Answer
$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{1-n^3}{70-4n^2}$
and $\lim\limits_{n \to \infty} \dfrac{1-n^3}{70-4n^2}=\dfrac{\infty}{\infty}$
WenNeed to apply L-Hospital's rule:
$=\lim\limits_{n \to \infty} \dfrac{3n^2}{8n}$
Again using L-Hospital's rule.
This implies , we get, $= \lim\limits_{n \to \infty} \dfrac{6n}{8}$
or, $=\infty$
Thus, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
