Evaluate each limit if it exists. If the limit does not exist, explain why. Use the given graphs of $f$ and $g$
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\begin{equation}
\begin{aligned}
\text{(a) }& \lim\limits_{x \rightarrow 2} \quad [f(x)+g(x)] &
\text{(b) }& \lim\limits_{x \rightarrow 1} \quad [f(x)+g(x)]\\
\text{(c) }& \lim\limits_{x \rightarrow 0} \quad [f(x)g(x)] &
\text{(d) }& \lim\limits_{x \rightarrow -1} \quad \frac{f(x)}{g(x)}\\
\text{(e) }& \lim\limits_{x \rightarrow 2} \quad [x^3f(x)] &
\text{(f) }& \lim\limits_{x \rightarrow 2} \quad \sqrt{3 + f(x)}
\end{aligned}
\end{equation}
$
a.) $ \lim\limits_{x \rightarrow 2} \quad [f(x)+g(x)] $
$
\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow 2} \quad [f(x)+g(x)] & = \lim\limits_{x \rightarrow 2} f(x) + \lim\limits_{x \rightarrow 2} g(x) && \text{(Substitute the values of }f(x) \text{ and } g(x))\\
\lim\limits_{x \rightarrow 2} \quad [f(x)+g(x)] & = 2 + 0 && \text{(Simplify)} \\
\end{aligned}
\end{equation}\\
\boxed{\lim\limits_{x \rightarrow 2} \quad [f(x)+g(x)] = 2 }
$
b.) $ \lim\limits_{x \rightarrow 1} \quad [f(x)+g(x)] $
$
\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow 1} \quad [f(x)+g(x)] & = \lim\limits_{x \rightarrow 1} f(x) + \lim\limits_{x \rightarrow 1} g(x) && \text{(Substitute the values of }f(x) \text{ and } g(x))\\
\lim\limits_{x \rightarrow 1} \quad f(x) & = 1 \\
\lim\limits_{x \rightarrow 1} \quad g(x) & = \text{Does not exist}\\
\end{aligned}
\end{equation}\\
\fbox{The given equation does not exist because the left and the right limits of the function $g(x)$ are different.}
$
c.) $ \lim\limits_{x \rightarrow 0} \quad [f(x)g(x)] $
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\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow 0} \quad [f(x)g(x)] &= \lim\limits_{x \rightarrow 0} f(x) \cdot \lim\limits_{x \rightarrow 0} g(x) && \text{(Substitute the values of }f(x) \text{ and } g(x))\\
\lim\limits_{x \rightarrow 0} \quad [f(x)g(x)] &= (0)(1.5) && \text{(Simplify)}
\end{aligned}
\end{equation}\\
\boxed{\lim\limits_{x \rightarrow 0} \quad [f(x)g(x)] = 0}
$
d.) $ \lim\limits_{x \rightarrow -1} \quad \frac{f(x)}{g(x)} $
$
\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow -1} \quad \displaystyle \frac{f(x)}{g(x)} &= \frac{\lim\limits_{x \rightarrow -1} \quad f(x)}{\lim\limits_{x \rightarrow -1} \quad g(x)} && \text{(Substitute the values of }f(x) \text{ and } g(x))\\
\lim\limits_{x \rightarrow -1} \quad \displaystyle \frac{f(x)}{g(x)} &= \frac{-1}{0} && \text{(Does not exist)}\\
\end{aligned}
\end{equation}\\
\boxed{\text{The limit does not exist, the function is undefined because the denominator is zero.}}
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e.) $ \lim\limits_{x \rightarrow 2} \quad [x^3f(x)] $
$
\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow 2} \quad [x^3f(x)] &= \lim\limits_{x \rightarrow 2} x^3 \cdot \lim\limits_{x \rightarrow 2} f(x)\\
\lim\limits_{x \rightarrow 2} \quad [x^3f(x)] &= (2)^3(2)\\
\end{aligned}
\end{equation}\\
\boxed{\lim\limits_{x \rightarrow 2} \quad [x^3f(x)] = 16}
$
f.) $ \lim\limits_{x \rightarrow 2} \quad \sqrt{3 + f(x)} $
$
\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow 2} \quad \sqrt{3+f(x)} &&& \text{(Substitute the value of } f(x))\\
\lim\limits_{x \rightarrow 2} \quad \sqrt{3+f(x)} &= \sqrt{3+2} && \text{(Simplify)}
\end{aligned}
\end{equation}\\
\boxed{\lim\limits_{x \rightarrow 2} \quad \sqrt{3+f(x)} = \sqrt{5}}
$
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