3. $lim_{x\rightarrow-infty}-(0.25)^{-x + 1}+5$

Explanation:

Step1: Rewrite the exponential term

We know that (a^{-b}=\frac{1}{a^{b}}), so ((0.25)^{-x + 1}=(\frac{1}{4})^{-x + 1}=4^{x - 1}=\frac{4^{x}}{4}).

Step2: Find the limit

As (x\to-\infty), for the exponential function (y = a^{x}) where (a>1) (here (a = 4)), (\lim_{x\to-\infty}4^{x}=0). So (\lim_{x\to-\infty}-(0.25)^{-x + 1}+5=-\lim_{x\to-\infty}\frac{4^{x}}{4}+5). Since (\lim_{x\to-\infty}4^{x}=0), then (-\lim_{x\to-\infty}\frac{4^{x}}{4}+5=- \frac{0}{4}+5).

Answer:

5