Answer
(a) -6
(b) -8
(c) 2
(d) -6
(e) does not exist because the denominator is 0
(f) 0
Work Step by Step
It is given that
$\lim\limits_{x \to 2}$$f(x)=4$
$\lim\limits_{x \to 2}$$g(x)=-2$
$\lim\limits_{x \to 2}$$h(x)=0$
(a) $\lim\limits_{x \to 2}$ $[f(x) + 5g(x)]$
Apply the addition law and the constant law.
$\lim\limits_{x \to 2}$$f(x) + 5$$\lim\limits_{x \to 2}$$g(x)$
Substitute.
$(4) + 5(-2)$
Clean up and simplify
$4-10$
$$-6$$
(b) $\lim\limits_{x \to 2}$$[g(x)]^3$
Apply the power law.
$(\lim\limits_{x \to 2}g(x))^3$
Substitute.
$(-2)^3$
$$-8$$
(c) $\lim\limits_{x \to 2}\sqrt {f(x)}$
Apply root law.
$\sqrt {\lim\limits_{x \to 2}f(x)}$
Substitute.
$\sqrt 4$
$$2$$
(d) $\lim\limits_{x \to 2}\frac{3f(x)}{g(x)}$
Apply the division law and the constant law.
$\frac{3\lim\limits_{x \to 2}f(x)}{\lim\limits_{x \to 2}g(x)}$
Substitute.
$\frac{3(4)}{(-2)}$
$$-6$$
(e) $\lim\limits_{x \to 2}\frac{g(x)}{h(x)}$
Apply the division law.
$\frac{\lim\limits_{x \to 2}g(x)}{\lim\limits_{x \to 2}h(x)}$
Substitute.
$\frac{-2}{0}$
The limit does not exist because the denominator is 0.
(f)$\lim\limits_{x \to 2}\frac{g(x)h(x)}{f(x)}$
Apply the multiplication law and the division law.
$\frac{\lim\limits_{x \to 2}g(x)*\lim\limits_{x \to 2}h(x)}{\lim\limits_{x \to 2}f(x)}$
Substitute.
$\frac{(-2)(0)}{(4)}$
Simplify.
$\frac{0}{4}$
$$0$$
