Práctica 3 ejercicio 5c

EJERCICIO 5: ...

(c) $a_n=\dfrac{\sqrt{n^3}+2}{n^2-1}$

Solución:

$\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty} \dfrac{\sqrt{n^3}+2}{n^2-1} = \lim\limits_{n\to \infty}\dfrac{\sqrt{n^3} \left(1+\dfrac{2}{\sqrt{n^3}}\right)}{n^2\left(1-\dfrac{1}{n^2}\right)} = $

$\lim\limits_{n\to \infty}\dfrac{\sqrt{n}^3 \left(1+\dfrac{2}{\sqrt{n^3}}\right)}{\sqrt{n}^4\left(1-\dfrac{1}{n^2}\right)} = \lim\limits_{n\to \infty}\dfrac{\left(1+\dfrac{2}{\sqrt{n^3}}\right)}{\sqrt{n}\left(1-\dfrac{1}{n^2}\right)} =$

$\lim\limits_{n\to \infty}\dfrac{\overbrace{\left(1+\dfrac{2}{\sqrt{n^3}}\right)}^{\to 1}}{\underbrace{\sqrt{n}}_{\to +\infty}\underbrace{\left(1-\dfrac{1}{n^2}\right)}_{\to 1}} = 0$