Question

1. $lim_{x

ightarrow-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