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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
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\lhead{\color{blue} \large Math 208}
\chead{\color{black} \Large Limits of Sequences}
\rhead{ page   \ \thepage}
\cfoot{}
\rfoot{\small   Last revised:  April 18, 2017}
\lfoot{\small   \copyright \;   Hidegkuti, 2013}
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\begin{document}


\begin{enumerate}
\item In each case, decide whether the sequence is convergent or divergent.
\ If convergent, find its limit.$\medskip \medskip \medskip $%
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a) \ $a_{n}=5-0.1^{n}\medskip \medskip $

b) \ $a_{n}=\dfrac{2-n}{7+3n}\medskip \medskip $

c) \ $a_{n}=1^{n}+\left( -1\right) ^{n}\medskip \medskip $

d) \ $a_{n}=\sqrt{\dfrac{2n}{n+1}}\medskip \medskip $

e) \ $a_{n}=\dfrac{\sin n}{n}\medskip \medskip $

f) \ $a_{n}=\dfrac{3^{n+1}}{2^{2n+1}}\medskip \medskip $

g) \ $a_{n}=\dfrac{3^{n-1}}{2^{n+3}}\medskip \medskip $

h) \ $a_{n}=n\pi \cos \left( \dfrac{n\pi }{2}\right) \medskip \medskip $

i) \ $a_{n}=8^{1/n}\medskip \medskip $

j) \ $a_{n}=\dfrac{\ln n}{\ln 2n}\medskip \medskip $

k) \ $a_{n}=\left( \dfrac{n+1}{2n}\right) \left( 1-\dfrac{1}{n}\right)
\medskip \medskip $

l) \ $a_{n}=\dfrac{n!}{10^{n}}\medskip \medskip $

m) $\ a_{n}=\dfrac{5^{n}}{n!}\medskip \medskip $

n) \ $a_{n}=\left( -1\right) ^{n}\dfrac{\sin n}{n^{2}}\medskip \medskip $

o) \ $a_{n}=\lim\limits_{n\rightarrow \infty }\sqrt[n]{2n}$ \ \ 
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\item Assume that each of the following sequences are convergent and find
the value of the limit.%
%TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}%
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a) \ $a_{1}=8$ \ and $a_{n+1}=\sqrt{a_{n}+6}\medskip \medskip $

b) \ $a_{1}=5$ and $a_{n+1}=\sqrt{2a_{n}-1}\medskip \medskip $

c) \ $a_{1}=\sqrt{30}$ \ and\ \ $a_{n+1}=\sqrt{30+a_{n}}\medskip \medskip $

d) \ $a_{1}=2$ \ and $a_{n+1}=\dfrac{1}{2+a_{n}}\medskip \medskip $

e) \ $a_{1}=1$ and $a_{n+1}=\dfrac{1}{3}\left( a_{n}+12\right) \medskip
\medskip $

f) \ $a_{1}=2$ \ \ and \ $a_{n+1}=\dfrac{a_{n}+1}{2a_{n}+1}\medskip \medskip 
$

g) \ $a_{1}=\sqrt{5}$ \ \ and \ $a_{n+1}=\sqrt{5a_{n}}\medskip \medskip $

h) \ $a_{1}=1$ \ and \ $a_{n+1}=\dfrac{a_{n}^{2}+4}{a_{n}+2}\medskip
\medskip $

i) \ $a_{1}=10$ \ and \ $a_{n+1}=0.6a_{n}+8\medskip \medskip $ \ 
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\item Assume that $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}%
\right) ^{n}=e$ and $\lim\limits_{n\rightarrow -\infty }\left( 1+\dfrac{1}{n}%
\right) ^{n}=e$ and compute each of the following limits.$\medskip \medskip $%
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a) \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n+2}\medskip \medskip $

b) \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{2n}\medskip \medskip $

c) \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{5}{n}\right)
^{n}\medskip \medskip $

d) \ $\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right) ^{n}$ $%
\medskip \medskip $\ 
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\pagebreak
\end{enumerate}

\begin{center}
{\LARGE Answers\bigskip }
\end{center}

\begin{enumerate}
\item 
%TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}%
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a) \ converges to $5\medskip $

b) \ converges to $-\dfrac{1}{3}\medskip $

c) \ diverges$\medskip $

d) \ \ converges to $\sqrt{2}\medskip $

e) \ converges to $0\medskip $

f) \ converges to $0\medskip $

g) \ diverges$\medskip $

h) \ diverges$\medskip $

i) \ converges to $1\medskip $

j) \ converges to $1\medskip $

k) \ converges to $\dfrac{1}{2}\medskip $

l) \ diverges to infinity$\medskip $

m) $\ $converges to $0\medskip $\ \ 

n) $\ $converges to $1\medskip $ \ 
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\item a) \ $3$ \ \ \ \ \ \ b) \ $1$ \ \ \ \ \ c) \ $6$ \ \ \ \ \ d) \ $\sqrt{%
2}-1$ \ \ \ \ e) \ $6$ \ \ \ \ \ \ \ f) \ $\dfrac{\sqrt{2}}{2}$ \ \ \ g) \ $%
5 $ \ \ \ \ h) \ $2$ \ \ \ \ i) \ $20$

\item a) \ $e$ \ \ \ \ \ b) \ $e^{2}$ \ \ \ \ \ c) \ $e^{5}$ \ \ \ \ \ \ d)
\ $\dfrac{1}{e}\medskip $\pagebreak
\end{enumerate}

\begin{center}
{\LARGE Solutions\bigskip }
\end{center}

\begin{enumerate}
\item In each case, decide whether the sequence is convergent or divergent.
\ If convergent, find its limit.$\medskip \medskip \medskip $

a) \ $a_{n}=5-0.1^{n}\medskip \medskip $

Solution: \ We separate the two parts via the difference rule.%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }5-0.1^{n}=\lim\limits_{n\rightarrow \infty
}5-\lim\limits_{n\rightarrow \infty }0.1^{n}
\end{equation*}%
The limit of the constant sequence is that same value. \ Since $0.1<1$, $\
0.1^{n}$ approaches zero as $n$ gets large.%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }5-0.1^{n}=\lim\limits_{n\rightarrow \infty
}5-\lim\limits_{n\rightarrow \infty }0.1^{n}=5-0=5
\end{equation*}%
$\medskip \medskip $

b) \ $a_{n}=\dfrac{2-n}{7+3n}\medskip \medskip $

Solution: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{2-n}{7+3n}=\lim\limits_{n%
\rightarrow \infty }\dfrac{\dfrac{2}{n}-1}{\dfrac{7}{n}+3}=\dfrac{%
\lim\limits_{n\rightarrow \infty }\dfrac{2}{n}-\lim\limits_{n\rightarrow
\infty }1}{\lim\limits_{n\rightarrow \infty }\dfrac{7}{n}+\lim\limits_{n%
\rightarrow \infty }3}=\dfrac{0-1}{0+3}=-\dfrac{1}{3}
\end{equation*}

c) \ $a_{n}=1^{n}+\left( -1\right) ^{n}\medskip \medskip $

Solution: \ The first few terms of the sequence are: $%
2,0,2,0,2,0,2,0,2,0.... $This sequence diverges.\medskip \medskip

d) \ $a_{n}=\sqrt{\dfrac{2n}{n+1}}\medskip \medskip $

Solution: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{2n}{n+1}=\lim\limits_{n\rightarrow
\infty }\dfrac{2}{1+\dfrac{1}{n}}=\dfrac{\lim\limits_{n\rightarrow \infty }2%
}{\lim\limits_{n\rightarrow \infty }1+\lim\limits_{n\rightarrow \infty }%
\dfrac{1}{n}}=\dfrac{2}{1+0}=2
\end{equation*}%
Since $f\left( x\right) =\sqrt{x}$ is continuous at $x=2$, 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\sqrt{\dfrac{2n}{n+1}}=\sqrt{%
\lim\limits_{n\rightarrow \infty }\dfrac{2n}{n+1}}=\sqrt{2}
\end{equation*}%
$\medskip \medskip $

e) \ $a_{n}=\dfrac{\sin n}{n}\medskip \medskip $

Solution: \ Tis sequence converges to zero by the sandwich rule.%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( -\dfrac{1}{n}\right) =0\text{ \ \
and \ }\lim\limits_{n\rightarrow \infty }\left( \dfrac{1}{n}\right) =0
\end{equation*}%
Since $-1\leq \sin n\leq 1$, \ \ $-\dfrac{1}{n}\leq \dfrac{\sin n}{n}\leq 
\dfrac{1}{n}$. \ Thus $\lim\limits_{n\rightarrow \infty }\dfrac{\sin n}{n}=0$%
.$\medskip \medskip $

f) \ $a_{n}=\dfrac{3^{n+1}}{2^{2n+1}}\medskip \medskip $

Solution: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{3^{n+1}}{2^{2n+1}}%
=\lim\limits_{n\rightarrow \infty }\dfrac{3\cdot 3^{n}}{2\cdot 2^{2n}}=%
\dfrac{3}{2}\lim\limits_{n\rightarrow \infty }\dfrac{3^{n}}{\left(
2^{2}\right) ^{n}}=\dfrac{3}{2}\lim\limits_{n\rightarrow \infty }\dfrac{3^{n}%
}{4^{n}}=\dfrac{3}{2}\lim\limits_{n\rightarrow \infty }\left( \dfrac{3}{4}%
\right) ^{n}
\end{equation*}

Since $\dfrac{3}{4}<1$, $\left( \dfrac{3}{4}\right) ^{n}$ approaches $1$ as $%
n$ gets large. \ Thus%
\begin{equation*}
\dfrac{3}{2}\lim\limits_{n\rightarrow \infty }\left( \dfrac{3}{4}\right)
^{n}=\dfrac{3}{2}\cdot 0=0
\end{equation*}%
$\medskip \medskip $

g) \ $a_{n}=\dfrac{3^{n-1}}{2^{n+3}}\medskip \medskip $

Solution: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{3^{n-1}}{2^{n+3}}%
=\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{1}{3}\cdot 3^{n}}{8\cdot
2^{n}}=\dfrac{1}{24}\lim\limits_{n\rightarrow \infty }\dfrac{3^{n}}{2^{n}}=%
\dfrac{1}{24}\lim\limits_{n\rightarrow \infty }\left( \dfrac{3}{2}\right)
^{n}
\end{equation*}%
Since $\dfrac{3}{2}>1$, $\left( \dfrac{3}{2}\right) ^{n}$ diverges to
infinity.$\medskip \medskip $

h) \ $a_{n}=n\pi \cos \left( \dfrac{n\pi }{2}\right) \medskip \medskip $

Solution: $\ \cos \left( \dfrac{n\pi }{2}\right) $ is $1$ or $-1$, depending
on whether $n$ is even or odd. \ $n\pi $ diverges to infinity as $n$ gets
large. \ Thus $a_{n}=n\pi \cos \left( \dfrac{n\pi }{2}\right) $ has larger
and larger absolute value but also alternating sings. \ A sequence like that
diverges.$\medskip \medskip $

i) \ $a_{n}=8^{1/n}\medskip \medskip $

Solution: $\ f\left( x\right) =8^{x}$ is continuous on $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$.%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }8^{1/n}=8^{\lim\limits_{n\rightarrow
\infty }\left( 1/n\right) }=8^{0}=1
\end{equation*}%
$\medskip \medskip $

j) \ $a_{n}=\dfrac{\ln n}{\ln 2n}\medskip \medskip $

Solution: \ $\lim\limits_{n\rightarrow \infty }\ln n=\infty $ and so $%
\lim\limits_{n\rightarrow \infty }\dfrac{1}{\ln n}=\allowbreak 0$%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{\ln n}{\ln 2n}=\lim\limits_{n%
\rightarrow \infty }\dfrac{\ln n}{\ln 2+\ln n}=\lim\limits_{n\rightarrow
\infty }\dfrac{1}{\dfrac{\ln 2}{\ln n}+1}=\dfrac{\lim\limits_{n\rightarrow
\infty }1}{\lim\limits_{n\rightarrow \infty }\dfrac{\ln 2}{\ln n}%
+\lim\limits_{n\rightarrow \infty }1}=\dfrac{1}{0+1}=1
\end{equation*}%
$\ \medskip \medskip $\pagebreak

k) \ $a_{n}=\left( \dfrac{n+1}{2n}\right) \left( 1-\dfrac{1}{n}\right)
\medskip \medskip $

Solution: $\ $%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( \dfrac{n+1}{2n}\right) \left( 1-%
\dfrac{1}{n}\right) =\lim\limits_{n\rightarrow \infty }\left( \dfrac{1+%
\dfrac{1}{n}}{2}\right) \left( 1-\dfrac{1}{n}\right)
=\lim\limits_{n\rightarrow \infty }\left( \dfrac{1+\dfrac{1}{n}}{2}\right)
\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right) =\dfrac{1}{2}%
\cdot 1=\dfrac{1}{2}
\end{equation*}%
$\medskip \medskip $

l) \ $a_{n}=\dfrac{n!}{10^{n}}\medskip \medskip $

Solution: $\ $This sequence diverges to infinity. \ Imagine that $n$ is very
large and look at how $a_{n+1}$ is related to $a_{n}$. \ The denominator is
multiplied by $10$ (fixed) while the numerator is multiplied by $n+1$
(indefinitely large). \ 
\begin{equation*}
a_{n+1}=a_{n}\cdot \dfrac{n+1}{10}
\end{equation*}%
After $n=20$, every term is over twice as large as the previous term. \ That
is, in $n>20$, then 
\begin{equation*}
a_{21}=a_{20}\cdot \dfrac{21}{10}>a_{20}\cdot \dfrac{20}{10}=2a_{20}
\end{equation*}%
and 
\begin{equation*}
a_{22}=a_{21}\cdot \dfrac{22}{10}>2a_{21}>4a_{20}
\end{equation*}%
and%
\begin{equation*}
a_{20+k}>2^{k}a_{21}
\end{equation*}%
Then $a_{n+k}>2^{k}a_{n}$ and so this sequence diverges to infinity.$%
\medskip \medskip $

m) $\ a_{n}=\dfrac{5^{n}}{n!}\medskip \medskip $

Solution: \ This sequence converges to zero. \ Imagine that $n$ is very
large and look at how $a_{n+1}$ is related to $a_{n}$. \ The numerator is
multiplied by $5$ (fixed) while the denominator is multiplied by $n+1$
(indefinitely large). \ If $n>10,$ then 
\begin{equation*}
a_{11}=\dfrac{5}{11}a_{10}<\dfrac{5}{10}a_{10}=\dfrac{1}{2}a_{10}
\end{equation*}%
and%
\begin{equation*}
a_{12}=\dfrac{5}{12}a_{11}<\dfrac{5}{10}a_{11}=\dfrac{1}{2}a_{11}=\dfrac{1}{4%
}a_{10}
\end{equation*}%
Thus 
\begin{equation*}
a_{10+k}<\left( \dfrac{1}{2}\right) ^{k}a_{10}
\end{equation*}%
Consider now the constant zero sequence and $b_{n}=\left( \dfrac{1}{2}%
\right) ^{n}a_{10}$. \ Both sequences converge to zero and 
\begin{equation*}
0\leq a_{m}\leq \left( \dfrac{1}{2}\right) ^{m}a_{10}
\end{equation*}%
So $a_{n}$ converges to zero by the sandwich theorem.$\medskip \medskip $%
\pagebreak

n) \ $a_{n}=\left( -1\right) ^{n}\dfrac{\sin n}{n^{2}}\medskip \medskip $

Solution: $\ $Converges to zero by the sandwich theorem.$\medskip \medskip $

o) \ $a_{n}=\lim\limits_{n\rightarrow \infty }\sqrt[n]{2n}$\ $\medskip
\medskip $

Solution : 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\sqrt[n]{2n}=\lim\limits_{n\rightarrow
\infty }\left( 2n\right) ^{1/n}=\lim\limits_{n\rightarrow \infty }e^{\ln
\left( 2n\right) ^{1/n}}=\lim\limits_{n\rightarrow \infty }e^{\left(
1/n\right) \ln \left( 2n\right) }
\end{equation*}%
Looking at the exponent, 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}\ln \left( 2n\right)
=\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 2n\right) }{n}%
=\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 2x\right) }{x}
\end{equation*}%
This is an indeterminate of the form $\dfrac{\infty }{\infty }\ $so we can
apply l'H\^{o}pital's rule. \ 
\begin{equation*}
\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 2x\right) }{x}%
=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{x}}{1}=0
\end{equation*}%
Since $f\left( x\right) =e^{x}$ is continuous on $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$, we have that 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }e^{\left( 1/n\right) \ln \left( 2n\right)
}=e^{\lim\limits_{n\rightarrow \infty }\left( 1/n\right) \ln \left(
2n\right) }=e^{0}=1
\end{equation*}%
$\medskip \medskip $

\item See sequences - part 3

\item Assume that $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}%
\right) ^{n}=e$ and $\lim\limits_{n\rightarrow -\infty }\left( 1+\dfrac{1}{n}%
\right) ^{n}=e$ and compute each of the following limits.$\medskip \medskip $

a) \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n+2}\medskip \medskip $

Solution :%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n+2}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n}\left( 1+\dfrac{1}{n}\right) ^{2}=\lim\limits_{n\rightarrow \infty
}\left( 1+\dfrac{1}{n}\right) ^{n}\lim\limits_{n\rightarrow \infty }\left( 1+%
\dfrac{1}{n}\right) ^{2}=e\cdot 1=e
\end{equation*}%
$\medskip \medskip \medskip \medskip $

Solution 2: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n+2}=\lim\limits_{n\rightarrow \infty }e^{\ln \left( 1+\dfrac{1}{n}\right)
^{n+2}}=\lim\limits_{n\rightarrow \infty }e^{\left( n+2\right) \ln \left( 1+%
\dfrac{1}{n}\right) }=e^{\lim\limits_{n\rightarrow \infty }\left( n+2\right)
\ln \left( 1+\dfrac{1}{n}\right) }=e^{\lim\limits_{n\rightarrow \infty }%
\dfrac{\ln \left( 1+\dfrac{1}{n}\right) }{1/\left( n+2\right) }}
\end{equation*}

Consider now the exponent: \ 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{1}{n}\right) }{%
\dfrac{1}{n+2}}=\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{%
1}{x}\right) }{\dfrac{1}{x+2}}
\end{equation*}%
This is an indeterminant of the form $\dfrac{0}{0}$ and so we can apply l'H%
\^{o}pital's rule.%
\begin{eqnarray*}
\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{1}{x}\right) }{%
\dfrac{1}{x+2}} &=&\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{1+%
\dfrac{1}{x}}\cdot -\dfrac{1}{x^{2}}}{-\dfrac{1}{\left( x+2\right) ^{2}}}%
=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{x^{2}+x}}{\dfrac{1}{%
\left( x+2\right) ^{2}}}=\lim\limits_{x\rightarrow \infty }\dfrac{\left(
x+2\right) ^{2}}{x^{2}+x}=\lim\limits_{x\rightarrow \infty }\dfrac{x^{2}+4x+4%
}{x^{2}+x} \\
&=&\lim\limits_{x\rightarrow \infty }\dfrac{1+\dfrac{4}{x}+\dfrac{4}{x^{2}}}{%
1+\dfrac{1}{x}}=1
\end{eqnarray*}%
Thus the limit is $e$ since%
\begin{equation*}
e^{\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{1}{n}\right) 
}{1/\left( n+2\right) }}=e^{1}=e
\end{equation*}

b) \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{2n}\medskip \medskip $

Solution :%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{2n}=\lim\limits_{n\rightarrow \infty }\left( \left( 1+\dfrac{1}{n}\right)
^{n}\right) ^{2}=\left( \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{%
n}\right) ^{n}\right) ^{2}=e^{2}
\end{equation*}%
$\medskip \medskip $Solution 2:%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{2n}=\lim\limits_{n\rightarrow \infty }e^{\ln \left( 1+\dfrac{1}{n}\right)
^{2n}}=\lim\limits_{n\rightarrow \infty }e^{2n\ln \left( 1+\dfrac{1}{n}%
\right) }=e^{\lim\limits_{n\rightarrow \infty }2n\ln \left( 1+\dfrac{1}{n}%
\right) }
\end{equation*}

Looking at the exponent, 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }2n\ln \left( 1+\dfrac{1}{n}\right)
=\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{1}{n}\right) }{%
\dfrac{1}{2n}}=\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{1%
}{x}\right) }{\dfrac{1}{2x}}
\end{equation*}%
This is an indeterminant of the form $\dfrac{0}{0}$ and so we can apply l'H%
\^{o}pital's rule.%
\begin{equation*}
\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{1}{x}\right) }{%
\dfrac{1}{2x}}=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{1+\dfrac{1%
}{x}}\cdot -\dfrac{1}{x^{2}}}{-\dfrac{1}{2x^{2}}}=\lim\limits_{x\rightarrow
\infty }\dfrac{2}{1+\dfrac{1}{x}}=2
\end{equation*}%
Thus the sequence converges to $e^{2}$.

c) \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{5}{n}\right)
^{n}\medskip \medskip $

Solution :%
\begin{eqnarray*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{5}{n}\right) ^{n}
&=&\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~\dfrac{n}{5}~}%
\right) ^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~\dfrac{n}{%
5}~}\right) ^{n\cdot \tfrac{5}{5}}=\lim\limits_{n\rightarrow \infty }\left[
\left( 1+\dfrac{1}{~\dfrac{n}{5}~}\right) ^{\tfrac{n}{5}}\right] ^{5} \\
&=&\left[ \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~\dfrac{n}{5}~%
}\right) ^{\tfrac{n}{5}}\right] ^{5}=e^{5}
\end{eqnarray*}%
$\medskip \medskip $Solution 2:%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{5}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }e^{\ln \left( 1+\dfrac{5}{n}\right)
^{n}}=\lim\limits_{n\rightarrow \infty }e^{n\ln \left( 1+\dfrac{5}{n}\right)
}=e^{\lim\limits_{n\rightarrow \infty }n\ln \left( 1+\dfrac{5}{n}\right) }
\end{equation*}%
Looking at the exponent, 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }n\ln \left( 1+\dfrac{5}{n}\right)
=\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{5}{n}\right) }{%
\dfrac{1}{n}}=\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{5%
}{x}\right) }{\dfrac{1}{x}}
\end{equation*}%
This is an indeterminant of the form $\dfrac{0}{0}$ and so we can apply l'H%
\^{o}pital's rule.%
\begin{equation*}
\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{5}{x}\right) }{%
\dfrac{1}{x}}=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{1+\dfrac{5}{%
x}}\cdot -\dfrac{5}{x^{2}}}{-\dfrac{1}{x^{2}}}=\lim\limits_{x\rightarrow
\infty }\dfrac{5}{1+\dfrac{5}{x}}=5
\end{equation*}%
Thus the sequence converges to $e^{5}$.

d) \ $\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right) ^{n}$ $%
\medskip $

Solution : 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{-1}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{-n}\right)
^{n\cdot \tfrac{-1}{-1}}=\lim\limits_{n\rightarrow \infty }\left[ \left( 1+%
\dfrac{1}{-n}\right) ^{-n}\right] ^{-1}
\end{equation*}%
Define $m=-n$%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left[ \left( 1+\dfrac{1}{-n}\right) ^{-n}%
\right] ^{-1}=\left[ \lim\limits_{m\rightarrow -\infty }\left( 1+\dfrac{1}{m}%
\right) ^{m}\right] ^{-1}=e^{-1}
\end{equation*}%
Solution 2:%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }e^{\ln \left( 1-\tfrac{1}{n}\right)
^{n}}=\lim\limits_{n\rightarrow \infty }e^{n\ln \left( 1-\tfrac{1}{n}\right)
}=e^{\lim\limits_{n\rightarrow \infty }n\ln \left( 1-\tfrac{1}{n}\right) }
\end{equation*}%
Looking at the exponent, 
\begin{equation*}
\lim\limits_{n\rightarrow \infty }n\ln \left( 1-\dfrac{1}{n}\right)
=\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 1-\dfrac{1}{n}\right) }{%
\dfrac{1}{n}}=\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1-\dfrac{1%
}{x}\right) }{\dfrac{1}{x}}
\end{equation*}%
This is an indeterminant of the form $\dfrac{0}{0}$ and so we can apply l'H%
\^{o}pital's rule.%
\begin{equation*}
\lim\limits_{x\rightarrow \infty }\dfrac{\ln \left( 1-\dfrac{1}{x}\right) }{%
\dfrac{1}{x}}=\lim\limits_{x\rightarrow \infty }\dfrac{\dfrac{1}{1-\dfrac{1}{%
x}}\cdot \dfrac{1}{x^{2}}}{-\dfrac{1}{x^{2}}}=\lim\limits_{x\rightarrow
\infty }\dfrac{-1}{1-\dfrac{1}{x}}=-1
\end{equation*}%
Thus the sequence converges to $e^{-1}$.
\end{enumerate}

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