Answer
$2.1$
Work Step by Step
When $x\rightarrow-\infty$, we have:
The numerator is a constant: $k=4.2$
In the denominator,
$3^{2x}=$ has the form $k^{-\infty}=0 \implies k^{Big\ negative }= Small$
So, $2-3^{2x}\rightarrow(2-$ small $)\rightarrow 2$
Thus, we have:
$\displaystyle \lim_{x\rightarrow-\infty}\frac{4.2}{2-3^{2x}}=\frac{4.2}{2}=2.1$
