 # Find the limits of the following sequences ({\cos((2n+1)\frac{\pi}{2})})_{n=1}^{\infty} ({\frac{\pi^n}{e^{2n+1}}})_{n=1}^{\infty} ({\frac{n^2+3}{n^3+n^2-1}})_{n=1}^{\infty} ({n\sin\frac{\pi}{n}})_{n=1}^{\infty} Ethen Wong 2022-01-24 Answered
Find the limits of the following sequences
${\left(\left\{\mathrm{cos}\left(\left(2n+1\right)\frac{\pi }{2}\right)\right\}\right)}_{n=1}^{\mathrm{\infty }}$
${\left(\left\{\frac{{\pi }^{n}}{{e}^{2n+1}}\right\}\right)}_{n=1}^{\mathrm{\infty }}$
${\left(\left\{\frac{{n}^{2}+3}{{n}^{3}+{n}^{2}-1}\right\}\right)}_{n=1}^{\mathrm{\infty }}$
${\left(\left\{n\mathrm{sin}\frac{\pi }{n}\right\}\right)}_{n=1}^{\mathrm{\infty }}$
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1)${\left(\left\{\mathrm{cos}\left(\left(2n+1\right)\frac{\pi }{2}\right)\right\}\right)}_{n=1}^{\mathrm{\infty }}$
$\underset{n\to \mathrm{\infty }}{lim}\left(\mathrm{cos}\left(2n+1\right)\frac{\pi }{2}\right)$
$=\mathrm{cos}\left(\mathrm{\infty }\right)\frac{\pi }{2}$
$=\mathrm{cos}\mathrm{\infty }=O\ne$
$=\underset{n\to \mathrm{\infty }}{lim}\left[\mathrm{cos}\left(2n+1\right)\frac{\pi }{2}\right]=o\ne$
2)${\left(\left\{\frac{{\pi }^{n}}{{e}^{2n+1}}\right\}\right)}_{n=1}^{\mathrm{\infty }}$
$\underset{n\to \mathrm{\infty }}{lim}\frac{{\pi }^{n}}{{e}^{2n+1}}$
$\underset{n\to \mathrm{\infty }}{lim}\frac{{\pi }^{n}}{{\left({e}^{2}\right)}^{n}×e}$
$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{{\pi }^{n}}{{e}^{2}}\right)}^{n}×\frac{1}{e}$
$={\left(\frac{\pi }{{e}^{2}}\right)}^{\mathrm{\infty }}×\frac{1}{e}=0$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{{\pi }^{n}}{{e}^{2n+1}}=0$

###### Not exactly what you’re looking for? bemolizisqt
(c)${\left(\left\{\frac{{n}^{2}+3}{{n}^{3}+{n}^{2}-1}\right\}\right)}_{n=1}^{\mathrm{\infty }}$
$\underset{n\to \mathrm{\infty }}{lim}n\mathrm{sin}\frac{\pi }{n}$
$\frac{\mathrm{sin}\frac{\pi }{n}}{\frac{1}{n}}$
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\mathrm{sin}\frac{\pi }{n}}{\frac{\pi }{n}}\right)$
$=\pi ×1$
$=\pi$
$\left(\because \underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{sin}\frac{1}{x}}{\frac{1}{x}}\right)=1$
$=\underset{n\to \mathrm{\infty }}{lim}\left(n\mathrm{sin}\frac{\pi }{n}\right)=\pi$
(d)${\left(\left\{n\mathrm{sin}\frac{\pi }{n}\right\}\right)}_{n=1}^{\mathrm{\infty }}$
$\underset{n\to \mathrm{\infty }}{lim}\frac{{n}^{2}+3}{{n}^{3}+{n}^{2}-1}$
$\underset{n\to \mathrm{\infty }}{lim}\frac{{n}^{2}\left(1+\frac{3}{{n}^{2}}\right)}{{n}^{3}\left(1+\frac{1}{n}-\frac{1}{{n}^{3}}\right)}$
$\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{n}\left(1+\frac{3}{{n}^{2}}\right)}{1+\frac{1}{n}-\frac{1}{{n}^{3}}}$
$\frac{\frac{1}{\mathrm{\infty }}\left(1+\frac{3}{\mathrm{\infty }}\right)}{1+\frac{1}{\mathrm{\infty }}-\frac{1}{\mathrm{\infty }}}$
$=\frac{0×\left(1+0\right)}{1+0-0}$
$=0$