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}
$