Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 1

Determine the limits that exist. If the limits does not exist, explain why. Given that
$\quad \lim\limits_{x \rightarrow 2} f(x) = 4 \qquad \lim\limits_{x \rightarrow 2} g(x) = -2 \qquad \lim\limits_{x \rightarrow 2} h(x) = 0$


$\begin{equation} \begin{aligned} \text{(a) }& \lim\limits_{x \to 2} \quad [f(x) + 5g(x)] & \text{(b) }& \lim\limits_{x \to 2} \quad [g(x)]^3\\ \text{(c) }& \lim\limits_{x\to 2} \quad \sqrt{f(x)} & \text{(d) }& \lim\limits_{x\to 2} \quad \frac{3f(x)}{g(x)}\\ \text{(e) }& \lim\limits_{x\to 2} \quad \frac{g(x)}{h(x)} & \text{(f) }& \lim\limits_{x\to 2} \quad \frac{g(x)h(x)}{f(x)} \end{aligned} \end{equation}$



a.) $\lim\limits_{x \to 2} \quad [f(x) + 5g(x)]$

$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 2} \quad[f(x) + 5g(x)] & = \lim\limits_{x \rightarrow 2} f(x) + \lim\limits_{x \rightarrow 2} 5 g(x) && \text{(Substitute the given values.)}\\ \lim\limits_{x \rightarrow 2} \quad[f(x) + 5g(x)] & = 4 +5(-2) & &\text{(Simplify.)}\\ \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \rightarrow 2} \quad[f(x) + 5g(x)] = -6 }$


b.)$\lim\limits_{x \to 2} \quad [g(x)]^3$

$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 2} \quad [g(x)]^3 &&& \text{(Substitute the given value.)}\\ \lim\limits_{x \rightarrow 2} \quad [g(x)]^3 &= (-2)^3 && \text{(Simplify)}\\ \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \rightarrow 2} \quad [g(x)]^3 = -8}$


c.) $\lim\limits_{x\to 2} \quad \sqrt{f(x)}$

$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 2} \quad \sqrt{f(x)} &&& \text{(Substitute the given value)}\\ \lim\limits_{x \rightarrow 2} \quad \sqrt{f(x)} &= \sqrt{4} && \text{(Simplify)}\\ \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \rightarrow 2} \quad \sqrt{f(x)} = 2}$


d.) $\lim\limits_{x\to 2} \quad \frac{3f(x)}{g(x)}$

$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 2} \quad \displaystyle\frac{3f(x)}{g(x)} &&& \text{(Substitute the given values)}\\ \lim\limits_{x \rightarrow 2} \quad \displaystyle\frac{3f(x)}{g(x)} &= \frac{3(4)}{-2} && \text{(Simplify)}\\ \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \rightarrow 2} \quad \displaystyle\frac{3f(x)}{g(x)} =-6}$


e.) $\lim\limits_{x\to 2} \quad \frac{g(x)}{h(x)}$

$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{g(x)}{h(x)}& \qquad \qquad \qquad \text{(Substitute the given values)}\\ \lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{g(x)}{h(x)}& = \frac{-2}{0}\\ \end{aligned} \end{equation}\\ \boxed{\text{Limit does not exist, the function is undefined because denominator is zero.} }$


f.) $\lim\limits_{x\to 2} \quad \frac{g(x)h(x)}{f(x)}$

$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{g(x)h(x)}{f(x)} & && \text{(Substitute the given values)}\\ \lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{g(x)h(x)}{f(x)} &= \frac{(-2)(0)}{4} && \text{(Simplify)} \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{g(x)h(x)}{f(x)} = 0}$