Answer
$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
Work Step by Step
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.
