Answer
$\infty$
Work Step by Step
Our aim is to compute the value of $\lim\limits_{x \to -\infty} \dfrac{8+0.5^{x}}{2-3^{2x}}$.
Here, we need to use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, this implies that:
$\lim\limits_{x \to -\infty} \dfrac{8+0.5^{x}}{2-3^{2x}}= \dfrac{8+0.5^{-\infty}}{2-3^{2(-\infty)}}\\=\dfrac{+\infty}{2-0}\\=\infty$
