(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 42372, 1422]*) (*NotebookOutlinePosition[ 44502, 1504]*) (* CellTagsIndexPosition[ 44174, 1487]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Take It to the Limit", "Title"], Cell[TextData[StyleBox["Chapter 1, Sections 1, 2, & 4", FontFamily->"Arial", FontSize->16, FontWeight->"Bold"]], "Text"], Cell["\<\ In executing this module, you can expect to hear some beeps informing you of \ an error in the evaluation. This will happen whenever you try to divide by 0, \ as we do in certain evaluations.\ \>", "Text"], Cell[CellGroupData[{ Cell["Introduction", "Section"], Cell["\<\ OBJECTIVE: To interpret the limit concept graphically and numerically.\ \>", "Text"], Cell["\<\ In this module, you will explore some limits by graphing the functions \ involved and by creating tables of values for the functions. You will \ explore, in some detail, functions that have indeterminate forms at the limit \ point, and you will come to appreciate more fully the importance of formal \ proofs in mathematics. \ \>", "Text"], Cell[CellGroupData[{ Cell["Technology Guidelines", "Subsection", CellDingbat->"\[LightBulb]"], Cell[TextData[{ StyleBox["NOTE: If you have just finished a module, restart ", CellFrame->True, Background->None], StyleBox["Mathematica", CellFrame->True, FontSlant->"Italic", Background->None], StyleBox[ " before executing a new module.\nTO OPEN CELLS, put your cursor on the \ right cell bracket and double click.", CellFrame->True, Background->None], "\nTO STOP AN EXECUTION\n\tSelect the ", StyleBox["Kernel", FontSlant->"Italic"], " pull-down menu and click on ", StyleBox["Abort Evaluation.\n", FontSlant->"Italic"], "ORDER OF EXECUTION\n\tExecute cells in the order given. Do not skip any \ Input cells within a given notebook.\nSAVING NOTEBOOKS\n\tYou can save \ anytime to any directory you choose, and it is wise to save often.\n\t\ However, before you do your final save, delete all your output by selecting \ the \n\t ", StyleBox["Delete All Output", FontSlant->"Italic"], " selection under the ", StyleBox["Kernel", FontSlant->"Italic"], " pull-down menu.\nEXPERIENCING MAJOR PROBLEMS\n\tSave if appropriate, then \ shut down ", StyleBox["Mathematica", FontSlant->"Italic"], " and start it up again." }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Part I: What is ", Cell[BoxData[ \(TraditionalForm\`0\/0\)]], " ? " }], "Section"], Cell[TextData[StyleBox["Section 1.1, Exercise 54.", FontWeight->"Bold"]], "Text"], Cell[TextData[{ "When we try to evaluate a function like ", Cell[BoxData[ \(TraditionalForm\`\(x\ \((1 - cos\ x)\)\)\/\(x - sin\ x\)\)]], " at ", Cell[BoxData[ \(TraditionalForm\`\(x = 0, \)\)]], " we get ", Cell[BoxData[ \(TraditionalForm\`0\/0\)]], ", but what does this mean? The answer to the question \"What is ", Cell[BoxData[ \(TraditionalForm\`0\/0\)]], "?\" depends on whether the numerator or the denominator of the rational \ function we are trying to evaluate approaches 0 faster.\nThis function is not \ defined at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], " because its denominator is 0 there; however, the numerator is also 0 at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ". To better understand the situation, let's investigate what happens to \ the values of ", Cell[BoxData[ \(TraditionalForm\`\(x\ \((1 - cos\ x)\)\)\/\(x - sin\ x\)\)]], " when ", StyleBox["x", FontSlant->"Italic"], " is very near 0. We begin by graphing the function." }], "Text"], Cell[BoxData[{ \(Off[General::spell]\ \), "\n", \(Off[General::spell1]\ \), "\n", \(Clear[x, f]\), "\n", \(f[x_] := x \((1 - Cos[x])\)/\((x - Sin[x])\)\), "\n", \(\(Plot[f[x], {x, \(-20\), 20}, AxesLabel -> {x, y}, AxesOrigin -> {0, 0}];\)\)}], "Input"], Cell[TextData[{ "The graph seems to indicate that this function is approaching 3 as ", StyleBox["x", FontSlant->"Italic"], " approaches 0. To see that this is the case, let's compute some values of \ the function for ", StyleBox["x", FontSlant->"Italic"], " near 0. The ", StyleBox["Table[ ]", FontWeight->"Bold"], " command provides a way to do this, and we use it to create a list of \ entries, {", StyleBox["x", FontSlant->"Italic"], ", f[", StyleBox["x", FontSlant->"Italic"], "], f[x]", Cell[BoxData[ \(TraditionalForm\`\(-3\)\)]], "}, for values of ", StyleBox["x", FontSlant->"Italic"], " ranging from ", Cell[BoxData[ \(TraditionalForm\`\(-1\)\)]], " to 1 in increments of 0.1. Don't be surprised by the error messages that \ are produced when you execute the next command; they are to remind you that ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is not defined at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], "." }], "Text"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h1", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb1"], Cell[BoxData[ \(values = Table[{x, f[x], f[x] - 3}, {x, \(-1. \), 1. , .1}]; TableForm[values, TableHeadings -> {None, {"\", "\", "\"}}]\)], \ "Input"], Cell[TextData[{ "How close to 0 must I hold ", StyleBox["x", FontSlant->"Italic"], " in order to guarantee that ", Cell[BoxData[ \(TraditionalForm\`f \((x)\)\)]], " stays within one ", Cell[BoxData[ \(TraditionalForm\`1\/100\)]], Cell[BoxData[ FormBox[ StyleBox["th", FontSlant->"Italic"], TraditionalForm]]], " of 3? You will explore that in the You Try It: Part I section. " }], "Text"], Cell[CellGroupData[{ Cell["Continuity Issues: Section 1.4", "Subsubsection"], Cell[TextData[{ "You might notice from the graph of ", StyleBox["f(x)", FontSlant->"Italic"], " that it appears to be continuous. For a function to be continuous, the \ limit as ", StyleBox["x", FontSlant->"Italic"], " approaches 0 must exist and this condition seems to be met. However, \ continuity requires that the function be defined when ", StyleBox["x", FontSlant->"Italic"], " is 0, and here we have a problem. If we evaluate ", Cell[BoxData[ \(TraditionalForm\`f(0)\)]], ", the result is an indeterminate form (", Cell[BoxData[ \(TraditionalForm\`\(0\ \)\/0\)]], " is such a form)." }], "Text"], Cell[BoxData[ \(f[0]\)], "Input"], Cell[TextData[{ "Is ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " continuous at ", StyleBox["x", FontSlant->"Italic"], " = 0? Why? \n\nThe graph of ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is deceiving. Mathematically, there is a hole in the graph at the point \ ", Cell[BoxData[ \(TraditionalForm\`\((0, 3)\)\)]], " even though we can't see it on the ", StyleBox["Mathematica", FontSlant->"Italic"], " graph of ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], ". As a result, we might be falsely led to believe that ", StyleBox["f(x)", FontSlant->"Italic"], " is continuous at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ". \n\nNow consider the function, ", Cell[BoxData[ FormBox[ RowBox[{\(g(x)\), "=", RowBox[{"{", RowBox[{GridBox[{ {\(\(x\ \((1 - cos\ x)\)\)\/\(x - sin\ x\)\)}, {"3"} }], GridBox[{ {\(if\ x \[NotEqual] 0\)}, {\(if\ x = 0\)} }]}]}]}], TraditionalForm]]], " .\n\nWhat is ", Cell[BoxData[ \(TraditionalForm\`g(0)\)]], "? What is ", Cell[BoxData[ \(TraditionalForm\`\(\(\(lim\)\(\ \)\)\+\(x \[Rule] 0\)\) \(g(x)\)\)]], "? Is this function continuous at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], "?\n\nThe discontinuity in ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], " is said to be \"removable\" since we can remove it by forming a new \ continuous function, ", Cell[BoxData[ \(TraditionalForm\`g(x)\)]], ", that is identical to ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " when ", Cell[BoxData[ \(TraditionalForm\`x \[NotEqual] 0\)]], " and is equal to 3 when ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ". " }], "Text"], Cell["\<\ For further exploration of continuous and discontinuous functions, refer to \ \"Continuous and Discontinuous Curves,\" a JAVA applet included in this \ supplement.\ \>", "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["You Try It: Part I", "Section"], Cell[CellGroupData[{ Cell["Take The Epsilon Challenge", "Subsection"], Cell[TextData[{ "We have a challenge for you. Here it is.\n\nDetermine how close ", StyleBox["x ", FontSlant->"Italic"], "must be to 0 so that the values of ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " will be within a specified distance from 3. More specifically, we \ challenge you with a small positive number, let's call it \[CurlyEpsilon], \ and ask you to try to determine a value for \[Delta] so that when ", Cell[BoxData[ \(TraditionalForm\`\(-\[Delta]\) < x < \[Delta]\)]], " with ", Cell[BoxData[ \(TraditionalForm\`x \[NotEqual] 0\)]], ", the values of ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " will satisfy the inequality,", Cell[BoxData[ \(TraditionalForm \`\[VerticalSeparator] f(x) - 3 \[VerticalSeparator] \( < \[CurlyEpsilon]\)\)]], ", that is, the distance between the value of ", Cell[BoxData[ \(TraditionalForm\`\(f(x)\ \)\)]], "and 3 will be less than \[CurlyEpsilon]. For example, if ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon] = 0.1\)]], ", the table of values in the cell above shows that ", Cell[BoxData[ \(TraditionalForm\`\(-0.1\) < f(x) - 3 < 0.1\)]], " is not satisfied when ", Cell[BoxData[ \(TraditionalForm\`\[Delta] = 1\)]], ". You need to find a smaller \[Delta] that will keep ", StyleBox["x", FontSlant->"Italic"], " closer to zero so that you can beat our \[CurlyEpsilon] challenge.\n\nTo \ help you with this, we include a command in the next cell that generates a \ list of ordered pairs ", Cell[BoxData[ \(TraditionalForm \`\((x, \ \( \[VerticalSeparator] \(f(x) - 3 \[VerticalSeparator] \)\)) \)\)]], " so that you can look at the magnitude of the differences between the \ values of ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " and 3 to see if the \[Delta] you picked meets our \[CurlyEpsilon] \ challenge. In addition, we added to each ordered pair the result of a test to \ see if ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is within a distance \[CurlyEpsilon] of ", Cell[BoxData[ \(TraditionalForm\`3\)]], ". We start with ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon] = 0.01\)]], ", and a value for \[Delta] that is too big. You should try different \ values for \[Delta], re-evaluating the commands in the cell each time, until \ you find a \[Delta] that works. After you find a \[Delta] that works for ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon] = 0.01\)]], " try to find \[Delta]-values that work for ", Cell[BoxData[ \(TraditionalForm\`\[CurlyEpsilon] = 0.0001, \ 0.0000001, \ 10\^\(-10\), \ \( . \ . \ . \)\)]], " ." }], "Text"], Cell[BoxData[{ \(f[x_] := x \((1 - Cos[x])\)/\((x - Sin[x])\)\), "\n", \(\(\[Delta] = 0.0025;\)\), "\n", \(\(\[CurlyEpsilon] = 0.01;\)\), "\n", \(\(L = 3;\)\), "\n", \(\(valuestest = Table[{x, Abs[f[x] - L], Abs[f[x] - L] < \[CurlyEpsilon]}, {x, \(-\[Delta]\), \[Delta], \ \[Delta]/10}];\)\), "\n", \(TableForm[valuestest, TableHeadings \[Rule] {None, {"\", "\<|f(x)-L|\>", "\<|f(x)-L|<\ \[CurlyEpsilon]\>"}}]\n\)}], "Input"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h2", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb2"], Cell[TextData[{ "The formal definition of the limit says that if you are able meet our \ \[CurlyEpsilon] challenge, no matter how small we make our \[CurlyEpsilon], \ and if you are able to prove that you can do it, then you can say \ unequivocally that ", Cell[BoxData[ \(TraditionalForm\`\(\(lim\ \)\+\(x \[Rule] 0\)\) \(f(x)\) = 3\)]], ". We aren't going to ask you to prove this limit, but your instructor may \ ask you to prove some simpler ones. One thing you can do, however, is use ", StyleBox["Mathematica", FontSlant->"Italic"], "'s ", StyleBox["Limit[ ]", FontWeight->"Bold"], " command to see if it verifies our conjecture that ", Cell[BoxData[ \(TraditionalForm\`\(\(lim\ \)\+\(x \[Rule] 0\)\) \(f(x)\) = 3\)]], "." }], "Text"], Cell[BoxData[ \(Limit[f[x], x \[Rule] 0]\)], "Input"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h3", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb3"] }, Closed]], Cell[CellGroupData[{ Cell["What's Up? (and Down?)", "Subsection"], Cell[TextData[{ "We can get more insight into what is happening to the function ", Cell[BoxData[ \(TraditionalForm\`\(x\ \((1 - cos\ x)\)\)\/\(x - sin\ x\)\)]], " as ", StyleBox["x", FontSlant->"Italic"], " gets close to 0 by looking at the numerator and denominator of the \ function. In the following cell, we plot the numerator and denominator as \ separate functions of ", StyleBox["x", FontSlant->"Italic"], ". " }], "Text"], Cell[TextData[{ "We set the viewing window with the ", StyleBox["PlotRange", FontWeight->"Bold"], " in the option for the ", StyleBox["Plot[ ]", FontWeight->"Bold"], " command so that the relation between the values of the numerator and \ denominator can be easily seen. You can zoom in closer by letting xb and yb \ (terms in red) get smaller and smaller. Your bounds for ", StyleBox["y", FontWeight->"Bold"], " will have to be much, much smaller than your bounds for ", StyleBox["x", FontWeight->"Bold"], "; otherwise you will not be able to see your graphs." }], "Text"], Cell[BoxData[ RowBox[{\(num\ = \ x \((1 - Cos[x])\)\), ";", "\n", \(den\ = \((x - Sin[x])\)\), ";", "\n", RowBox[{"xb", "=", StyleBox[".1", FontColor->RGBColor[1, 0, 0]]}], StyleBox[";", FontColor->RGBColor[1, 0, 0]], "\n", RowBox[{"yb", "=", StyleBox[ SuperscriptBox[ StyleBox["10", FontColor->GrayLevel[0]], RowBox[{"-", StyleBox["4", FontColor->RGBColor[1, 0, 0]]}]], FontColor->RGBColor[1, 0, 0]]}], StyleBox[";", FontColor->RGBColor[1, 0, 0]], "\n", \(Plot[{num, den}, {x, \(-xb\), xb}, PlotRange \[Rule] {{\(-xb\), xb}, {\(-yb\), yb}}, PlotStyle -> {RGBColor[0, 1, 0], RGBColor[0, 0, 1]}]\), ";"}]], "Input"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h4", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb4"], Cell[TextData[{ "Which curve do you think is the graph of the numerator function, and which \ is the denominator? Why? See if you can differentiate between the two curves \ as ", StyleBox["x", FontSlant->"Italic"], " gets closer and closer to 0. " }], "Text"], Cell[TextData[{ "Use the graphs above to find an approximate numeric relation between the \ value of the numerator and the value of the denominator for ", StyleBox["x", FontSlant->"Italic"], " values that are close to 0, Based upon the relationship you find, explain \ why the value of the limit is not surprising." }], "Text"], Cell[TextData[{ "Try using the ", StyleBox["Table[ ] ", FontWeight->"Bold"], "and ", StyleBox["TableForm[ ] ", FontWeight->"Bold"], "commands to generate a table with values of ", StyleBox["x", FontWeight->"Bold"], ", ", StyleBox["num", FontWeight->"Bold"], ", ", StyleBox["den", FontWeight->"Bold"], ", and ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["num", FontWeight->"Bold"], StyleBox["den", FontWeight->"Bold"]], TraditionalForm]]], " in each row, for values of ", StyleBox["x", FontSlant->"Italic"], " close to 0. Comment on what is happening to the values of ", StyleBox["num", FontWeight->"Bold"], ", ", StyleBox["den", FontWeight->"Bold"], ", and ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["num", FontWeight->"Bold"], StyleBox["den", FontWeight->"Bold"]], TraditionalForm]]], " as ", StyleBox["x", FontSlant->"Italic"], " gets closer to 0." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Try another ", Cell[BoxData[ \(TraditionalForm\`0\/0\)]], " expression" }], "Subsection"], Cell[BoxData[ FormBox[ RowBox[{\(Find\ the\ limit\ as\ x\), " ", "\[Rule]", " ", RowBox[{"1", " ", "of", " ", "the", " ", "function", " ", FormBox[\(\(Log[x]\ \)\/\(1 - 2\^\(x - 1\)\)\), "TraditionalForm"]}]}], TextForm]], "Text"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h5", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb5"], Cell[TextData[{ "Start by trying to evaluate the function at ", StyleBox["x", FontSlant->"Italic"], " = 1." }], "Text"], Cell[BoxData[{ \(f[x_] := Log[x]/\((1 - 2\^\(x - 1\))\)\), "\n", \(f[1]\)}], "Input"], Cell["\<\ I guess we had better graph the function to see what is happening.\ \>", "Text"], Cell[BoxData[ \(\(Plot[f[x], {x, .5, 1.5}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input"], Cell[TextData[{ "How can you determine the precise limit? Try looking at some values of the \ function for ", StyleBox["x", FontSlant->"Italic"], " near 1." }], "Text"], Cell[BoxData[ \(Table[{x, f[x]}, {x, .9, 1.1, .01}] // TableForm\)], "Input"], Cell["\<\ To investigate this limit, try some explorations like those in Part I.\ \>", "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Part II: What is ", Cell[BoxData[ \(TraditionalForm\`0\^0\)]], " ?" }], "Section"], Cell[TextData[{ "How does a function such as ", Cell[BoxData[ \(TraditionalForm\`\(( | x | )\)\^x\)]], " behave as ", StyleBox["x ", FontSlant->"Italic"], "gets closer and closer to 0? The answer is not obvious, since numbers to \ the 0 power go to 1, but 0 to ant power is 0. We will start by trying out \ the", StyleBox[" Limit", FontWeight->"Bold"], " command." }], "Text"], Cell[BoxData[{ \(Clear[x, g]\), "\n", \(g[x_] := Abs[x]\^x\), "\n", \(Limit[g[x], x -> 0]\)}], "Input"], Cell[TextData[{ "Is this correct? We might look at the graph of the function. Note what \ happens when ", StyleBox["x", FontSlant->"Italic"], " is 0 on the graph." }], "Text"], Cell[BoxData[ \(\(Plot[g[x], {x, \(-2\), 2}, AxesLabel -> {x, g[x]}];\)\)], "Input"], Cell[CellGroupData[{ Cell["Continuity Issues: Section 1.4", "Subsubsection"], Cell[TextData[{ "If we ask for ", Cell[BoxData[ \(TraditionalForm\`g(0)\)]], ", the result indicates that it is an indeterminate form (", Cell[BoxData[ \(TraditionalForm\`0\^0\)]], " is such a form)." }], "Text"], Cell[BoxData[ \(g[0]\)], "Input"], Cell[TextData[{ "The graph of ", Cell[BoxData[ \(TraditionalForm\`\(g(x)\ \)\)]], "appears to be continuous at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ". Is it? Why?" }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Part III: One-Sided Limits", "Section"], Cell[TextData[StyleBox["Section1.2, Exercise #32", FontWeight->"Bold"]], "Text"], Cell[TextData[{ " Consider the function ", Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ SqrtBox[ RowBox[{"2", StyleBox["x", FontSlant->"Plain"]}]], RowBox[{ StyleBox["(", FontSlant->"Italic"], RowBox[{ StyleBox["x", FontSlant->"Italic"], "-", "1"}], ")"}]}], RowBox[{"|", " ", RowBox[{ StyleBox["x", FontSlant->"Plain"], "-", "1"}], "|"}]], TraditionalForm]]], ". This function appears to have problems at ", StyleBox["x", FontSlant->"Italic"], " = 1. Let's begin by looking at the graph." }], "Text"], Cell[BoxData[{ \(h[x_] := \(\@\(2 x\)\) \((x - 1)\)/Abs[x - 1]\), "\n", \(\(Plot[h[x], {x, 0, 3}, AxesLabel -> {x, y}];\)\t\)}], "Input"], Cell[TextData[{ "The vertical line at ", StyleBox["x", FontSlant->"Italic"], " = 1 really shouldn't be there. The lower piece is not really connected to \ the upper piece. ", StyleBox["Mathematica", FontSlant->"Italic"], " graphs a function by plotting a series of points on the function and then \ connects them with straight lines, sometimes when they really shouldn't be \ connected.\n\n", "What is happening at ", StyleBox["x", FontSlant->"Italic"], " = 1? Is ", Cell[BoxData[ \(TraditionalForm\`h(x)\)]], " defined at ", StyleBox["x", FontSlant->"Italic"], " = 1?" }], "Text"], Cell[BoxData[ \(h[1]\)], "Input"], Cell[TextData[{ "Let's try the ", StyleBox["Limit[ ] ", FontWeight->"Bold"], "command", ". Because of the jump at ", Cell[BoxData[ \(TraditionalForm\`x = 1\)]], ", we will use one-sided limits. ", StyleBox["Direction\[Rule]1", FontWeight->"Bold"], " indicates that we are approaching the value specified from values that \ are smaller, whereas ", StyleBox["Direction\[Rule] ", FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(-1\), FontWeight->"Bold"], TraditionalForm]]], " indicates that we are approaching the value specified from values that \ are larger. " }], "Text"], Cell[BoxData[ ButtonBox[ ButtonBox[ RowBox[{ StyleBox["\[MathematicaIcon]", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontSize->14, FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]]}], ButtonStyle->"Paste"], ButtonData:>"h6", ButtonStyle->"Hyperlink"]], "Input", Evaluatable->False, CellTags->"hb6"], Cell[BoxData[{ \(Limit[\(\@\(2 x\)\) \((x - 1)\)/\(-\((x - 1)\)\), x -> 1, Direction -> 1, Analytic -> True]\), "\n", \(Limit[\(\@\(2 x\)\) \((x - 1)\)/\((x - 1)\), x -> 1, Direction -> \(-1\), Analytic -> True]\)}], "Input"], Cell[TextData[{ "This result verifies what is shown in the graph. That is, ", Cell[BoxData[ \(TraditionalForm \`\(\(lim\ \)\+\(x \[Rule] \(1\^-\)\)\) \(h(x)\) = \(-\@2\)\)]], " and ", Cell[BoxData[ \(TraditionalForm \`\(\(lim\ \)\+\(x \[Rule] \(1\^+\)\)\) \(h(x)\) = \@2\)]], ". 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Maybe, if we zoom in close enough to ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ", we will see a jump. How can we tell for sure whether there is or is not \ a jump at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], "?" }], "Text"], Cell[BoxData[ \(Limit[x\ Log[x], x -> 1, Direction -> \(-1\), Analytic -> True]\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Part IV: What a Difference a Power Makes! ", "Section"], Cell[TextData[{ "In your coursework, you have undoubtedly studied the limit of the ", Cell[BoxData[ \(TraditionalForm\`sinx\/x\)]], " as ", StyleBox["x", FontSlant->"Italic"], " \[Rule] 0 and have found that its value is 1. What happens to the limit \ if we change the power of ", StyleBox["x", FontSlant->"Italic"], " in the denominator to values that are near but not equal to 1? We begin \ by graphing three of these functions and see what the graphs suggest. The \ function ", Cell[BoxData[ \(TraditionalForm\`sinx\/x\)]], " will be graphed in ", StyleBox["red", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], ", ", Cell[BoxData[ \(TraditionalForm\`\(sin\ x\)\/x\^0.9\)]], " in ", StyleBox["green", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], StyleBox[",", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], "and ", Cell[BoxData[ \(TraditionalForm\`\(sin\ x\)\/x\^1.1\)]], " in ", StyleBox["blue", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". 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Without being more careful, we might conclude from \ the graphs above that in all three cases the ratio of ", StyleBox["num", FontWeight->"Bold"], " to ", StyleBox["den", FontWeight->"Bold"], " is a constant and that each limit should be approaching some finite \ number. To see what is really happening, you need to look at several graphs \ of the numerator and denominator functions, each one zoomed by a factor of ", Cell[BoxData[ \(TraditionalForm\`1\/100\)]], ". Go back to the preceding cell and re-execute the commands with the \ exponent of ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["b", FontWeight->"Bold", FontSlant->"Plain"], StyleBox["=", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(-7\), FontWeight->"Plain", FontSlant->"Plain"]}], StyleBox[",", FontWeight->"Plain", FontSlant->"Plain"], StyleBox[\(-9\), FontWeight->"Plain", FontSlant->"Plain"], StyleBox[",", FontWeight->"Plain", FontSlant->"Plain"], StyleBox[\(-11\), FontWeight->"Plain", FontSlant->"Plain"], StyleBox[",", FontWeight->"Plain", FontSlant->"Plain"], StyleBox[" ", FontWeight->"Plain", FontSlant->"Plain"], StyleBox[\( . \ . \ . \), FontWeight->"Plain", FontSlant->"Plain"]}], TraditionalForm]]], " (term in red in the input cell). What does this tell you about the \ ratio of ", StyleBox["num", FontWeight->"Bold"], " to ", StyleBox["den", FontWeight->"Bold"], " in each of the three limits as ", StyleBox["x", FontSlant->"Italic"], " approaches 0?" }], "Text"], Cell[TextData[{ "Wait a minute! How do we know for sure that the pattern illustrated in the \ graphs above will continue if we zoom in even closer. Maybe things will \ \"settle down\" and the ratios will, in fact, approach constant values. What \ about the function ", Cell[BoxData[ \(TraditionalForm\`\(x\ \((1 - cos\ x)\)\)\/\(x - sin\ x\)\)]], " from Part I? Maybe, if we zoom in close enough to 0, the graphs of ", StyleBox["num", FontWeight->"Bold"], " and ", StyleBox["den ", FontWeight->"Bold"], "will start to look like the corresponding graphs for one of the three \ functions considered in this Part. How can we be sure of the situation? The \ answer is in the formal definition of the limit and proof. As mathematicians, \ we are not fully satisfied until we can prove our conjectures about the \ behavior of the functions as ", StyleBox["x", FontSlant->"Italic"], " approaches 0. Then, there is no doubt." }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["\[MathematicaIcon]", FontWeight->"Bold", FontColor->RGBColor[0.792981, 0.777356, 0.144533], FontVariations->{"CompatibilityType"->0}], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["About", FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.500008]], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0.500008, 0, 0.500008]] }], "Section", CellDingbat->None], Cell[TextData[{ "In ", StyleBox["Mathematica", FontSlant->"Italic"], ", the ", StyleBox["Table[expression, iterator] ", FontWeight->"Bold"], "command provides a convenient way to generate a list. 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