Límites de Secuencias Racionales I | math4u.vsb.cz

Enviado por vladimir.arzt el Mié, 08/28/2024 - 11:15

Project ID:

7100000061

Accepted:

SubArea:

Límite de una sucesión

Tipo:

Layout:

Question:

Relaciona las sucesiones dadas con sus límites (finitos o infinitos).

Question 1:

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

Answer 1:

- \frac{3}{2}

Question 2:

{(\frac{- 3 n + 2 n^{5}}{3^{5} - 3 n^{3} + 2 n})}_{n = 1}^{\infty}

Answer 2:

\frac{2}{3}

Question 3:

{(\frac{2 n^{6}}{(3 n^{4} - 2) (3 - n^{2})})}_{n = 1}^{\infty}

Answer 3:

- \frac{2}{3}

Question 4:

{(\frac{2 - 3 n}{3 - 2 n})}_{n = 1}^{\infty}

Answer 4:

\frac{3}{2}

Question 5:

{(\frac{3 n^{3}}{n - 2 n^{2}})}_{n = 1}^{\infty}

Answer 5:

$- \infty$

Question 6:

{(\frac{(3 n^{4} - 2) (2 + n^{3})}{(2 n^{4} - 3) (3 + n^{2})})}_{n = 1}^{\infty}

Answer 6:

\infty

Question 7:

{(\frac{(3 n^{4} - 2) (2 + n)}{(2 n^{4} - 3) (3 - n^{2})})}_{n = 1}^{\infty}

Answer 7:

0

Question 8:

{(\frac{3 n^{2}}{(3 n + 3) 2 n})}_{n = 1}^{\infty}

Answer 8:

\frac{1}{2}

Question 9:

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

Answer 9:

- \frac{1}{2}