Answer
$$ - 4$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } \frac{{1 - 4x}}{{x + 1}} \cr
& = \frac{{1 - 4\left( \infty \right)}}{{\left( \infty \right) + 1}} = \frac{\infty }{\infty } \cr
& {\text{Divide the numerator and denominator by }}x \cr
& \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{1}{x} - \frac{{4x}}{x}}}{{\frac{x}{x} + \frac{1}{x}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{1}{x} - 4}}{{1 + \frac{1}{x}}} \cr
& {\text{By properties of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{1}{x}} \right) - \mathop {\lim }\limits_{x \to - \infty } \left( 4 \right)}}{{\mathop {\lim }\limits_{x \to - \infty } \left( 1 \right) + \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{1}{x}} \right)}} \cr
& {\text{Evaluating}} \cr
& = \frac{{\frac{1}{{ - \infty }} - 4}}{{1 + \frac{1}{{ - \infty }}}} = \frac{{0 - 4}}{{1 - 0}} \cr
& = - 4 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to - \infty } \frac{{1 - 4x}}{{x + 1}} = - 4 \cr} $$
