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Question Number 98938 by john santu last updated on 17/Jun/20

Answered by mathmax by abdo last updated on 17/Jun/20

$$={\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{{\mathrm{n}}^2(3+\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}})(2+\frac{5}{\mathrm{n}})}{{\mathrm{n}}^2(7+\frac{2}{\mathrm{n}})(5-\frac{\mathrm{cosn}}{\mathrm{n}})}=\frac{3\times 2}{7\times 5}=\frac{6}{35}$$

Answered by bramlex last updated on 17/Jun/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{2n+5}{7n+2}.\underset{n\to \mathrm{\infty }}{\lim }\frac{3+\frac{{(-1)}^n}{n}}{5-\left(\frac{\cos n}{n}\right)}=$$$$\frac{2}{7}\times \frac{3}{5}=\frac{6}{35}$$

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