## Thomas' Calculus 13th Edition

$\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is divergent.
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{1-n^3}{70-4n^2}$ and $\lim\limits_{n \to \infty} \dfrac{1-n^3}{70-4n^2}=\dfrac{\infty}{\infty}$ ($\bf{Indeterminate form}$) Now, apply L-Hospital's rule. $\lim\limits_{n \to \infty} \dfrac{3n^2}{8n}=\dfrac{\infty}{\infty}$ ($\bf{Indeterminate form}$) Need to apply L-Hospital's rule again. This implies , we get, $\lim\limits_{n \to \infty} \dfrac{6n}{8}=\infty$ Hence, $\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is divergent.