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\pagestyle{fancy}
\lhead{\color{blue} \large Lecture Notes}
\chead{\color{black} \Large Limits at Infinity - Part1}
\rhead{ page \ \thepage}
\cfoot{}
\lfoot{\small \copyright $\;$ Hidegkuti, Powell, 2009}
\rfoot{\small Last revised: April 14, 2015}
\textwidth 7.5in
\textheight 9.7in
\setlength{\headheight}{26pt}
\setlength{\parindent}{0in}
\begin{document}
\begin{center}
{\LARGE Sample Problems}\bigskip
\end{center}
\begin{enumerate}
\item (Monomials) \ Compute each of the following limits.\medskip
a) $\ \lim\limits_{x\rightarrow \infty }3x^{4}$\medskip\ \ \ \ \ \ \ \ \ \
c) $\ \lim\limits_{x\rightarrow \infty }\left( -2x^{5}\right) $ \ \ \ \ \ \
\ e) \ $\lim\limits_{x\rightarrow \infty }\left( -\dfrac{2}{3}x^{6}\right) $
\ \ \ \ \ \ \ g) \ $\lim\limits_{x\rightarrow \infty }4x^{3}$
b) \ $\lim\limits_{x\rightarrow -\infty }3x^{4}$ \ \ \ \ \ \ \ d) \ $%
\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}\right) $\medskip\ \ \ \ \
\ f) \ $\lim\limits_{x\rightarrow -\infty }\left( -\dfrac{2}{3}x^{6}\right) $
\ \ \ \ \ \ h) \ $\lim\limits_{x\rightarrow -\infty }4x^{3}$
\item (Exponential Functions) \ \ Compute each of the following
limits.\medskip
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a) $\ \lim\limits_{x\rightarrow \infty }2^{x}$
b) $\ \lim\limits_{x\rightarrow -\infty }2^{x}$
c) \ $\lim\limits_{x\rightarrow \infty }\left( \dfrac{2}{3}\right) ^{x}$
d) \ $\lim\limits_{x\rightarrow -\infty }\left( \dfrac{2}{3}\right) ^{x}$
e) $\ \lim\limits_{x\rightarrow \infty }\dfrac{2^{x+3}}{3^{x+1}}$
f) $\ \lim\limits_{x\rightarrow -\infty }\dfrac{2^{x+3}}{3^{x+1}}$
g) $\ \lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+1}}{3^{x-1}}$
h) $\ \lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+1}}{3^{x-1}}$ \
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\item (Basic Functions) \ Compute each of the following limits.\medskip
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a) \ $\ \lim\limits_{x\rightarrow \infty }\dfrac{1}{x}$\medskip\ \ \ \ \ \ \
\ \ \ \
b) \ $\ \lim\limits_{x\rightarrow -\infty }\dfrac{1}{x}$\medskip\ \ \ \ \ \
\ \
c) $\ \lim\limits_{x\rightarrow \infty }\dfrac{-5}{2x^{3}}$ \ \ \ \ \ \ \ \
d) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{3x-2}{x}$\medskip\
e) \ $\lim\limits_{x\rightarrow \infty }\sqrt{x}$\medskip
f) \ $\lim\limits_{x\rightarrow -\infty }\sqrt{x}$\medskip
g) \ $\lim\limits_{x\rightarrow \infty }\log _{3}x$\medskip
h) \ $\lim\limits_{x\rightarrow -\infty }\log _{3}x$\medskip
i) \ $\lim\limits_{x\rightarrow \infty }7$\medskip
j) \ $\lim\limits_{x\rightarrow -\infty }7$\medskip
k*) \ $\lim\limits_{x\rightarrow \infty }\left( 3^{x+2}-3^{x}\right) $
l) \ $\lim\limits_{x\rightarrow -\infty }\left( 3^{x+2}-3^{x}\right) $
m*) \ $\lim\limits_{x\rightarrow \infty }\left( \log _{10}3x-\log
_{10}x\right) $
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\bigskip \bigskip \bigskip
\end{enumerate}
\begin{center}
{\LARGE Practice Problems}\bigskip
\end{center}
\begin{enumerate}
\item Compute each of the following limits.\medskip
a) $\ \lim\limits_{x\rightarrow \infty }\left( -\dfrac{3}{8}x^{15}\right) $
\ \ \ \ \ \ \ c) $\ \lim\limits_{x\rightarrow \infty }\dfrac{1}{3}x^{8}$\ \
\ \ \ \ \ \ \medskip\ \ \ e) $\ \lim\limits_{x\rightarrow \infty }4x^{9}$ \
\ \ \ \ \ \ \ \ \ \ g) \ $\lim\limits_{x\rightarrow \infty }\left(
-7x^{10}\right) $
b) $\ \lim\limits_{x\rightarrow -\infty }\left( -\dfrac{3}{8}x^{15}\right) $
\ \ \ \ \ d) $\ \lim\limits_{x\rightarrow -\infty }\dfrac{1}{3}x^{8}$\ \ \ \
\medskip\ \ \ \ \ f) $\ \ \lim\limits_{x\rightarrow -\infty }4x^{9}\ \ \ \ \
\ \ \ \ \ $h)$\ \ \lim\limits_{x\rightarrow -\infty }\left( -7x^{10}\right) $
\item Compute each of the following limits.\medskip
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a) $\ \lim\limits_{x\rightarrow \infty }\dfrac{2^{3x-1}}{5^{x-1}}$\medskip
\medskip
b) $\ \lim\limits_{x\rightarrow -\infty }\dfrac{2^{3x-1}}{5^{x-1}}$\medskip
\medskip
c) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}}{5^{x-1}}$\medskip
\medskip
d) $\ \lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+3}}{5^{x-1}}$\medskip
\medskip
e) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}}{4^{x-1}}$\medskip
\medskip
f) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+3}}{4^{x-1}}$\medskip
\medskip
g) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x+3}\cdot 3^{x-1}}{7^{x-2}}
$\medskip \medskip
h) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}\cdot 3^{x-1}}{7^{x-2}%
}$ \
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\pagebreak
\item Compute each of the following limits.\medskip
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a) $\ \lim\limits_{x\rightarrow \infty }\dfrac{3}{x^{5}}$\medskip
b) $\ \lim\limits_{x\rightarrow -\infty }\dfrac{3}{x^{5}}$\medskip
c) \ $\lim\limits_{x\rightarrow \infty }\dfrac{5x-3}{x}$\medskip
d) $\ \lim\limits_{x\rightarrow -\infty }\dfrac{5x-3}{x}$\medskip
e) $\ \lim\limits_{x\rightarrow \infty }e^{x}$\medskip\
f) $\ \lim\limits_{x\rightarrow -\infty }e^{x}$\medskip\
g) \ $\lim\limits_{x\rightarrow \infty }\left( 5x-\dfrac{2}{x+3}\right) $%
\medskip
h)\ $\ \lim\limits_{x\rightarrow -\infty }\left( 5x-\dfrac{2}{x+3}\right) $%
\medskip
i) \ $\lim\limits_{x\rightarrow \infty }\dfrac{5x-3}{x}$\medskip
j) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{5x-3}{x}$\medskip
k) \ $\lim\limits_{x\rightarrow \infty }\log _{0.2}x$\medskip
l) \ $\lim\limits_{x\rightarrow -\infty }\log _{0.2}x$\medskip
m) \ $\lim\limits_{a\rightarrow \infty }\dfrac{-3a^{5}+2a-5}{a^{2}}$\medskip
n) $\lim\limits_{a\rightarrow -\infty }\dfrac{-3a^{5}+2a-5}{a^{2}}$\medskip
o) \ $\lim\limits_{x\rightarrow \infty }\left( \ln 5x-\ln x\right) $\medskip
p) \ $\lim\limits_{x\rightarrow -\infty }\left( \ln 5x-\ln x\right) $\medskip
q) \ $\lim\limits_{x\rightarrow \infty }\dfrac{-4x^{8}+x^{3}-x+7}{x^{4}}$%
\medskip
r) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{-4x^{8}+x^{3}-x+7}{x^{4}}$%
\medskip
s) \ $\lim\limits_{x\rightarrow \infty }\left( 2^{x+1}-2^{x}\right) $\medskip
t) \ $\lim\limits_{x\rightarrow -\infty }\left( 2^{x+1}-2^{x}\right) $%
\medskip
u*) \ $\lim\limits_{x\rightarrow \infty }\left( \log _{2}x-\log _{2}\left(
x+1\right) \right) $\medskip
v*) \ $\lim\limits_{x\rightarrow \infty }\left( \log _{2}\left( \dfrac{1}{x}%
\right) \right) $\medskip
x*) \ $\lim\limits_{x\rightarrow \infty }\left( \dfrac{2^{x}+1}{2^{x}}%
\right) $\medskip\ \
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\ \
\item A company is introducing a new product. The marketing manager
determines that $t$ weeks after
an advertising campaign begins (where $t\geq 4$), $P(t)$ percent of the
potential market is aware of the burners, where%
\begin{equation*}
P(t)=75\dfrac{t^{2}-4t-1}{t^{2}}+3\text{ \ \ \ \ \ \ }t\geq 4
\end{equation*}
a) \ What percent of the potential market knows about the product after $7$
weeks?
b) \ What happens to the percentage $P(t)$ in the long run? \ \pagebreak
\end{enumerate}
\begin{center}
{\LARGE Sample Problems - Answers}\bigskip
\end{center}
\begin{enumerate}
\item a) $\ \infty $\medskip\ \ \ \ b) \ $\infty $ \ \ \ c) $\ -\infty $ \ \
\ \ d) \ $\infty $\ \ \ \ \ e) \ $-\infty $ \ \ \ \ f) \ $-\infty $ \ \ \ g)
\ $\infty $ \ \ \ \ h) \ $-\infty $
\item a) $\ \infty $ \ \ \ \ b) $\ 0$ \ \ \ c) \ $0$ \ \ \ \ d) \ $\infty $
\ \ \ \ e) $\ 0$ \ \ \ \ f) $\ \infty $ \ \ \ \ g) $\ \infty $ \ \ \ h) $\ 0$
\item a) \ $\ 0$\medskip\ \ \ \ \ \ \ \ b) \ $\ 0$\ \ \ \ \ \ c) $\ 0$ \ \ \
\ d) \ $3$\ \ \ \ \ e) \ $\infty $ \ \ \ \ \ f) \ $\func{undefined}$ \ \ \
g) \ $\infty $ \ \ \ \ f) \ $\func{undefined}$ \ \ \ \ h) \ $\func{undefined}
$ \ \
i) $\ 7$ \ \ \ \ j) \ $7$ \ \ \ k) \ $\infty $ \ \ \ \ l) \ $0$ \ \ \ \ m) \
$\log _{10}3$\medskip
\bigskip \bigskip
\end{enumerate}
\begin{center}
{\LARGE Practice Problems - Answers}\bigskip \bigskip
\end{center}
\begin{enumerate}
\item a) $\ -\infty $ \ \ \ \ b) $\ \infty $ \ \ \ \ \ c) $\
\lim\limits_{x\rightarrow \infty }\dfrac{1}{3}x^{8}=\infty $\ \ \ \ \ \ \ d)
$\ \infty $\ \ \ \medskip\ \ \ e) $\ \infty $ \ \ \ \ \ \ \ \ \ \ f) $%
-\infty $ \ \ \ \ \ \ \ g) \ $-\infty $ \ \ \ \ \ \ h)$\ \ -\infty $
\item a) $\ \infty \qquad $b) $\ 0\qquad $c) \ $0$\qquad d) $\ \infty \qquad
$e) \ $32\qquad $f) \ $32\qquad $g) \ $\allowbreak 0\qquad $h) \ $\infty $ \
\item a) $\ 0\qquad $b) $\ 0\qquad $c) \ $5\qquad $d) $\ 5\qquad $e) $\
\infty \qquad $f) $\ 0\qquad $g) \ $\infty \qquad $h)\ $\ -\infty \qquad $i)
\ $5\qquad $j) \ $5\qquad $k) \ $-\infty $
l) \ $\func{undefined}\qquad $m) \ $-\infty \qquad $n) $\ \infty $\qquad o)
\ $\ln 5\qquad $p) \ $\ln 5\qquad $q) $-\infty \qquad $r) \ $-\infty $
\qquad s) \ $\infty \qquad $t) \ $0$
u*) \ $0\qquad $v*) \ $-\infty \qquad $x*) \ $1$
\item a) \ $P\left( 7\right) =\dfrac{1797}{49}\approx 36.\,\allowbreak 673$
\ \ \ \ b) \ $\lim\limits_{t\rightarrow \infty }P(t)=78$ \ \ - this means
that on the long run, eventually, about $78\%$ of the market will be aware
of this product.\medskip
\end{enumerate}
\pagebreak
\begin{center}
{\LARGE Sample Problems - Solutions}\bigskip \bigskip
\end{center}
\begin{enumerate}
\item (Monomials) \ \ Compute each of the following limits.
a) $\ \lim\limits_{x\rightarrow \infty }3x^{4}$\medskip \newline
Solution: \ Since the limit we are asked for is as $x$ approaches infinity,
we should think of $x$ as a very large positive number. \ Then $3x^{4}$ is
very large, and also positive because it is the product of five positive
numbers. \
\begin{equation*}
3x^{4}=\underset{\text{positive}}{3}\cdot \underset{\text{positive}}{x}\cdot
\underset{\text{positive}}{x}\cdot \underset{\text{positive}}{x}\cdot
\underset{\text{positive}}{x}
\end{equation*}%
So the answer is $\infty $. \ We state the answer: $\lim\limits_{x%
\rightarrow \infty }3x^{4}=\infty $.\medskip \medskip \medskip
b) \ $\lim\limits_{x\rightarrow -\infty }3x^{4}$\medskip \newline
Solution: \ Since the limit we are asked for is as $x$ approaches negative
infinity, we should think of $x$ as a very large negative number. \ Then $%
3x^{4}$ is very large, and also positive because it is the product of one
positive and four negative numbers.
\begin{equation*}
3x^{4}=\underset{\text{positive}}{3}\cdot \underset{\text{negative}}{x}\cdot
\underset{\text{negative}}{x}\cdot \underset{\text{negative}}{x}\cdot
\underset{\text{negative}}{x}
\end{equation*}%
\ So the answer is $\infty $. \ We state the answer: $\lim\limits_{x%
\rightarrow -\infty }3x^{4}=\infty $\medskip \medskip \medskip
c) $\ \lim\limits_{x\rightarrow \infty }\left( -2x^{5}\right) $\medskip
\newline
Solution: \ Since the limit we are asked for is as $x$ approaches infinity,
we should think of $x$ as a very large positive number. \ Then $-2x^{5}$ is
very large, and also negative because it is the product of one negative and
five positive numbers. \
\begin{equation*}
-2x^{5}=\underset{\text{negative}}{-2}\cdot \underset{\text{positive}}{x}%
\cdot \underset{\text{positive}}{x}\cdot \underset{\text{positive}}{x}\cdot
\underset{\text{positive}}{x}\cdot \underset{\text{positive}}{x}
\end{equation*}%
So the answer is $-\infty $. \ We state the answer: $\lim\limits_{x%
\rightarrow \infty }\left( -2x^{5}\right) =-\infty $\medskip \medskip
\medskip
d) \ $\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}\right) $\medskip
\newline
Solution: \ Since the limit we are asked for is as $x$ approaches negative
infinity, we should think of $x$ as a very large negative number. \ Then $%
-2x^{5}$ is very large, and also positive because it is the product of six
negative numbers. \
\begin{equation*}
-2x^{5}=\underset{\text{negative}}{-2}\cdot \underset{\text{negative}}{x}%
\cdot \underset{\text{negative}}{x}\cdot \underset{\text{negative}}{x}\cdot
\underset{\text{negative}}{x}\cdot \underset{\text{negative}}{x}
\end{equation*}%
So the answer is $\infty $. \ We state the answer: $\lim\limits_{x%
\rightarrow -\infty }\left( -2x^{5}\right) =\infty $
e) \ $\lim\limits_{x\rightarrow \infty }\left( -\dfrac{2}{3}x^{6}\right) $%
\medskip \newline
Solution: \ Since the limit we are asked for is as $x$ approaches infinity,
we should think of $x$ as a very large positive number. \ Then $-\dfrac{2}{3}%
x^{6}$ is very large, and also negative because it is the product of one
negative and six positive numbers. \
\begin{equation*}
-\dfrac{2}{3}x^{6}=\underset{\text{negative}}{-\dfrac{2}{3}}\cdot \underset{%
\text{positive}}{x}\cdot \underset{\text{positive}}{x}\cdot \underset{\text{%
positive}}{x}\cdot \underset{\text{positive}}{x}\cdot \underset{\text{%
positive}}{x}\cdot \underset{\text{positive}}{x}
\end{equation*}%
So the answer is $-\infty $. \ We state the answer: \ $\lim\limits_{x%
\rightarrow \infty }\left( -\dfrac{2}{3}x^{6}\right) =-\infty $\medskip
\medskip \medskip
f) \ $\lim\limits_{x\rightarrow -\infty }\left( -\dfrac{2}{3}x^{6}\right) $%
\medskip \newline
Solution: \ Since the limit we are asked for is as $x$ approaches negative
infinity, we should think of $x$ as a very large negative number. \ \ Then $-%
\dfrac{2}{3}x^{6}$ is very large, and also negative because it is the
product of seven negative numbers. \
\begin{equation*}
-\dfrac{2}{3}x^{6}=\underset{\text{negative}}{-\dfrac{2}{3}}\cdot \underset{%
\text{negative}}{x}\cdot \underset{\text{negative}}{x}\cdot \underset{\text{%
negative}}{x}\cdot \underset{\text{negative}}{x}\cdot \underset{\text{%
negative}}{x}\cdot \underset{\text{negative}}{x}
\end{equation*}%
\ So the answer is $-\infty $. \ We state the answer: $\lim\limits_{x%
\rightarrow -\infty }-\dfrac{2}{3}x^{6}=-\infty $\medskip \medskip \medskip
g) \ $\lim\limits_{x\rightarrow \infty }4x^{3}$\medskip \newline
Solution: \ Since the limit we are asked for is as $x$ approaches infinity,
we should think of $x$ as a very large positive number. \ Then $4x^{3}$ is
very large, and also positive because it is the product of four positive
numbers. \
\begin{equation*}
4x^{3}=\underset{\text{positive}}{4}\cdot \underset{\text{positive}}{x}\cdot
\underset{\text{positive}}{x}\cdot \underset{\text{positive}}{x}
\end{equation*}%
So the answer is $\infty $. \ We state the answer: $\lim\limits_{x%
\rightarrow \infty }4x^{3}=\infty $\medskip \medskip \medskip
h) \ $\lim\limits_{x\rightarrow -\infty }4x^{3}$\medskip \newline
Solution: \ Since the limit we are asked for is as $x$ approaches negative
infinity, we should think of $x$ as a very large negative number. \ \ Then $%
4x^{3}$ is very large, and also negative because it is the product of one
positive and three negative numbers. \
\begin{equation*}
4x^{3}=\underset{\text{positive}}{4}\cdot \underset{\text{negative}}{x}\cdot
\underset{\text{negative}}{x}\cdot \underset{\text{negative}}{x}
\end{equation*}%
\ So the answer is $-\infty $. \ We state the answer: $\lim\limits_{x%
\rightarrow -\infty }4x^{3}=-\infty $\medskip \medskip
\item (Exponential Functions) \ Compute each of the following limits.
Let \ $a>0.$ \ Then the limit of the exponential function $f\left( x\right)
=a^{x}$ \ is as follows.%
\begin{eqnarray*}
\text{Case 1. \ \ \ If\ \ }a &>&1\text{, then\ \ \ }\lim\limits_{x%
\rightarrow \infty }a^{x}=\infty \text{ \ \ and \ }\lim\limits_{x\rightarrow
-\infty }a^{x}=0 \\
\text{Case 2. \ \ \ If\ \ }0 &<&a<1\text{, then\ \ \ }\lim\limits_{x%
\rightarrow \infty }a^{x}=0\text{ \ \ and \ }\lim\limits_{x\rightarrow
-\infty }a^{x}=\infty
\end{eqnarray*}
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