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\lhead{\color{blue} \Large Lecture Notes}
\chead{\color{black} \LARGE Limits at Infinity - Part 2}
\rhead{\large page   \ \thepage}
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\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2010}
\rfoot{\small  Last revised: August 15, 2010}
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\begin{center}
{\LARGE Sample Problems}\bigskip
\end{center}

Compute each of the following limits. \ Show all steps, using correct
notation.\bigskip 
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\begin{enumerate}
\item $\lim\limits_{x\rightarrow \infty }2^{x}$\bigskip

\item $\lim\limits_{x\rightarrow -\infty }2^{x}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\left( \dfrac{2}{3}\right) ^{x}$%
\bigskip

\item $\lim\limits_{x\rightarrow -\infty }\left( \dfrac{2}{3}\right) ^{x}$%
\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x+3}}{3^{x+1}}$ \bigskip

\item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{x+3}}{3^{x+1}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+1}}{3^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+1}}{3^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\left( 3^{x+1}-3^{x}\right) $%
\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{3\sqrt{x}+2}{5\sqrt{x}+1}$%
\bigskip

\item $\lim\limits_{x\rightarrow \infty }\left( \sqrt{2x-1}-\sqrt{2x}\right) 
$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{2x%
}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{5}-\dfrac{1}{5-%
\dfrac{1}{x}}\right) $\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{\cos x}{x}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x%
}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$%
\bigskip

\item $\lim\limits_{x\rightarrow -\infty }\left( \sqrt{3x-1}-\sqrt{3x+1}%
\right) $\bigskip
\end{enumerate}

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\bigskip

\begin{center}
{\LARGE Practice Problems}\bigskip
\end{center}

Compute each of the following limits. \ Show all steps, using correct
notation.\bigskip 
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\begin{enumerate}
\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{3x-1}}{5^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{3x-1}}{5^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}}{5^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+3}}{5^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}}{4^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+3}}{4^{x-1}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x+3}\cdot 3^{x-1}}{7^{x-2}%
}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}\cdot 3^{x-1}}{%
7^{x-2}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\left( \sqrt{x+1}-\sqrt{x}\right) $%
\bigskip

\item $\lim\limits_{x\rightarrow \infty }\left( \dfrac{1}{\sqrt{x+1}}-\dfrac{%
1}{\sqrt{x}}\right) $\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{x}{\sqrt{x-1}+\sqrt{x+1}}$%
\bigskip

\item $\lim\limits_{x\rightarrow -\infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{%
2x}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{a}-\dfrac{1}{a-%
\dfrac{1}{x}}\right) $\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{0.5^{x}+0.5^{-x}}{%
0.5^{x}-0.5^{-x}}$\bigskip

\item $\lim\limits_{x\rightarrow \infty }\dfrac{\sin x-\cos x}{\sqrt{x^{2}+1}%
}$
\end{enumerate}

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\begin{center}
{\LARGE Sample Problems - Answers}\bigskip
\end{center}

1.) $\ \infty $\ \ \ \ \ \ \ 2.) $\ 0$ \ \ \ \ \ 3.) \ $0$ \ \ \ \ \ 4.) \ $%
\infty $ \ \ \ \ \ 5.) \ $0$ \ \ \ \ \ 6.) \ $\infty $ \ \ \ \ \ 7.) \ $%
\infty $ \ \ \ \ \ \ 8.) \ $0$ \ \ \ \ \ 9.) \ $\infty $ \ \bigskip

10.) \ $\dfrac{3}{5}$ \ \ \ \ \ 11.) \ $0$ \ \ \ \ \ 12.) \ $\dfrac{1}{1-%
\sqrt{2}}=-1-\sqrt{2}$ \ \ \ \ \ 13.) \ $-\dfrac{1}{25}$ \ \ \ \ 14.) \ $0$
\ \ \ \ 15.) \ $1$ \ \ \ 16.) \ $1$\bigskip

17.)\ $\ \func{undefined}$\bigskip \bigskip

\begin{center}
{\LARGE Practice Problems - Answers}\bigskip \bigskip
\end{center}

1.) $\ \infty $ \ \ \ \ \ \ \ 2.) $\ 0$ \ \ \ \ \ 3.) $\ 0$ \ \ \ \ \ \ 4.) $%
\ \infty $ \ \ \ \ \ 5.) $\ 32$ \ \ \ \ \ 6.) $\ 32$ \ \ \ \ \ \ 7. ) $\ 0$
\ \ \ \ \ \ 8.) $\ \infty $ \ \ \ \ \ 9.) $\ 0$\bigskip

10.) \ $0$ \ \ \ \ \ 11.) $\ \infty $ \ \ \ \ \ \ 12.) $\ \func{undefined}$
\ \ \ 13.) \ $-\dfrac{1}{a^{2}}$ \ \ \ \ \ 14.) \ $-1$ \ \ \ \ \ 15.) \ $0$%
\bigskip \bigskip \bigskip \bigskip

\begin{center}
{\LARGE Sample Problems - Solutions}\bigskip \bigskip
\end{center}

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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1.) $\ \lim\limits_{x\rightarrow \infty }2^{x}$ \ \ and\ \ \ \ \ \ 2.) $\
\lim\limits_{x\rightarrow -\infty }2^{x}$\newline
Solution: \ Since $2>1,$ \ these limits are $\infty $ and $0,$ i.e. $%
\lim\limits_{x\rightarrow \infty }2^{x}=\infty $ and $\lim\limits_{x%
\rightarrow -\infty }2^{x}=0$.\bigskip

3.) \ $\lim\limits_{x\rightarrow \infty }\left( \dfrac{2}{3}\right) ^{x}$ \
\ and\ \ \ \ 4.) \ $\lim\limits_{x\rightarrow -\infty }\left( \dfrac{2}{3}%
\right) ^{x}$\newline
Solution: \ Since $\dfrac{2}{3}<1,$ \ these limits are $0$ and $\infty $,
i.e. $\lim\limits_{x\rightarrow \infty }\left( \dfrac{2}{3}\right)
^{x}=\allowbreak 0$ and $\lim\limits_{x\rightarrow -\infty }\left( \dfrac{2}{%
3}\right) ^{x}=\infty $.\bigskip

5.) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x+3}}{3^{x+1}}$\newline
Solution: \ We start by re-writing the exponential expressions. \ The goal
is to bring it into a form where there is only one exponential expression
involving $x.$%
\begin{equation*}
\dfrac{2^{x+3}}{3^{x+1}}=\dfrac{2^{x}\cdot 2^{3}}{3^{x}\cdot 3^{1}}=\dfrac{%
2^{x}\cdot 8}{3^{x}\cdot 3}=\dfrac{8}{3}\left( \dfrac{2}{3}\right) ^{x}
\end{equation*}%
\begin{equation*}
\text{Thus \ \ }\lim\limits_{x\rightarrow \infty }\dfrac{2^{x+3}}{3^{x+1}}%
=\lim\limits_{x\rightarrow \infty }\dfrac{8}{3}\left( \dfrac{2}{3}\right)
^{x}=\dfrac{8}{3}\lim\limits_{x\rightarrow \infty }\left( \dfrac{2}{3}%
\right) ^{x}=0\text{ \ \ \ \ \ \ \ \ \ \ since }\dfrac{2}{3}<1
\end{equation*}%
\bigskip 6.) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{x+3}}{3^{x+1}}$%
\newline
Solution: \ $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{x+3}}{3^{x+1}}%
=\lim\limits_{x\rightarrow -\infty }\dfrac{2^{x+3}}{3^{x+1}}%
=\lim\limits_{x\rightarrow -\infty }\dfrac{8}{3}\left( \dfrac{2}{3}\right)
^{x}=\dfrac{8}{3}\lim\limits_{x\rightarrow -\infty }\left( \dfrac{2}{3}%
\right) ^{x}=\infty $\bigskip

7.) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+1}}{3^{x-1}}$\newline
Solution: \ We start by re-writing the exponential expressions. \ The goal
is to bring it into a form where there is only one exponential expression
involving $x.$%
\begin{equation*}
\dfrac{2^{2x+1}}{3^{x-1}}=\dfrac{2^{2x}\cdot 2^{1}}{\dfrac{3^{x}}{3^{1}}}=%
\dfrac{\left( 2^{2}\right) ^{x}\cdot 2}{3^{x}\cdot \dfrac{1}{3}}=\dfrac{%
4^{x}\cdot 6}{3^{x}}=6\left( \dfrac{4}{3}\right) ^{x}
\end{equation*}%
\begin{equation*}
\text{Thus \ \ }\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+1}}{3^{x-1}}%
=\lim\limits_{x\rightarrow \infty }6\left( \dfrac{4}{3}\right)
^{x}=6\lim\limits_{x\rightarrow \infty }\left( \dfrac{4}{3}\right)
^{x}=\infty \text{ \ \ \ since }\dfrac{4}{3}>1
\end{equation*}%
\bigskip 8.) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+1}}{3^{x-1}}$%
\newline
Solution: \ $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+1}}{3^{x-1}}%
=\lim\limits_{x\rightarrow -\infty }6\left( \dfrac{4}{3}\right)
^{x}=6\lim\limits_{x\rightarrow -\infty }\left( \dfrac{4}{3}\right) ^{x}=0$%
\bigskip

9.) $\ \lim\limits_{x\rightarrow \infty }\left( 3^{x+1}-3^{x}\right) $%
\newline
Solution: \ $\lim\limits_{x\rightarrow \infty }\left( 3^{x+1}-3^{x}\right)
=\lim\limits_{x\rightarrow \infty }\left( 3^{x}\cdot 3-3^{x}\right)
=\lim\limits_{x\rightarrow \infty }\left( 3\cdot 3^{x}-3^{x}\right)
=\lim\limits_{x\rightarrow \infty }\left( 2\cdot 3^{x}\right)
=2\lim\limits_{x\rightarrow \infty }3^{x}=\infty $\bigskip

10.) $\ \lim\limits_{x\rightarrow \infty }\dfrac{3\sqrt{x}+2}{5\sqrt{x}+1}$%
\newline
Solution: \ Since $\lim\limits_{x\rightarrow \infty }\sqrt{x}=\infty ,$
clearly $\lim\limits_{x\rightarrow \infty }\dfrac{1}{\sqrt{x}}=0.$ \ We will
use this fact; we factor out $\sqrt{x}$ from both numerator and denominator.%
\begin{equation*}
\lim\limits_{x\rightarrow \infty }\dfrac{3\sqrt{x}+2}{5\sqrt{x}+1}%
=\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}\left( 3+\dfrac{2}{\sqrt{x}%
}\right) }{\sqrt{x}\left( 5+\dfrac{1}{\sqrt{x}}\right) }=\lim\limits_{x%
\rightarrow \infty }\dfrac{3+\dfrac{2}{\sqrt{x}}}{5+\dfrac{1}{\sqrt{x}}}=%
\dfrac{3}{5}
\end{equation*}%
\pagebreak

11.) $\ \lim\limits_{x\rightarrow \infty }\left( \sqrt{2x-1}-\sqrt{2x}%
\right) $\newline
Solution: \ We will transform this expression by multiplying it by $1$,
written as a fraction with numerator and denominator both being the
conjugate of the expression.%
\begin{eqnarray*}
\lim\limits_{x\rightarrow \infty }\left( \sqrt{2x-1}-\sqrt{2x}\right)
&=&\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{2x-1}-\sqrt{2x}}{1}\cdot 
\dfrac{\sqrt{2x-1}+\sqrt{2x}}{\sqrt{2x-1}+\sqrt{2x}} \\
&=&\lim\limits_{x\rightarrow \infty }\dfrac{\left( 2x-1\right) -\left(
2x\right) }{\sqrt{2x-1}+\sqrt{2x}}=\lim\limits_{x\rightarrow \infty }\dfrac{%
-1}{\sqrt{2x-1}+\sqrt{2x}}=0
\end{eqnarray*}%
\bigskip

12.) \ $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{%
2x}}$\newline
Solution: \ We factor out $\sqrt{x}$ \ from both numerator and denominator.%
\begin{eqnarray*}
\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{2x}}
&=&\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x}\left( \dfrac{%
\sqrt{x+1}}{\sqrt{x}}-\sqrt{2}\right) }=\lim\limits_{x\rightarrow \infty }%
\dfrac{1}{\sqrt{\dfrac{x+1}{x}}-\sqrt{2}} \\
&=&\lim\limits_{x\rightarrow \infty }\dfrac{1}{\sqrt{1+\dfrac{1}{x}}-\sqrt{2}%
}=\dfrac{1}{1-\sqrt{2}}=\dfrac{1}{1-\sqrt{2}}\cdot \dfrac{1+\sqrt{2}}{1+%
\sqrt{2}}=\dfrac{1+\sqrt{2}}{-1} \\
&=&-1-\sqrt{2}
\end{eqnarray*}%
\bigskip

13.) $\ \lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{5}-\dfrac{1}{5-%
\dfrac{1}{x}}\right) $\newline
Solution: \ We just need to simplify the complex fraction. \ As it turns
out, this problem boils down to a type we have already seen.%
\begin{eqnarray*}
x\left( \dfrac{1}{5}-\dfrac{1}{5-\dfrac{1}{x}}\right) &=&x\left( \dfrac{1}{5}%
-\dfrac{1}{~~\dfrac{5x-1}{x}~~}\right) =x\left( \dfrac{1}{5}-\dfrac{x}{5x-1}%
\right) =x\left( \dfrac{\left( 5x-1\right) -5x}{5\left( 5x-1\right) }\right)
\\
&=&x\left( \dfrac{5x-1-5x}{5\left( 5x-1\right) }\right) =x\dfrac{-1}{25x-5}=%
\dfrac{-x}{25x-5}
\end{eqnarray*}%
Thus%
\begin{equation*}
\lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{5}-\dfrac{1}{5-\dfrac{1}{%
x}}\right) =\lim\limits_{x\rightarrow \infty }\dfrac{-x}{25x-5}%
=\lim\limits_{x\rightarrow \infty }\dfrac{x\left( -1\right) }{x\left( 25-%
\dfrac{5}{x}\right) }=-\dfrac{1}{25}
\end{equation*}

\pagebreak

14.) \ $\lim\limits_{x\rightarrow \infty }\dfrac{\cos x}{x}$\newline
Solution: \ This problem can be solved by the sandwich principle. \ Consider
the limits $\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}$ \ and $%
\lim\limits_{x\rightarrow \infty }\left( -\dfrac{1}{x}\right) $. \ These
limits are both zero. \ \ Furthermore, since%
\begin{eqnarray*}
-1 &\leq &\cos x\leq 1\text{ \ \ \ \ \ \ \ \ \ \ for all }x\text{, we also
have} \\
-\dfrac{1}{x} &\leq &\dfrac{\cos x}{x}\leq \dfrac{1}{x}\text{ \ \ \ \ \ \
for all positive }x
\end{eqnarray*}%
Our function $f\left( x\right) =\dfrac{\cos x}{x}$ \ is 'locked' between $%
g\left( x\right) =\dfrac{1}{x}$ and $h\left( x\right) =-\dfrac{1}{x}.$ \
Since these both approach zero, so must the function $f\left( x\right) =%
\dfrac{\cos x}{x}.$ \ Thus $\lim\limits_{x\rightarrow \infty }\dfrac{\cos x}{%
x}=0$.\bigskip

15.) \ $\lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{%
x}}$\newline
Solution: \ We will factor out $x$ from both numerator and denominator, and
use the fact that $\lim\limits_{x\rightarrow \infty }\dfrac{1}{x^{k}}=0$ for
all positive integers $k$.%
\begin{equation*}
\lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}}%
=\lim\limits_{x\rightarrow \infty }\dfrac{x\left( 1+\dfrac{1}{x^{2}}\right) 
}{x\left( 1-\dfrac{1}{x^{2}}\right) }=\lim\limits_{x\rightarrow \infty }%
\dfrac{1+\dfrac{1}{x^{2}}}{1-\dfrac{1}{x^{2}}}=1
\end{equation*}%
\bigskip

16.) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$%
\newline
Solution: \ First, $\lim\limits_{x\rightarrow \infty }2^{x}=\infty $ (and so 
$2^{x}$ is large) and $\lim\limits_{x\rightarrow \infty }2^{-x}=0$ \ (and so 
$2^{-x}$ is small). \ With that in mind, this limit is similar to \ $%
\lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}}$. \
The solution also will be similar. \ \ We will factor out $2^{x}$ \ from
both numerator and denominator.%
\begin{equation*}
\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}%
=\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+\dfrac{1}{2^{x}}}{2^{x}-%
\dfrac{1}{2^{x}}}=\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}\left( 1+%
\dfrac{1}{\left( 2^{x}\right) ^{2}}\right) }{2^{x}\left( 1-\dfrac{1}{\left(
2^{x}\right) ^{2}}\right) }=\lim\limits_{x\rightarrow \infty }\dfrac{1+%
\dfrac{1}{2^{2x}}}{1-\dfrac{1}{2^{2x}}}=1
\end{equation*}%
\bigskip

17.)\ $\ \lim\limits_{x\rightarrow -\infty }\left( \sqrt{3x-1}-\sqrt{3x+1}%
\right) $\newline
Solution: \ When $x\rightarrow -\infty ,$ then we may assume it is negative.
\ Then the expressions under the square root are negative and the function
is not defined. \ Thus, there is no limit at negative infinity. \ The answer
is: $\func{undefined}$.\bigskip

\bigskip

\bigskip

\vspace{1in}

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