53 lines
2.8 KiB
TeX
53 lines
2.8 KiB
TeX
\section{Limit Theorems}
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\begin{definition}
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Let $A \subseteq \R$, let $f:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. We say that $f$ is \textbf{bounded on a neighborhood of $c$} if there exists a $\delta$-neighborhood $V_\delta(c)$ of $c$ and a constant $M > 0$ such that we have $|f(x)| \leq M$ for all $x \in A \cap V_\delta (c)$.
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\end{definition}
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\begin{theorem}
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If $A \subseteq \R$ and $f:A \rightarrow \R$ has a limit at $c \in \R$, then $f$ is bounded on some neighborhood of $c$.
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\end{theorem}
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\begin{theorem}
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Let $A \subseteq \R$ and let $f$ and $g$ be functions defined on $A$ to $\R$. We define the \textbf{sum} $f+g$, the \textbf{difference} $f-g$, and the \textbf{product} $fg$ on $A$ to $\R$ to be the functions given by
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\[(f+g)(x):=f(x)+g(x),\]
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\[(f-g)(x):=f(x)-g(x),\]
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\[(fg)(x):=f(x)g(x)\]
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for all $x \in A$. Further, if $b \in \R$, we define the \textbf{multiple} $bf$ to be the function given by
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\[(bf)(x) := bf(x)\ \ \ \text{for all}\ \ \ x \in A\]
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Finally, if $h(x)\neq 0$ for $x \in A$, we define the \textbf{quotient} $f/h$ to be the function given by
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\[\left( \frac{f}{h}\right)(x) := \frac{f(x)}{h(x)}\ \ \ \ \text{for all}\ \ \ \ x \in A\]
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\end{theorem}
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\begin{theorem}
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let $A \subseteq \R$, let $f$ and $g$ be functions on $A$ to $\R$, and let $c \in \R$ be a cluster point of $A$. Further, let $b \in \R$.
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\begin{enumerate}
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\item If $\lim\limits_{x\to c} f = L$ and $\lim\limits_{x\to c} g = M$, then
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\[\lim\limits_{x\to c} (f+g) = L+M,\]
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\[\lim\limits_{x\to c} (f-g)=L-M,\]
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\[\lim\limits_{x\to c} (fg) = LM,\]
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\[\lim\limits_{x\to c} (bf) = bL.\]
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\item If $h: A \rightarrow \R$, if $h(x) \neq 0$ for all $x \in A$, and if $\lim\limits_{x\to c} h = H \neq 0$, then
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\[\lim\limits_{x\to c} \left( \frac{f}{h} \right)= \frac{L}{H}\]
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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Let $A \subseteq \R$, let $f: A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. If
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\[a \leq f(x) \leq b\ \ \ \ \text{for all}\ \ \ \ x \in A,\ x \neq c,\]
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and if $\lim\limits_{x\to c} f$ exists, then $a \leq \lim\limits_{x\to c} f \leq b$.
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\end{theorem}
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\begin{theorem}[\textbf{Squeeze Theorem}]
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Let $A \subseteq \R$, let $f,g,h:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. If
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\[f(x) \leq g(x) \leq h(x)\ \ \ \ \text{for all}\ \ \ \ x \in A,\ x \neq c,\]
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and if $\lim\limits_{x\to c} f = L = \lim\limits_{x\to c} h$, then $\lim\limits_{x\to c} g =L$.
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\end{theorem}
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\begin{theorem}
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Let $A \subseteq \R$, let $f:A \rightarrow \R$ and let $c \in \R$ be a cluster point of $A$. If
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\[\lim\limits_{x\to c} f > 0\ \ \ \left[\textit{respectively, } \lim\limits_{x\to c} f < 0\right],\]
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then there exists a neighborhood $V_\delta (c)$ of $c$ such that $f(x) > 0$ [respectively, $f(x) < 0$] for all $x \in A \cap V_\delta (c),\ x \neq c$.
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\end{theorem}
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