OBJECTIVES

When you finish this section, you should be able to:

1. Use the \(\epsilon \)-\(\delta \) definition of a limit (p. 132)

“ \(x\) must be within \(\dfrac{\epsilon^{\vphantom{\vrule height2pt width0pt depth0pt}} }{3_{\vphantom{\vrule height1.7pt width0pt depth0pt}}}\) of \(2\), but not equal to \(2\), to guarantee that \(f(x)\) is within \(\epsilon \) of \(7\)”

NOW WORK

Problem 41.

spanDEFINITIONspan Limit of a Function

Let \(f\,\) be a function defined everywhere in an open interval containing \(c\), except possibly at \(c\). Then the limit as \({x}\) approaches \({c}\) of \({f(x)}\) is \({L}\), written \[\bbox[5px, border:1px solid black, #F9F7ED]{ \lim\limits_{x\rightarrow c}f(x)=L} \]

if given any number \(\epsilon >0\), there is a number \(\delta >\) \(0\) so that \[\bbox[5px, border:1px solid black, #F9F7ED]{ \hbox{whenever\(\,\ \ 0 \lt \vert x-c\vert \lt \delta \) then \(\vert f(x)-L\vert \lt \epsilon \)}} \]

Using the \(\epsilon \)-\(\delta\) Definition of a Limit

Use the \(\epsilon \)-\(\delta \) definition of a limit to prove \( \lim\limits_{x\rightarrow -1} (1-2x) =3\).

Solution Given any \(\epsilon \gt0\), we must show there is a number \(\delta \gt0\) so that \begin{equation*} \hbox{whenever }\quad 0 \lt \vert x-(-1) \vert \lt \delta \qquad \hbox{then }\vert ( 1-2x) -3\vert \lt \epsilon \end{equation*}

The idea is to find a connection between \(\vert x-(-1)\vert =|x+1|\ \) and \(\vert ( 1-2x) -3\vert \). Since \begin{equation*} \vert (1-2x)-3\vert =\vert -2x-2\vert = \vert -2(x+1)\vert = \vert -2\vert \cdot \vert x+1\vert =2\vert x+1\vert \end{equation*}

we see that for any \(\epsilon >0,\) \begin{equation*} \hbox{whenever}\quad \vert x-(-1)\vert =\vert x+1\vert \lt \dfrac{\epsilon }{2}\qquad \hbox{ then }\vert (1-2x)-3\vert \lt \epsilon \hbox{ } \end{equation*}

That is, given any \(\epsilon \gt 0\) there is a \(\delta,\delta =\dfrac{\epsilon}{2}\), so that whenever \(0 \lt \vert x-(-1)\vert \lt \delta\), we have \(\vert (1-2x)-3\vert \lt \epsilon \). This proves that \(\lim\limits_{x\rightarrow -1}(1-2x)=3\).

NOW WORK

Problem 21.

Using the \(\epsilon \)-\(\delta\) Definition of a Limit

Use the \(\epsilon \)-\(\delta \) definition of a limit to prove that:

1. \(\lim\limits_{x\rightarrow c}A=A\), where \(A\) and \(c\) are real numbers
2. \(\lim\limits_{x\rightarrow c}x=c\), where \(c\) is a real number

Solution

(a) \(f(x)=A\) is the constant function whose graph is a horizontal line. Given any \(\epsilon >0\), we must find \(\delta >0\) so that whenever \(0 \lt \vert x-c\vert \lt \delta ,\) then \(\left\vert f(x)-A\right\vert \lt \epsilon \).

Since \(\left\vert A-A\right\vert =0,\) then \(\left\vert f(x)-A\right\vert \lt \epsilon \) no matter what positive number \(\delta \) is used. That is, any choice of \(\delta \) guarantees that whenever \(0 \lt \vert x-c\vert \lt \delta \), then \(\left\vert f(x)-A\right\vert \lt \epsilon \).

(b) \(f( x) =x\) is the identity function. Given any \( \epsilon >0\), we must find \(\delta \) so that whenever \( 0 \lt \vert x-c\vert \lt \delta \), then \(\vert f(x)-c \vert =\vert x-c\vert \lt \epsilon \). The easiest choice is to make \(\delta =\epsilon \). That is, whenever \(0 \lt \vert x-c\vert \lt \delta =\epsilon \), then \(\vert f(x)-c \vert \hspace{-6.5pt}\underset{\underset{\color{#0066A7}{\scriptsize\hbox{f(x) = x}}}{\color{#0066A7}{\uparrow}}}{=}\hspace{-6.5pt} \vert x-c\vert \lt \epsilon \).

• The limit of a function in no way depends on the value of the function at \(c\).
• In general, the size of \(\delta \) depends on the size of \(\epsilon\).
• For any \(\epsilon \), if a suitable \(\delta \) has been found, any smaller positive number will also work for \(\delta \). That is, \(\delta \) is not uniquely determined when \(\epsilon \) is given.

Using the \(\epsilon \)-\(\delta \) Definition of a Limit

Prove: \(\lim\limits_{x\rightarrow 2}x^{2}=4\).

Solution Given any \(\epsilon >0\), we must show there is a number \(\delta >0\) so that \begin{equation*} \hbox{whenever }\quad 0 \lt \vert x-2\vert \lt \delta \qquad \hbox{ then } \vert x^{2}-4\vert \lt \epsilon \end{equation*}

To establish a connection between \(\vert x^{2}-4\vert \) and \( \vert x-2\vert \), we write \(\left\vert x^{2}-4\right\vert \) as \begin{equation*} \left\vert x^{2}-4\right\vert =\left\vert \,(x+2)(x-2)\right\vert =\left\vert x+2\right\vert \cdot \vert x-2\vert \end{equation*}

Now, if we can find a number \(K\) for which \(\left\vert x+2\right\vert \lt K\), then we can choose \(\delta =\dfrac{\epsilon }{K}\). To find \(K\), we restrict \( x\) to some interval centered at \(2\). For example, suppose the distance between \(x\) and \(2\) is less than \(1\). Then \begin{equation*} \begin{array}{rcl@{\qquad}l} \\[-5pc] &&\vert x-2\vert \lt 1 & \\[3pt] -1&\lt&x-2 \lt 1 & \\[3pt] 1&\lt&x \lt 3 & {\color{#0066A7}{\hbox{Simplify.}}} \\ 1+2&\lt&x+2 \lt 3+2 & {\color{#0066A7}{\hbox{Add 2 to each part. }}} \\ 3&\lt&x+2 \lt 5 & \end{array} \end{equation*}

In particular, we have \(\vert x+2\vert \lt 5\). It follows that whenever \(\vert x-2\vert \lt 1\), \begin{equation*} \vert x^{2}-4\vert =\left\vert x+2\right\vert \cdot \vert x-2\vert \lt 5\,\vert x-2\vert \hbox{ } \end{equation*}

If \(\vert x-2\vert \lt \delta =\dfrac{\epsilon }{5}\), then \( \left\vert \hbox{ }x^{2}-4\right\vert \lt 5\,\vert x-2\vert \lt 5\cdot \dfrac{\epsilon }{5}=\epsilon \), as desired.

But before choosing \(\delta =\dfrac{\epsilon }{5}\), we must remember that there are two constraints on \(\vert x-2\vert \). Namely, \begin{equation*} \vert x-2\vert \lt 1\qquad \hbox{and}\qquad \vert x-2\vert \lt \dfrac{\epsilon }{5} \end{equation*}

To ensure that both inequalities are satisfied, we select \(\delta \) to be the smaller of the numbers 1 and \(\dfrac{\epsilon }{5}\), abbreviated as \( \delta =\min \left\{ 1,\dfrac{\epsilon }{5}\right\} \). Now, \[ \hbox{whenever } \vert x-2\vert \lt \delta =\min \left\{ 1,\dfrac{\epsilon }{5}\right\} \hbox{ then } \vert x^{2}-4\vert \lt \epsilon \]

proving \(\lim\limits_{x\rightarrow 2}x^{2}=4\).

NOW WORK

Problem 23.

Using the \(\epsilon \)-\(\delta \) Definition of a Limit

Prove \(\lim\limits_{x\rightarrow c}\dfrac{1}{x}=\dfrac{1}{c}\), where \(c\gt0\).

Solution The domain of \(f(x)=\dfrac{1}{x}\) is \(\{x|x\neq 0\} \).

For any \(\epsilon \gt0\), we need to find a positive number \(\delta \) so that whenever \(0 \lt \vert x-c\vert \lt \delta \), then \(\left\vert \dfrac{1}{x}-\dfrac{1}{c}\right\vert \lt \epsilon \). For \(x\neq 0\), and \(c\gt0\), we have \begin{equation*} \left\vert \frac{1}{x}-\frac{1}{c}\right\vert =\left\vert \frac{c-x}{xc} \right\vert =\frac{\vert c-x\vert }{\vert x\vert \cdot \vert c\vert }=\dfrac{|x-c|}{c\vert x\vert} \end{equation*}

The idea is to find a connection between \begin{equation*} \vert x-c\vert\qquad \hbox{ and }\qquad \dfrac{|x-c|}{c\vert x\vert } \end{equation*}

We proceed as in Example 3. Since we are interested in \(x\) near \(c\), we restrict \(x\) to a small interval around \(c\), say, \(\vert x-c\vert \lt \dfrac{c}{2}\). Then, \begin{equation*} \begin{array}{rcl@{\qquad}l} -\dfrac{c}{2}&\lt&x-c \lt \dfrac{c}{2}\\ \dfrac{c}{2}&\lt&x \lt \dfrac{3c}{2} & {\color{#0066A7}{\hbox{Add \(c\) to each expression.}}} \end{array} \end{equation*}

Since \(c \gt 0,\) then \(x\gt\dfrac{c}{2}\gt0,\) and \(\dfrac{1}{x} \lt \dfrac{2}{c}\). Now \begin{equation*} \left\vert \dfrac{1}{x}-\dfrac{1}{c}\right\vert =\dfrac{|x-c|}{c|x|} \lt \dfrac{2 }{c^{2}}\cdot |x-c|\qquad {\color{#0066A7}{\hbox{Substitute \(\dfrac{1}{\vert x\vert } \lt \dfrac{2}{c}\).}}} \end{equation*}

We can make \(\left\vert \dfrac{1}{x}-\dfrac{1}{c}\right\vert \lt\epsilon \) by choosing \(\delta =\dfrac{c^{2}}{2}\epsilon .\) Then \begin{equation*} \hbox{whenever }\vert x-c\vert \lt\delta =\dfrac{c^{2}}{2} \epsilon ,\hbox{ we have }\left\vert \dfrac{1}{x}-\dfrac{1}{c} \right\vert \lt \dfrac{2}{c^2} \cdot |x-c| \lt \dfrac{2}{c^2}\cdot \left(\dfrac{c^2}{2}\cdot \epsilon\right)=\epsilon \end{equation*}

But remember, there are two restrictions on \(|x-c|\). \begin{equation*} \vert x-c\vert \lt\dfrac{c}{2}\qquad \hbox{ and }\qquad \vert x-c\vert \lt\dfrac{c^{2}}{2}\cdot \epsilon \end{equation*}

So, given any \(\epsilon \gt0\), we choose \(\delta =\min \! \left(\! \dfrac{c}{2},\dfrac{c^{2}}{2}\cdot \epsilon\!\! \right) \). Then whenever \(0\lt\vert x-c\vert \lt\delta \), we have \(\left\vert \dfrac{1}{x}-\dfrac{1}{c} \right\vert \lt\epsilon .\) This proves \(\lim\limits_{x\rightarrow c}\dfrac{1 }{x}=\dfrac{1}{c},\) \(c\gt0\).

NOW WORK

Problem 25.

Using the \(\epsilon \)-\(\delta \) Definition of a Limit

Use the \(\epsilon \)-\(\delta \) definition of a limit to prove the statement \(\lim\limits_{x\rightarrow 3}(4x-5)\neq 10\).

Solution We use a proof by contradiction.* Assume \( \lim\limits_{x\rightarrow 3}(4x-5)=10\) and choose \(\epsilon =1\). (Any smaller positive number \(\epsilon \) will also work.) Then there is a number \(\delta >0\), so that \begin{equation*} \hbox{whenever }\quad 0\lt\vert x-3\vert \lt\delta\qquad \hbox{then}\quad \vert ( 4x-5) -10\vert \lt1\ \end{equation*}

We simplify the right inequality. \begin{equation*} \begin{array}{rcl} \vert (4x-5)-10 \vert &=& \vert 4x-15 \vert \lt1 \\[3pt] -1&\lt&4x-15\lt 1 \\[3pt] 14&\lt&4x\lt 16 \\[3pt] 3.5&\lt&x\lt 4 \end{array} \end{equation*}

According to our assumption, whenever \(0\lt\left\vert x-3\right\vert \lt\delta ,\) then \(3.5\lt x\lt 4\). Regardless of the value of \(\delta\), the inequality \( 0\lt \left\vert x-3\right\vert \lt \delta \) is satisfied by a number \(x\) that is less than \(3\). This contradicts the fact that \(3.5\lt x\lt 4\). The contradiction means that \(\lim\limits_{x\rightarrow 3}(4x-5)\neq 10.\)

NOW WORK

Problem 33.

Using the \(\epsilon \)-\(\delta \) Definition of a Limit

The Dirichlet function is defined by \(f(x)=\left\{ \begin{array}{l@{\quad}l} 1 & \hbox{if }x \hbox{ is rational} \\[3pt] 0 & \hbox{if }x \hbox{ is irrational} \end{array} \right. .\)

Prove \(\lim\limits_{x\rightarrow c}\) \(f( x) \) does not exist for any \(c.\)

Solution We use a proof by contradiction. That is, we assume that \(\lim\limits_{x\rightarrow c} f(x)\) exists and show that this leads to a contradiction.

Assume \(\lim\limits_{x\rightarrow c}f(x)=L\) for some number \(c\). Now if we are given \(\epsilon =\dfrac{1}{2}\) (or any smaller positive number), then there is a positive number \(\delta \), so that \begin{equation*} \hbox{whenever }\quad 0\lt\vert x-c\vert \lt\delta\qquad \hbox{ then }\quad \vert f( x) -L \vert \lt\dfrac{1}{2} \end{equation*}

Suppose \(x_{1}\) is a rational number satisfying \(0\lt \vert x_{1}-c\vert \lt\delta \), and \(x_{2}\) is an irrational number satisfying \(0\lt\vert x_{2}-c \vert \lt\delta \). Then from the definition of the function \(f\), \begin{equation*} f (x_{1}) =1\qquad \hbox{and}\qquad f( x_{2}) =0 \end{equation*}

Using these values in the inequality \(\vert f(x)-L\vert \lt\epsilon \), we get \begin{equation*} \begin{array}[t]{rcl} \vert f(x_{1})-L\vert &=&\vert 1-L\vert \lt\dfrac{1}{2}\\[5pt] -\dfrac{1}{2}&\lt&1-L\lt\dfrac{1}{2} \\[5pt] -\dfrac{3}{2}&\lt&-L\lt-\dfrac{1}{2} \\[5pt] \dfrac{1}{2}&\lt&L\lt\dfrac{3}{2} \end{array} \quad \hbox{ and }\quad \begin{array}[t]{rcl} \vert f(x_{2})-L\vert &=& \vert 0-L\vert \lt\dfrac{1}{2} \\[5pt] -\dfrac{1}{2}&\lt&-L\lt\dfrac{1}{2} \\[5pt] \dfrac{1}{2}&\gt&L\gt-\dfrac{1}{2} \\[5pt] -\dfrac{1}{2}&\lt&L\lt\dfrac{1}{2} \end{array} \end{equation*}

From the left inequality, we have \(L\gt\dfrac{1}{2}\), and from the right inequality, we have \(L\lt\dfrac{1}{2}\). Since it is impossible for both inequalities to be satisfied, we conclude that \(\lim\limits_{x\rightarrow c}f(x)\) does not exist.

Using the \(\epsilon \)-\(\delta \) Definition of a Limit

Prove that if \(\lim\limits_{x\rightarrow c}f(x)\gt0\), then there is an open interval around \(c\), for which \(f(x)\gt0\) everywhere in the interval except possibly at \(c\).

Solution Suppose \(\lim\limits_{x\rightarrow c}f(x)=L\gt0\). Then given any \(\epsilon \gt0\), there is a \(\delta \gt0\) so that \begin{equation*} \hbox{whenever }\quad 0\lt\vert x-c\vert \lt\delta \qquad \hbox{ then}\quad \vert f(x)-L\vert \lt\epsilon \ \end{equation*}

If \(\epsilon =\dfrac{L}{2}\), then from the definition of limit, there is a \(\delta \gt0\) so that \begin{equation*} \hbox{whenever } 0\lt\vert x-c\vert \lt\delta\quad \hbox{ then } \vert f(x)-L\vert \lt\dfrac{L}{2}\quad \hbox{ or equivalently, } \dfrac{L}{2}\lt f(x)\lt\dfrac{3L}{2} \end{equation*} Since \(\dfrac{L}{2}\gt0\), the last statement proves our assertion that \(f(x)\gt0\) for all \(x\) in the interval \((c-\delta ,c+\delta)\).

spanDEFINITIONspan Limit at Infinity

Let \(f\) be a function defined on an open interval \((b,\infty )\). Then \(f\) has a limit at infinity \begin{equation*} \lim_{x\rightarrow \infty }f(x)=L \end{equation*}

where \(L\) is a real number, if given any \(\epsilon \gt 0\), there is a positive number \(M\) so that whenever \(x \gt M\), then \(\vert f(x)-L\vert \lt \epsilon\).

If \(f\) is a function defined on an open interval \((-\infty , a)\), then \begin{equation*} \lim\limits_{x\rightarrow -\infty }f(x)=L \end{equation*}

if given any \(\epsilon \gt 0\), there is a negative number \(N\) so that whenever \(x \lt N\), then \(\vert f(x)-L\vert \lt \epsilon \).

spanDEFINITIONspan Infinite Limit

Let \(f\) be a function defined everywhere on an open interval containing \(c\), except possibly at \(c\). Then \(f(x)\) becomes unbounded in the positive direction (has an infinite limit) as \(x\) approaches \(c\), written \begin{equation*} \lim_{x\rightarrow c}f(x)=\infty \end{equation*}

if, for every positive number \(M\), a positive number \(\delta \) exists so that \[ \hbox{whenever } 0\lt\vert x-c\vert \lt \delta \hbox{ then } f(x) \gt M. \]

Similarly, \(f(x)\) becomes unbounded in the negative direction (has an infinite limit) as \(x\) approaches \(c\), written \begin{equation*} \lim\limits_{x\rightarrow c}f(x)=-\infty \end{equation*}

if, for every negative number \(N\), a positive number \(\delta \) exists so that \[ \hbox{whenever } 0\lt\vert x-c\vert \lt \delta \hbox{ then } f(x) \lt N. \]

spanDEFINITIONspan Infinite Limit at Infinity

Let \(f\) be a function defined on an open interval \((b,\,\infty )\). Then \(f\) has an infinite limit at infinity \begin{equation*} \lim_{x\rightarrow \infty }f(x)=\infty \end{equation*}

if for any positive number \(M\), there is a corresponding positive number \(N\) so that whenever \(x \gt N\), then \(f(x) \gt M.\)