2.6E: Exercises for Section 2.5
- Page ID
- 25931
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In exercises 1 - 4, write the appropriate \( ε − δ\) definition for each of the given statements.
1) \(\displaystyle \lim_{x →a}f(x)=N\)
2) \(\displaystyle \lim_{t →b}g(t)=M\)
- Answer
- For every \( ε >0\), there exists a \( δ >0\), so that if \(0 <|t −b| < δ\), then \(|g(t) −M| < ε\)
3) \(\displaystyle \lim_{x →c}h(x)=L\)
4) \(\displaystyle \lim_{x →a} φ(x)=A\)
- Answer
- For every \( ε >0\), there exists a \( δ >0\), so that if \(0 <|x −a| < δ\), then \(| φ(x) −A| < ε\)
The following graph of the function \(f\) satisfies \(\displaystyle \lim_{x →2}f(x)=2\). In the following exercises, determine a value of \( δ >0\) that satisfies each statement.
5) If \(0 <|x −2| < δ\), then \(|f(x) −2| <1\).
6) If \(0 <|x −2| < δ\), then \(|f(x) −2| <0.5\).
- Answer
- \( δ ≤0.25\)
The following graph of the function \(f\) satisfies \(\displaystyle \lim_{x →3}f(x)= −1\). In the following exercises, determine a value of \( δ >0\) that satisfies each statement.
7) If \(0 <|x −3| < δ\), then \(|f(x)+1| <1\).
8) If \(0 <|x −3| < δ\), then \(|f(x)+1| <2\).
- Answer
- \( δ ≤2\)
The following graph of the function \(f\) satisfies \(\displaystyle \lim_{x →3}f(x)=2\). In the following exercises, for each value of \( ε\), find a value of \( δ >0\) such that the precise definition of limit holds true.
9) \( ε=1.5\)
10) \( ε=3\)
- Answer
- \( δ ≤1\)
[T] In exercises 11 - 12, use a graphing calculator to find a number \( δ\) such that the statements hold true.
11) \(\left|\sin(2x) −\frac{1}{2}\right| <0.1\), whenever \(\left|x −\frac{ π}{12}\right| < δ\)
12) \(\left|\sqrt{x −4} −2\right| <0.1\), whenever \(|x −8| < δ\)
- Answer
- \( δ <0.3900\)
In exercises 13 - 17, use the precise definition of limit to prove the given limits.
13) \(\displaystyle \lim_{x →2}\,(5x+8)=18\)
14) \(\displaystyle \lim_{x →3}\frac{x^2 −9}{x −3}=6\)
- Answer
- Let \( δ= ε\). If \(0 <|x −3| < ε\), then \(\left|\dfrac{x^2 −9}{x −3} - 6\right| = \left|\dfrac{(x+3)(x −3)}{x −3} - 6\right| = |x+3 −6|=|x −3| < ε\).
15) \(\displaystyle \lim_{x →2}\frac{2x^2 −3x −2}{x −2}=5\)
16) \(\displaystyle \lim_{x →0}x^4=0\)
- Answer
- Let \( δ=\sqrt[4]{ ε}\). If \(0 <|x| <\sqrt[4]{ ε}\), then \(\left|x^4-0\right|=x^4 < ε\).
17) \(\displaystyle \lim_{x →2}\,(x^2+2x)=8\)
In exercises 18 - 20, use the precise definition of limit to prove the given one-sided limits.
18) \(\displaystyle \lim_{x →5^ −}\sqrt{5 −x}=0\)
- Answer
- Let \( δ= ε^2\). If \(- ε^2 < x - 5 < 0,\) we can multiply through by \(-1\) to get \(0 <5-x < ε^2.\)
Then \(\left|\sqrt{5 −x} - 0\right|=\sqrt{5 −x} < \sqrt{ ε^2} = ε\).
19) \(\displaystyle \lim_{x →0^+}f(x)= −2\), where \(f(x)=\begin{cases}8x −3, & \text{if }x <0\\4x −2, & \text{if }x ≥0\end{cases}\).
20) \(\displaystyle \lim_{x →1^ −}f(x)=3\), where \(f(x)=\begin{cases}5x −2, & \text{if }x <1\\7x −1, & \text{if }x ≥1\end{cases}\).
- Answer
- Let \( δ= ε/5\). If \( − ε/5 < x - 1 <0,\) we can multiply through by \(-1\) to get \(0 <1-x < ε/5.\)
Then \(|f(x) −3|=|5x-2-3| = |5x −5| = 5(1-x),\) since \(x <1\) here.
And \(5(1-x) < 5( ε/5) = ε\).
In exercises 21 - 23, use the precise definition of limit to prove the given infinite limits.
21) \(\displaystyle \lim_{x →0}\frac{1}{x^2}= ∞\)
22) \(\displaystyle \lim_{x → −1}\frac{3}{(x+1)^2}= ∞\)
- Answer
- Let \( δ=\sqrt{\frac{3}{N}}\). If \(0 <|x+1| <\sqrt{\frac{3}{N}}\), then \(f(x)=\frac{3}{(x+1)^2} >N\).
23) \(\displaystyle \lim_{x →2} −\frac{1}{(x −2)^2}= − ∞\)
24) An engineer is using a machine to cut a flat square of Aerogel of area \(144 \,\text{cm}^2\). If there is a maximum error tolerance in the area of \(8 \,\text{cm}^2\), how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to \( δ\), \( ε\), \(a\), and \(L\)?
- Answer
- \(0.033 \text{ cm}, \, ε=8,\, δ=0.33,\,a=12,\,L=144\)
25) Use the precise definition of limit to prove that the following limit does not exist: \(\displaystyle \lim_{x →1}\frac{|x −1|}{x −1}.\)
26) Using precise definitions of limits, prove that \(\displaystyle \lim_{x →0}f(x)\) does not exist, given that \(f(x)\) is the ceiling function. (Hint: Try any \( δ <1\).)
- Answer
- Answers may very.
27) Using precise definitions of limits, prove that \(\displaystyle \lim_{x →0}f(x)\) does not exist: \(f(x)=\begin{cases}1, & \text{if }x\text{ is rational}\\0, & \text{if }x\text{ is irrational}\end{cases}\). (Hint: Think about how you can always choose a rational number \(0 <d\), >
28) Using precise definitions of limits, determine \(\displaystyle \lim_{x →0}f(x)\) for \(f(x)=\begin{cases}x, & \text{if }x\text{ is rational}\\0, & \text{if }x\text{ is irrational}\end{cases}\). (Hint: Break into two cases, \(x\) rational and \(x\) irrational.)
- Answer
- \(0\)
29) Using the function from the previous exercise, use the precise definition of limits to show that \(\displaystyle \lim_{x →a}f(x)\) does not exist for \(a ≠0\)
For exercises 30 - 32, suppose that \(\displaystyle \lim_{x →a}f(x)=L\) and \(\displaystyle \lim_{x →a}g(x)=M\) both exist. Use the precise definition of limits to prove the following limit laws:
30) \(\displaystyle \lim_{x →a}(f(x) −g(x))=L −M\)
- Answer
- \(f(x) −g(x)=f(x)+( −1)g(x)\)
31) \(\displaystyle \lim_{x →a}[cf(x)]=cL\) for any real constant \(c\) (Hint: Consider two cases: \(c=0\) and \(c ≠0\).)
32) \(\displaystyle \lim_{x →a}[f(x)g(x)]=LM\). (Hint: \(|f(x)g(x) −LM|= |f(x)g(x) −f(x)M +f(x)M −LM| ≤|f(x)||g(x) −M| +|M||f(x) −L|.)\)
- Answer
- Answers may vary.