Answer
$$\lim_{x\to-\infty}\frac{x^2-7x}{x+1}=-\infty$$
Work Step by Step
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$
$$A=\lim_{x\to-\infty}\frac{x^2-7x}{x+1}$$
Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$ in this case:
$$A=\lim_{x\to-\infty}\frac{x-7}{1+\frac{1}{x}}$$
$$A=\frac{\lim_{x\to-\infty}(x-7)}{1+0}=\lim_{x\to-\infty}(x-7)$$
As $x\to-\infty$, $(x-7)$ approaches $-\infty$. Therefore, $$A=-\infty$$
