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Question-98938




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

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