}}{\mathrm{7}×\mathrm{5}}\:=\frac{\mathrm{6}}{\mathrm{35}} \\ $$
Answered by bramlex last updated on 17/Jun/20
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}{n}+\mathrm{5}}{\mathrm{7}{n}+\mathrm{2}}\:.\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}}{\mathrm{5}−\left(\frac{\mathrm{cos}\:{n}}{{n}}\right)}\:= \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}\:×\:\frac{\mathrm{3}}{\mathrm{5}}\:=\:\frac{\mathrm{6}}{\mathrm{35}}\: \\ $$