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%TCIDATA{<META NAME="Title" CONTENT="Problem Set 1 - long - Math 207 - Spring 2011">}
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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}
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\newtheorem{exercise}[theorem]{Exercise}
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\lhead{\color{blue} \Large Lecture Notes}
\lfoot{\small   \copyright $\;$  Hidegkuti,  Powell,  2012}
\cfoot{}
\chead{\LARGE Sequences  2}
\rfoot{\small Last revised: March 21, 2013}
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Theorem: \ (The Continuous Function Theorem for Sequences) \ \ Let $\left\{
a_{n}\right\} $ be a sequence of real numbers. \ 

\qquad If $a_{n}\rightarrow L$ and if $f$ is a function that is continuous
at $L$ and defined at all $a_{n}$, then $f\left( a_{n}\right) \rightarrow
f\left( L\right) $.%
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\medskip

Example 1. \ a) $\ \sqrt{\dfrac{n+1}{n}}\rightarrow 1$\bigskip

proof: \ $\dfrac{n+1}{n}\rightarrow 1$ and $f\left( x\right) =\sqrt{x}$ is
continuous at $x=1$. \ Thus $\sqrt{\dfrac{n+1}{n}}\rightarrow \sqrt{1}=1$%
\bigskip

\qquad b) \ $\ln \left( \dfrac{n^{2}-1}{n^{2}+1}\right) \rightarrow \ln 1=0$%
\bigskip

\qquad c) \ $2^{1/n}\rightarrow 2^{0}=1$\bigskip

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Theorem: \ Suppose that $f\left( x\right) $ is a function defined for all $%
x\geq n_{0}$ for some $n_{0}\in 
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\mathbb{N}
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$ and that $\left\{ a_{n}\right\} $ is a sequence of real numbers such that $%
a_{n}=f\left( n\right) $ for all $n\geq n_{0}$ \ Then 
\begin{equation*}
\text{if }\lim\limits_{x\rightarrow \infty }f\left( x\right) =L\text{, then }%
\lim\limits_{n\rightarrow \infty }a_{n}=L
\end{equation*}

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\bigskip

Then we can apply L'H%
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pital's rule to find limits of sequences.\bigskip

Example 2. \ a) \ $\lim\limits_{n\rightarrow \infty }\dfrac{\ln n}{n}%
=0\medskip $

proof: \ $\lim\limits_{n\rightarrow \infty }\dfrac{\ln n}{n}%
=\lim\limits_{x\rightarrow \infty }\dfrac{\ln x}{x}=\lim\limits_{x%
\rightarrow \infty }\dfrac{\dfrac{1}{x}}{1}=0$\bigskip

b) \ $\lim\limits_{n\rightarrow \infty }\left( \dfrac{n+1}{n-1}\right)
^{n}=e^{2}$\bigskip

proof: \ $\ln a_{n}=\ln \left( \dfrac{n+1}{n-1}\right) ^{n}=n\ln \left( 
\dfrac{n+1}{n-1}\right) =\dfrac{\ln \left( \dfrac{n+1}{n-1}\right) }{\dfrac{1%
}{n}}\medskip $

\qquad This is an indeterminate of the type $\dfrac{0}{0}\medskip \medskip $

\qquad $\lim\limits_{n\rightarrow \infty }\ln
a_{n}=\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( \dfrac{n+1}{n-1}%
\right) }{\dfrac{1}{n}}=\lim\limits_{n\rightarrow \infty }\dfrac{-\dfrac{2}{%
n^{2}-1}}{-\dfrac{1}{n^{2}}}=\lim\limits_{n\rightarrow \infty }\left( -%
\dfrac{2}{n^{2}-1}\right) \left( -n^{2}\right) =\lim\limits_{n\rightarrow
\infty }\dfrac{2n^{2}}{n^{2}-1}=2\bigskip $

$\qquad \ln a_{n}\rightarrow 2$ and $f\left( x\right) =e^{x}$ is continuous
on $%
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$. \ thus $a_{n}=e^{\ln a_{n}}\rightarrow e^{2}$\bigskip

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Commonly Occurring Limits. \ In each of the following, $x$ is a fixed
number.\bigskip 
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\begin{enumerate}
\item $\lim\limits_{n\rightarrow \infty }\dfrac{\ln n}{n}=0$

\item $\lim\limits_{n\rightarrow \infty }\sqrt[n]{n}=1$

\item $\lim\limits_{n\rightarrow \infty }x^{1/n}=1$ \ \ \ $\ \ x>0$

\item $\lim\limits_{n\rightarrow \infty }x^{n}=0$ \ \ \ \ \ $-1<x<1$

\item $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{x}{n}\right)
^{n}=e^{x}$ \ \ \ \ (any $x$)

\item $\lim\limits_{n\rightarrow \infty }\dfrac{x^{n}}{n!}=0$ \ \ \ \ \ (any 
$x$) \ \bigskip
\end{enumerate}

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\bigskip

proof. \ 1) was done using L'H%
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\^{o}%
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pital's rule\bigskip

\qquad 2) \ $\lim\limits_{n\rightarrow \infty }\sqrt[n]{n}=1$

$\qquad $proof: $\ \lim\limits_{n\rightarrow \infty }\sqrt[n]{n}%
=\lim\limits_{n\rightarrow \infty }n^{1/n}=\lim\limits_{x\rightarrow \infty
}x^{1/x}$\qquad We will think of $x^{1/x}$ as $e^{\ln \left( x^{1/x}\right)
} $

$\qquad \lim\limits_{x\rightarrow \infty }\ln \left( x^{1/x}\right)
=\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}\ln
x=\lim\limits_{x\rightarrow \infty }\dfrac{\ln x}{x}=0$\bigskip

$\qquad f\left( x\right) =e^{x}$ is continuous on $%
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$

$\qquad \ln \left( x^{1/x}\right) \rightarrow 0$ and so $e^{\ln \left(
x^{1/x}\right) }\rightarrow e^{0}$ \ \ and so\ \ \ $x^{1/x}\rightarrow 1$%
\bigskip

\qquad 3) \ $\lim\limits_{n\rightarrow \infty }a^{1/n}=1$ \ \ \ $\ \ a>0$%
\bigskip

proof: \ We will prove that $\lim\limits_{x\rightarrow \infty }a^{1/x}=1$

\begin{equation*}
\lim\limits_{x\rightarrow \infty }a^{1/x}=\lim\limits_{x\rightarrow \infty
}e^{\ln a^{1/x}}=\lim\limits_{x\rightarrow \infty }e^{1/x\ln a}=e^{0}=1
\end{equation*}%
\bigskip \bigskip

\qquad 4) \ $\lim\limits_{n\rightarrow \infty }x^{n}=0$ \ for all$\ \ x<1$%
\bigskip

proof: \ Let $\varepsilon >0$ be given. \ Since (by limit 3) \ $%
\lim\limits_{n\rightarrow \infty }\varepsilon ^{1/n}=1$ and $\left\vert
x\right\vert <1$, there exists $N$ such that for all $n>N$, 
\begin{equation*}
\varepsilon ^{1/n}>\left\vert x\right\vert \text{ \ and so \ }\varepsilon
>\left\vert x\right\vert ^{n}\geq x^{n}
\end{equation*}%
\bigskip

\qquad 5.) \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{x}{n}\right)
^{n}=e^{x}$ \ \ \ \ for any $x$\bigskip

proof: \ We will present two methods. \ In the first method, we will use the
fact that $e=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n}$.%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{x}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~\dfrac{n}{x}~}%
\right) ^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~\dfrac{n}{%
x}~}\right) ^{\dfrac{n}{x}\cdot x}=\lim\limits_{n\rightarrow \infty }\left[
\left( 1+\dfrac{1}{~\dfrac{n}{x}~}\right) ^{\dfrac{n}{x}}\right] ^{x}=\left[
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~\dfrac{n}{x}~}\right)
^{\dfrac{n}{x}}\right] ^{x}=e^{x}
\end{equation*}

\bigskip

\qquad Method 2: \ We will use the identity $x=e^{\ln x}$\ \ and L'H%
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\^{o}%
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pital's rule.%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{x}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }e^{\ln \left( 1+\tfrac{x}{n}\right)
^{n}}=e^{\lim\limits_{n\rightarrow \infty }\ln \left( 1+\tfrac{x}{n}\right)
^{n}}=e^{\lim\limits_{n\rightarrow \infty }n\ln \left( 1+\tfrac{x}{n}\right)
}=e^{\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 1+\tfrac{x}{n}%
\right) }{\tfrac{1}{n}}}=e^{M}
\end{equation*}%
The exponent is now an indeterminate of the type $\dfrac{0}{0}$, so we may
apply L'H%
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\^{o}%
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pital's rule. \ We differentiate with respect to $n$. \ 
\begin{equation*}
M=\lim\limits_{n\rightarrow \infty }\dfrac{\ln \left( 1+\dfrac{x}{n}\right) 
}{\dfrac{1}{n}}=\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{1}{1+\dfrac{x%
}{n}}\cdot \left( -\dfrac{x}{n^{2}}\right) }{-\dfrac{1}{n^{2}}}%
=\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{1}{\dfrac{n+x}{n}}\cdot
\left( -\dfrac{x}{n^{2}}\right) }{-\dfrac{1}{n^{2}}}=\lim\limits_{n%
\rightarrow \infty }\dfrac{n}{n+x}\cdot \left( -\dfrac{x}{n^{2}}\right)
\cdot \left( -n^{2}\right) =\lim\limits_{n\rightarrow \infty }\dfrac{nx}{n+x}
\end{equation*}%
We may apply L'H%
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\^{o}%
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pital's rule again or just divide both numerator and denominator by $n$ to
see that this limit $M=x$:%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\dfrac{nx}{n+x}=\lim\limits_{n\rightarrow
\infty }\dfrac{x}{1+\dfrac{x}{n}}=x
\end{equation*}%
and so the entire limit is $e^{M}=e^{x}$.

\bigskip

\qquad 6.) \ $\lim\limits_{n\rightarrow \infty }\dfrac{x^{n}}{n!}=0$ \ \ \ \
\ (any $x$)\bigskip

proof: \ Let us first assume that $x>0$. \ Let $M$ be an integer for which $%
\dfrac{x}{M}<1$.

\begin{eqnarray*}
\dfrac{x^{n}}{n!} &=&\dfrac{x\cdot x\cdot ...\cdot x}{1\cdot 2\cdot 3\cdot
...\cdot n}=\dfrac{x\cdot x\cdot ...\cdot x}{1\cdot 2\cdot 3\cdot ...\cdot
M\cdot \left( M+1\right) \cdot \left( M+2\right) \cdot ...\cdot n} \\
&\leq &\dfrac{x^{n}}{1\cdot 2\cdot 3\cdot ...\cdot M\cdot M\cdot M\cdot
...\cdot M}=\dfrac{x^{n}}{M!\cdot M^{n-M}} \\
&=&\dfrac{x^{n}}{M!\cdot M^{n-M}}\cdot \dfrac{M^{M}}{M^{M}}=\dfrac{M^{M}}{M!}%
\cdot \dfrac{x^{n}}{M^{n}}=\dfrac{M^{M}}{M!}\left( \dfrac{x}{M}\right) ^{n}
\end{eqnarray*}%
\bigskip

Since $\dfrac{M^{M}}{M!}$ is a constant and $\left( \dfrac{x}{M}\right)
^{n}\rightarrow 0$, so does $\dfrac{x^{n}}{n!}$.

If $x\leq 0$, then we use the sandwich theorem:%
\begin{equation*}
-\left\vert \dfrac{x^{n}}{n!}\right\vert \leq \dfrac{x^{n}}{n!}\leq
\left\vert \dfrac{x^{n}}{n!}\right\vert
\end{equation*}%
and since $\left\vert \dfrac{x^{n}}{n!}\right\vert \rightarrow 0$, we also
have $\dfrac{x^{n}}{n!}\rightarrow 0$.

\vspace{0.4in}

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