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Indeterminación en límites de sucesiones

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Indeterminación en límites de sucesiones

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Desafío sobre límites y formas indeterminadas

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jose de la hoz
jose de la hoz
Colombia

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Indeterminación en límites de sucesiones
 

Indeterminación en límites de sucesionesOnline version

Desafío sobre límites y formas indeterminadas

by jose de la hoz
1

1) Lim n→∞ (√(n+1) - √n). Forma indeterminada: ∞-∞. ¿Resultado correcto?

2

2) Lim n→∞ (n^2)/(n^2+2n). Forma ∞/∞. ¿Valor límite?

3

3) Lim n→∞ (ln(n))/n. Indeterminación?

4

4) Lim n→∞ (√(n+1) - √n) / (1/√n). ¿Qué forma resulta y límite?

5

5) Lim n→∞ (n+1) - n. ¿Forma indeterminada?

6

6) Lim n→∞ (n^2+n)/(n^2-n). Indeterminación?

7

7) Lim n→∞ (e^n)/(n^n). Indeterminación?

8

8) Lim n→∞ (√(n^2+n) - n). Indeterminación?

9

9) Lim n→∞ (ln(n^2+1) - 2 ln n). Indeterminación?

10

10) Lim n→∞ (n·(1+1/n)^n). Indeterminación?

Explicación

La diferencia de raíces se resuelve racionalizando: [(n+1)-n]/(√(n+1)+√n) = 1/(√(n+1)+√n) → 0.

Dividir por n^2: (1)/(1+2/n) → 1.

∞/∞ o 0/0 pueden aparecer; aquí crece más lento que n, así el límite es 0, pero se justifica comparando con 1/n.

Racionalizando: (1)/(√(n+1)+√n)·√n → 0.

Es una resta de términos que divergen, pero el resultado es 1 (deriva de telescopado).

Ambos términos dominan por n^2; cociente tiende a 1.

Formas ∞/∞ y resulta extremadamente pequeño; uso logaritmo puede mostrar límite 0.

√(n^2+n)=n√(1+1/n) ≈ n(1+1/(2n)) = n+1/2, resta → 1/2.

Se puede reescribir como ln((n^2+1)/n^2)=ln(1+1/n^2)→0.

(1+1/n)^n→e, entonces n·e diverge; forma ∞ con crecimiento exponencial.

 
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