You have the sequence s_n=\frac{3n-4n^3}{5n^2+2} and t_n=\cos(\frac{1}{n}). What conclusion can

You have the sequence ${s}_{n}=\frac{3n-4{n}^{3}}{5{n}^{2}+2}$ and ${t}_{n}=\mathrm{cos}\left(\frac{1}{n}\right)$. What conclusion can you make about each of the sequences?
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${s}_{n}=\frac{3n-4{n}^{3}}{5{n}^{2}+2}=\frac{\frac{3}{n}-4n}{5+\frac{2}{{n}^{2}}}$
$\underset{n\to \mathrm{\infty }}{lim}{s}_{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{3}{n}-4n}{5+\frac{2}{{n}^{2}}}$
$=\frac{0-\mathrm{\infty }}{5+0}=-\mathrm{\infty }$
$\underset{n\to \mathrm{\infty }}{lim}{s}_{n}$ does not exists definitely so. That ${s}_{n}$ is a divergent sequence.
${t}_{n}=\mathrm{cos}\left(\frac{1}{n}\right)$
$\underset{n\to \mathrm{\infty }}{lim}{t}_{n}=\underset{n\to \mathrm{\infty }}{lim}\mathrm{cos}\left(\frac{1}{n}\right)=\mathrm{cos}\left(\frac{1}{\mathrm{\infty }}\right)=\mathrm{cos}\left(0\right)=1$
So that ${t}_{n}$ converges to $1$