Answer
a) 20
b) 0
c) $+\infty$
d) $+\infty$
e) $\sqrt[3] -42$ or does not exist
f) $-6/7$
g) 7
h) $-7/12$
Work Step by Step
As we have been given that $\lim\limits_{x \to -\infty}f(x)$ = 7 and $\lim\limits_{x \to -\infty}g(x)$ = -6
a) $\lim\limits_{x \to -\infty}[2f(x)-g(x)]$ = $2\lim\limits_{x \to -\infty}f(x) - \lim\limits_{x \to -\infty}g(x)$
By putting values
= $2(7) - (-6)$
= 20
b) $\lim\limits_{x \to -\infty}[6f(x)+7g(x)]$ = $6\lim\limits_{x \to -\infty}f(x)$ + $7\lim\limits_{x \to -\infty}g(x)$
By putting values
= $6(7) + 7(-6)$
= 0
c) $\lim\limits_{x \to -\infty}[x^{2} + g(x)]$ = $\lim\limits_{x \to -\infty}x^{2} + \lim\limits_{x \to -\infty}g(x)$
By putting values
= $(-\infty)^{2}+(-6)$
= $+\infty$
d) $\lim\limits_{x \to -\infty}[x^{2}g(x)]$ = $(\lim\limits_{x \to -\infty}x^{2}) (\lim\limits_{x \to -\infty}g(x))$
By putting values
= $[(-\infty)^{2}(-6)]$
= $+\infty$
e) $\lim\limits_{x\to -\infty}\sqrt[3] (f(x)g(x))$ = $\sqrt[3] (\lim\limits_{x \to -\infty}f(x))(\lim\limits_{x \to -\infty}g(x))$
By putting values
=$\sqrt[3] (7)(-6)$
=$\sqrt[3] -42$ or does not exist
f) $\lim\limits_{x \to -\infty}\frac{g(x)}{f(x)}$ = $\frac{\lim\limits_{x \to -\infty}g(x)}{\lim\limits_{x \to -\infty}f(x)}$
By putting values
= $\frac{-6}{7}$
g) $\lim\limits_{x \to -\infty}[f(x) + \frac{g(x)}{x}]$ = $\lim\limits_{x \to -\infty}f(x) + \frac{\lim\limits_{x \to -\infty}g(x)}{\lim\limits_{x \to -\infty}x}$
By putting values
= $7+ \frac{(-6)}{(-\infty)}$
= 7 + 0
= 7
h) $\lim\limits_{x \to -\infty}\frac{xf(x)}{(2x+3)g(x)}$ = $\frac{(\lim\limits_{x \to -\infty}x)(\lim\limits_{x \to -\infty}f(x))}{(2\lim\limits_{x \to -\infty}x+3)(\lim\limits_{x \to -\infty}g(x))}$
By putting values
= $\frac{7}{(2+\frac{3}{-\infty})(-6)}$
= $\frac{7}{(2+0)(-6)}$
= $\frac{-7}{12}$
