Ethen Wong
2022-01-24
Answered

Find the limits of the following sequences

$\left(\left\{\mathrm{cos}\left((2n+1)\frac{\pi}{2}\right)\right\}\right)}_{n=1}^{\mathrm{\infty}$

$\left(\left\{\frac{{\pi}^{n}}{{e}^{2n+1}}\right\}\right)}_{n=1}^{\mathrm{\infty}$

$\left(\left\{\frac{{n}^{2}+3}{{n}^{3}+{n}^{2}-1}\right\}\right)}_{n=1}^{\mathrm{\infty}$

$\left(\left\{n\mathrm{sin}\frac{\pi}{n}\right\}\right)}_{n=1}^{\mathrm{\infty}$

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waijazar1

Answered 2022-01-25
Author has **13** answers

1)

2)

bemolizisqt

Answered 2022-01-26
Author has **16** answers

(c)$\left(\left\{\frac{{n}^{2}+3}{{n}^{3}+{n}^{2}-1}\right\}\right)}_{n=1}^{\mathrm{\infty}$

$\underset{n\to \mathrm{\infty}}{lim}n\mathrm{sin}\frac{\pi}{n}$

$\frac{\mathrm{sin}\frac{\pi}{n}}{\frac{1}{n}}$

$\underset{n\to \mathrm{\infty}}{lim}\left(\frac{\mathrm{sin}\frac{\pi}{n}}{\frac{\pi}{n}}\right)$

$=\pi \times 1$

$=\pi$

$(\because \underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{sin}\frac{1}{x}}{\frac{1}{x}})=1$

$=\underset{n\to \mathrm{\infty}}{lim}\left(n\mathrm{sin}\frac{\pi}{n}\right)=\pi$

(d)$\left(\left\{n\mathrm{sin}\frac{\pi}{n}\right\}\right)}_{n=1}^{\mathrm{\infty}$

$\underset{n\to \mathrm{\infty}}{lim}\frac{{n}^{2}+3}{{n}^{3}+{n}^{2}-1}$

$\underset{n\to \mathrm{\infty}}{lim}\frac{{n}^{2}(1+\frac{3}{{n}^{2}})}{{n}^{3}(1+\frac{1}{n}-\frac{1}{{n}^{3}})}$

$\underset{n\to \mathrm{\infty}}{lim}\frac{\frac{1}{n}(1+\frac{3}{{n}^{2}})}{1+\frac{1}{n}-\frac{1}{{n}^{3}}}$

$\frac{\frac{1}{\mathrm{\infty}}(1+\frac{3}{\mathrm{\infty}})}{1+\frac{1}{\mathrm{\infty}}-\frac{1}{\mathrm{\infty}}}$

$=\frac{0\times (1+0)}{1+0-0}$

$=0$

(d)

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