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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 4B Solutions, October 6, 2017}
\vskip 10pt
\noindent
Section 2.3
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\noindent
40.  By the Limit Sum Law, 
$$\lim_{x\to -4} (2f(x) + 3g(x)) = \lim_{x\to -4} 2f(x) + \lim_{x\to -4} 3g(x)$$
Then by the Limit Product Law and the given information, this equals
\begin{align*}
&= (\lim_{x\to -4} 2) \cdot (\lim_{x\to -4} f(x)) + (\lim_{x\to -4} 3)\cdot(\lim_{x\to -4}g(x))\\
&= 2\cdot 3 + 3 \cdot 1\\
&= 9.
\end{align*}
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\noindent
Section 2.4  
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\noindent
64.  Any graph $y = f(x)$ for which 
\begin{itemize}
\item $\lim_{x\to 2^-} f(x) = f(2)$, but $\lim_{x \to 2^+} f(x)$ is something
other than $f(2)$ (it could be a different number -- so a jump discontinuity at $x = 2$ or even an
infinite discontinuity from the right side),  and
\item $\lim_{x\to 3^+} f(x) = f(3)$, but $\lim_{x \to 3^-} f(x)$ is something
other than $f(3)$ (it could be a different number -- so a jump discontinuity at $x = 2$ or even an
infinite discontinuity from the right side).
\end{itemize}
is OK.  One example is shown in Figure 1.
  \begin{figure}
 \begin{center}
 \includegraphics[width=3.0in]{TwoPointFour64.png} 
 \caption{$y = f(x)$ for Question 2.4/64.}
\end{center}
 \end{figure}
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\noindent
66.  This one is similar to 64 -- any graph for which 
\begin{itemize}
\item $\lim_{x\to 1^+} f(x) = f(1)$, but $\lim_{x \to 1^-} f(x)$ is something
other than $f(1)$,  and
\item $\lim_{x\to 2^-} f(x) = f(2)$, but $\lim_{x \to 2^+} f(x)$ is something
other than $f(2)$,
\item $\lim_{x\to 3^-} f(x)$ and $\lim_{x\to 3^+} f(x)$ both different
from $f(3)$.  This could be because at least one of the one-sided limit does not exist (e.g. infinite
discontinuities), or because $f(x)$ has a removable singularity at $x = 3$, with
$f(3)$ different from the one-sided limits.)
\end{itemize}
is OK.  One example is shown in Figure 2.
  \begin{figure}
 \begin{center}
 \includegraphics[width=3.0in]{TwoPointFour66.png} 
 \caption{$y = f(x)$ for Question 2.4/66.}
\end{center}
 \end{figure}
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\noindent
Section 2.5/
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\noindent
8.  Factor the top and cancel $x - 8$ to get a determinate limit:
\begin{align*}
\lim_{x\to 8} \frac{x^3 - 64x}{x - 8} &= \lim_{x\to 8} \frac{x(x + 8)(x - 8)}{(x - 8)}\\
&= \lim_{x\to 8} x(x + 8) = 128.
\end{align*}
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\noindent
19.  Put the terms on the top over a common denominator, simplify the fraction, 
and work to cancel a factor of $h$ top and bottom:
\begin{align*}
\lim_{h\to 0} \frac{\frac{1}{(h+2)^2} - \frac{1}{4}}{h} = \lim_{h\to 0} \frac{\frac{4 - (h+2)^2}{4(h + 2)^2}}{h}\\
&= \lim_{h\to 0} \frac{4 - (h+2)^2}{4h(h+2)^2}\\
&= \lim_{h\to 0} \frac{4 - (h^2 + 4h + 4)}{4h(h+2)^2}\\
&= \lim_{h\to 0} \frac{-h(h + 4)}{4h(h+2)^2} \qquad {\rm (now\ cancel\ } h)\\
&= \lim_{h\to 0} \frac{-(h+4)}{4(h+2)^2}\\
&= \frac{-1}{4}.
\end{align*}
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\noindent
22.  Multiply by the conjugate radical top and bottom, simplify, cancel $x - 8$ and evaluate:
\begin{align*}
\lim_{x\to 8} \frac{\sqrt{x - 4} - 2}{x - 8} &=
\lim_{x\to 8} \frac{(\sqrt{x - 4} - 2)(\sqrt{x-4}+2)}{(x - 8)(\sqrt{x-4} + 2)}\\
&= \lim_{x\to 8} \frac{(x - 4) - 4}{(x-8)(\sqrt{x-4} + 2)}\\
&= \lim_{x\to 8} \frac{(x-8)}{(x-8)(\sqrt{x-4} + 2)}\\
&= \lim_{x\to 8} \frac{1}{\sqrt{x-4} + 2}\\
&= \frac{1}{4}.
\end{align*}
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1^+} f(x) = 1$.
This implies $\lim_{x\to 1} f(x)$ \emph{does not exist} (the one-sided limits are not
equal).
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