THEOREM A Limit Is Unique

If a function $f$ is defined on an open interval containing the number $c$, except possibly at $c$ itself, and if $\lim\limits_{x\rightarrow c}f( x) = L_{1}$ and $\lim\limits_{x\rightarrow c}f(x) = L_{2},$ then $L_{1} = L_{2}.$

Recall

An indirect proof (a proof by contradiction) begins by assuming the conclusion is false. Then we show that this assumption leads to a contradiction.

Proof

Assume that $L_{1}\neq L_{2}$. We will show that this assumption leads to a contradiction. By the definition of the limit of a function $f,$ $\lim\limits_{x\rightarrow c}f(x) = L_{1}$ if, for any given number $\varepsilon > 0$, there is a number $\delta _{1} > 0$, so that $$\begin{equation} {\rm whenever}\ 0 < \vert x-c\vert < \delta_{1}\quad {\rm then}\ \vert f(x)-L_{1}\vert < \varepsilon\tag{1} \end{equation}$$

Similarly, $\lim\limits_{x\rightarrow c}f(x) = L_{2}$ if, for any given number $\varepsilon > 0$, there is a number $\delta_{2} > 0$, so that $$\begin{equation} {\rm whenever}\ 0 < \vert x-c\vert < \delta_{2}\quad {\rm then }\ \vert f(x)-L_{2}\vert < \varepsilon\tag{2} \end{equation}$$

Now $$\begin{equation*} L_{1} - L_{2} = L_{1}-f(x) + f(x)-L_{2} \end{equation*}$$

so, by applying the Triangle Inequality, we get $$\begin{equation} \vert L_{1}-L_{2}\vert = \vert L_{1}-f(x)+f(x)-L_{2}\vert \leq \vert L_{1}-f(x)\vert + \vert f(x)-L_{2}\vert\tag{3} \end{equation}$$

For any given number $\varepsilon > 0$, let $\delta$ be the smaller of $\delta _{1}$ and $\delta _{2}.$ Then from (1)–(3), we can conclude that whenever $0 < \vert x-c \vert < \delta \leq \delta_{1}$ and $0 < \vert x-c\vert < \delta \leq \delta _{2}$, we have $$\begin{equation} \vert L_{1}-L_{2}\vert < \varepsilon +\varepsilon = 2 \varepsilon \tag{4} \end{equation}$$

In particular, (4) is true for $\varepsilon =\dfrac{1}{2}\vert L_{1}-L_{2}\vert >0$. (Remember $L_{1}\neq L_{2}$.) Then from (4), $$\begin{equation*} \vert L_{1}-L_{2}\vert < 2\varepsilon = \vert L_{1}-L_{2}\vert \end{equation*}$$

which is a contradiction. Therefore, $L_{1}=L_{2}$, and the limit, if it exists, is unique.

THEOREM Limit of a Sum

If $f$ and $g$ are functions for which $\lim\limits_{x\rightarrow c}f( x)$ and $\lim\limits_{x\rightarrow c}g(x)$ both exist, then $\lim\limits_{x\rightarrow c}\,[f(x)+g(x)]$ exists and $$\lim\limits_{x\rightarrow c}[f(x)+g(x)]=\lim\limits_{x\rightarrow c}f(x)+\lim\limits_{x\rightarrow c}g(x)$$

Proof

Suppose $\lim\limits_{x\rightarrow c}f(x)=L$ and $\lim\limits_{x\rightarrow c}g(x)=M.$ We need to show that for any number $\varepsilon > 0,$ there is a number $\delta > 0,$ so that $$\begin{equation*} {\rm whenever}\ 0 < \vert x-c\vert < \delta\quad {\rm then}\ \vert [ f(x)+g(x)] - [ L+M] \vert < \varepsilon \end{equation*}$$

Since $\lim\limits_{x\rightarrow c}f(x)$ $=L$, by the definition of a limit, given the number $\dfrac{\varepsilon }{2} > 0$, there is a number $\delta_{1} > 0$, so that $$\begin{equation*} {\rm whenever}\ 0 < \vert x-c\vert < \delta_{1}\quad {\rm then}\ \vert f(x)-L\vert < \dfrac{\varepsilon }{2} \end{equation*}$$

Since $\lim\limits_{x\rightarrow c}g(x)=M$, for this same number $\dfrac{\varepsilon }{2},$ there is a number $\delta _{2} > 0,$ so that $$\begin{equation*} {\rm whenever}\ 0 < \vert x-c\vert < \delta _{2}\quad {\rm then}\ \vert g(x)-M\vert < \dfrac{\varepsilon }{2} \end{equation*}$$

Let $\delta$ be the smaller of $\delta _{1}$ and $\delta _{2}$. Then $\delta \leq \delta _{1}$ and $\delta \leq \delta _{2}$. Using this $\delta$, $$\begin{eqnarray*} {\rm whenever}\ 0 & < \vert x-c\vert <\delta\quad {\rm then}\ \vert f(x)-L\vert < \dfrac{\varepsilon }{2} \\[4.5pt] {\rm whenever}\ 0 & < \vert x-c\vert < \delta\quad {\rm then}\ \vert g(x)-M\vert < \dfrac{\varepsilon }{2} \end{eqnarray*}$$

That is, whenever $0 < \vert x-c\vert < \delta,$ $$\begin{eqnarray*} \begin{array}{ll} \vert [ f(x)+g(x)] -[ L+M]\vert & = \vert [ f(x)-L] + [ g(x)-M] \vert \\[4pt] & \leq \vert f(x)-L\vert + \vert g(x)-M \vert\quad {\color{#0066A7}{\hbox{Use the Triangle Inequality.}}} \\[4pt] & < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon }{2} = \varepsilon \end{array}\end{eqnarray*}$$ So, $\lim\limits_{x\rightarrow c} [ f(x)+g(x)] =L+M$.

THEOREM Limit of a Product

If $f$ and $g$ are functions for which $\lim\limits_{x\rightarrow c}f(x)$ and $\lim\limits_{x\rightarrow c}g(x)$ both exist,
then $\lim\limits_{x\rightarrow c}\,[f(x)\cdot g(x)]$ exists and $$\begin{eqnarray*} \lim\limits_{x\rightarrow c}\,[f(x)\cdot g(x)]=\lim\limits_{x\rightarrow c}f(x)\cdot \lim\limits_{x\rightarrow c}g(x) \end{eqnarray*}$$

Proof

Suppose $\lim\limits_{x\rightarrow c}f(x) = L$ and $\lim\limits_{x\rightarrow c}g(x) = M.$ We need to show that for any number $\varepsilon > 0,$ there is a number $\delta > 0$, so that $$\begin{equation*} {\rm whenever}\ 0 < \vert x-c\vert < \delta\quad {\rm then}\ \vert f(x)\cdot g(x)-L\cdot M\vert < \varepsilon \end{equation*}$$

Subtracting and adding $f(x)\cdot M$ in the expression $f(x)\cdot g(x)-L\cdot M$ result in terms involving $g(x)-M$ and $f(x)-L$:

Since $\lim\limits_{x\rightarrow c}f(x)=L$ and $\lim\limits_{x\rightarrow c}g(x)=M$, then there is a number $\delta _{1} > 0$, so that whenever $0 < \vert x-c\vert <\delta _{1}$, then $$\begin{equation} \vert f(x)-L\vert < 1, \qquad \hbox{from which}\qquad \vert f(x)\vert < 1 + \vert L\vert\tag{6} \end{equation}$$

Also given a number $\varepsilon >0$, there is a number $\delta _{2}$, so that whenever $0 < \vert x-c\vert < \delta_{2}$, then $$\begin{equation} \vert g(x)-M\vert < \dfrac{\varepsilon }{1+ \vert L\vert +\vert M\vert }\tag{7} \end{equation}$$

Given a number $\varepsilon >0$, there is a number $\delta _{3}$, so that whenever $0 < \vert x-c \vert <\delta _{3}$, then $$\begin{equation} \vert f(x)-L\vert <\dfrac{\varepsilon }{1+\vert L\vert +\vert M\vert }\tag{8} \end{equation}$$

Choose $\delta$ to be the minimum of $\delta _{1},$ $\delta _{2},$ and $\delta _{3}$ and combine (5)–(8). Then for any given $\varepsilon >0,$ there is a $\delta >0$, so that whenever $0 < \vert x-c\vert <\delta$, we have $$\begin{eqnarray*} \vert f(x)\cdot g(x)-L\cdot M\vert & < & [ 1 + \vert L\vert ] \dfrac{\varepsilon }{1 + \vert L\vert + \vert M\vert } + \vert M\vert \dfrac{\varepsilon }{ 1 + \vert L\vert + \vert M\vert } \\[4pt] & < & [ 1 + \vert L\vert + \vert M\vert ] \dfrac{ \varepsilon }{1 + \vert L\vert + \vert M\vert } =\varepsilon \end{eqnarray*}$$

That is, $\lim\limits_{x\ c}\,[f(x)\cdot g(x)]=\lim\limits_{x\rightarrow c}f(x)\cdot \lim\limits_{x\rightarrow c}g(x).$

THEOREM Squeeze Theorem

If the functions $f$, $g$, and $h$ have the property that for all $x$ in an open interval containing $c$, except possibly at $c$ itself, $$f(x)\leq g(x)\leq h(x)$$ and if $$\lim_{x\rightarrow c}f(x) = \lim_{x\rightarrow c}h(x) = L$$ then $$\lim_{x\rightarrow c}g(x) = L$$

Proof

Since $\lim\limits_{x\rightarrow c}f(x) = \lim\limits_{x\rightarrow c}h(x) = L,$ then for any number $\varepsilon > 0$, there are positive numbers $\delta _{1}$ and $\delta _{2}$, so that $$\begin{eqnarray*} {\rm whenever}\ 0 & < \vert x-c \vert < \delta_{1}\quad \hbox{then } \vert f(x)-L \vert < \varepsilon\\ {\rm whenever}\ 0 & < \vert x-c\vert < \delta_{2}\quad \hbox{then } \vert h(x)-L \vert < \varepsilon \end{eqnarray*}$$

Choose $\delta$ to be the smaller of the numbers $\delta_{1}$ and $\delta_{2}$. Then $0 < \vert x-c\vert < \delta$ implies that both $\vert f(x)-L \vert < \varepsilon$ and $\vert h(x)-L \vert < \varepsilon$. In other words, $0 < \vert x-c \vert < \delta$ implies that both $$\begin{equation*} L - \varepsilon < f(x) < L + \varepsilon\qquad \hbox{and}\qquad L - \varepsilon < h(x) < L + \varepsilon \end{equation*}$$

Since $f(x)\leq g(x)\leq h(x)$ for all $x\neq c$ in the open interval, it follows that whenever $0 < \vert x-c\vert < \delta$ and $x$ is in the open interval, we have $$\begin{equation*} L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon \end{equation*}$$

Then for any given number $\varepsilon >0$, there is a positive number $\delta$, so that whenever $0 < \vert x-c\vert < \delta,$ then $L - \varepsilon < g(x) < L + \varepsilon$, or equivalently, $\vert g(x) -L\vert < \varepsilon.$ That is, $\lim\limits_{x\rightarrow c}g(x) = L.$