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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT 268 17 "Lab # 1:   Limits" }}
{PARA 259 "" 0 "" {TEXT -1 75 "In this lab you we will investigate lim
its.  These come in three flavors:  " }{TEXT 269 16 "Infinite Limits,
" }{TEXT -1 88 " when the function goes to positive or negative infini
ty as x goes to a finite number,  " }{TEXT 270 19 "Limits at Infinity,
" }{TEXT -1 135 " this is often referred to as \"end behavior\", and d
escribes the limit of the function as x goes to positive or negative i
nfinity, and  " }{TEXT 271 14 "Finite Limits," }{TEXT -1 72 " when a f
unction goes to a finite number as x goes to a finite number.  " }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 38 "Infinite Limits and Limits at Inf
inity" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Here we  plot the functio
n f(x) =  " }{XPPEDIT 18 0 "3*x/(x-2);" "6#*(\"\"$\"\"\"%\"xGF%,&F&F%
\"\"#!\"\"F)" }{TEXT -1 127 "  which has a discontinuity at x = 2. Spe
cifically, the function is undefined at this value.  We first define t
he function by  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "f := x \+
-> 3*x/(x-2);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG
f*6#%\"xG6\"6$%)operatorG%&arrowGF(,$*(\"\"$\"\"\"9$F/,&F0F/\"\"#!\"\"
F3F/F(F(F(" }}}{EXCHG {PARA 0 "" 1 "" {TEXT -1 136 "Now let's try to p
lot this with the plot command.  Try the following three options seque
ntially to get a better picture of this graph.  " }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 36 "plot(f(x),x=-4..8,labels=[\"x\",\"f\"]);" }
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "This shows some peculiar behavior
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2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 108 "The above sequence of graphs should make it cl
ear that the limit of f(x) as x approaches 2 from the left is " }}
{PARA 0 "" 0 "" {TEXT -1 20 "negative infinity (-" }{XPPEDIT 18 0 "inf
inity;" "6#%)infinityG" }{TEXT -1 82 ")  and the limit of f(x) as x ap
proaches 2 from the right is positive infinity  ( " }{XPPEDIT 18 0 "in
finity;" "6#%)infinityG" }{TEXT -1 121 ").  These are called infinite \+
limits and describe an infinite limit of the function as x approaches \+
a finite value (2).  " }}{PARA 0 "" 0 "" {TEXT -1 6 " Does " }
{XPPEDIT 18 0 "limit(f(x),x = 2);" "6#-%&limitG6$-%\"fG6#%\"xG/F)\"\"#
" }{TEXT -1 48 "  exist?  Try f (1.99999)  and  f (2.00001).    " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Maple has a command for finding th
ese limits. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "limit(f(x),
x=2);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%*undefinedG" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 250 "This \"undefined\" makes sense
 because the limit from the left is different from the limit from the \+
right.  We can determine these limits from the left and right by inclu
ding this in the limit command. The command below finds the limit from
 the left.  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "limit(f(x),
x=2,left);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%)infinityG!\"\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Similarly the limit from the right
 of x = 2 is found by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "l
imit(f(x),x=2,right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)infinityG" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "Therefore we say: \"the limit \+
of f as x approaches 2 from the left is negative infinity and the limi
t of f as x approaches 2 from the right is (positive) infinity. Becaus
e these two are not the same, \"the limit of f as x approaches 2 is un
defined\".  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 222 "It also appears that as x goes to negative and positive infini
ty, the function tends towards a finite value.  I initially guess this
 value to be 3.   I check this by increasing my domain and plotting th
e line y=3 as well. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plo
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Fa`mFa[n" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Cur
ve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 204 "It is now cle
ar that as x goes to positive or negative infinity, the function tends
 to 3.  These are called limits at infinity. You state: \"The limit as
 x approaches positive or negative infinity is 2\".  " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 79 "Problem number 1 asks for these same type
s of limits for a different function. " }}}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT -1 13 "Finite Limits" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "In \+
this section we try to reconcile the notion of the limit of a function
 with the formal definition by examining a simple example. " }}{PARA 
0 "" 0 "" {TEXT -1 31 "Consider the function:  f(x) = " }{XPPEDIT 18 
0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT 
-1 13 "I claim that " }{XPPEDIT 18 0 "limit(f(x),x = 3);" "6#-%&limitG
6$-%\"fG6#%\"xG/F)\"\"$" }{TEXT -1 10 "  =  9.   " }}{PARA 0 "" 0 "" 
{TEXT -1 46 "This should be fairly obvious because we know " }
{XPPEDIT 18 0 "3^2;" "6#*$\"\"$\"\"#" }{TEXT -1 53 " = 9 and therefore
 if x is \"close enough to\" 3, then " }{XPPEDIT 18 0 "x^2" "6#*$%\"xG
\"\"#" }{TEXT -1 238 " should be \"close to\" 9.  Mathematically speak
ing, the phrases \"close enouph to\" and \"close to\" are ambiguous.  \+
Therefore, proving that we can get \"close to\" 9 by choosing x-values
 \"close enough to\" 3 reduces to a problem with 2 numbers: " }
{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }{TEXT -1 58 " (epsilon) defin
es how \"close to\" 9 you want to get,  and " }{XPPEDIT 18 0 "delta" "
6#%&deltaG" }{TEXT -1 128 " (delta) defines what we mean by \"close en
ough\" to 3.   The following game illustrates a typical \"delta - epsi
lon\" limit proof. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 
"" {TEXT -1 0 "" }{TEXT 263 26 "The \"delta - epsilon\" game" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "(1)  I give you
 a small positive number " }{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }
{TEXT -1 11 " (epsilon)." }}{PARA 0 "" 0 "" {TEXT -1 37 "(2) You find \+
a small positive number " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT 
-1 19 " (delta) such that:" }}{PARA 0 "" 0 "" {TEXT -1 28 "           \+
  if x is within " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT -1 13 " \+
of 3  (ie.  " }{XPPEDIT 18 0 "abs(x-3);" "6#-%$absG6#,&%\"xG\"\"\"\"\"
$!\"\"" }{TEXT -1 5 "  <  " }{XPPEDIT 18 0 "delta;" "6#%&deltaG" }
{TEXT -1 53 "),  with the possible exclusion of x = 3, ( ie. 0 <  " }
{XPPEDIT 18 0 "abs(x-3);" "6#-%$absG6#,&%\"xG\"\"\"\"\"$!\"\"" }{TEXT 
-1 5 "  <  " }{XPPEDIT 18 0 "delta;" "6#%&deltaG" }{TEXT -1 2 ")," }}
{PARA 0 "" 0 "" {TEXT -1 33 "             then f(x) is within " }
{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 12 " of 9 (ie.  " }
{XPPEDIT 18 0 "abs(f(x)-9);" "6#-%$absG6#,&-%\"fG6#%\"xG\"\"\"\"\"*!\"
\"" }{TEXT -1 4 "  < " }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT 
-1 2 ") " }}{PARA 0 "" 0 "" {TEXT -1 27 "(3) If you can find such a " 
}{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT -1 15 " for any given " }
{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 65 ", you win the gam
e and the limit of f(x) as x approaches 3 is 9. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "With this accomplished, y
ou can force the function to get as close to 9 as you wish (with " }
{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 42 ") by choosing x's
 close enough to 3 (with " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT 
-1 143 "). This seems like an awful lot of work to show that a functio
n has a limit. However, consider what happens if you can't win the gam
e for some " }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 108 ". \+
This means that no matter how close x is to 3, there is a number betwe
en x and 3 where f(x) is more than  " }{XPPEDIT 18 0 "epsilon" "6#%(ep
silonG" }{TEXT -1 271 " away from 9. Think really really small and its
 clear that if this is the case, f(x) cannot possibly go to 9 as x goe
s to 3.   Therefore, winning the game described above constitutes the \+
definition of a finite limit.  Winning the \"delta - epsilon\" game fo
r an arbitrary  " }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 
135 " can be tricky, especially for nonlinear functions, so we satisfy
 ourselves with one successful round of the game for a given value of \+
" }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 2 ". " }}{PARA 0 "
" 0 "" {TEXT -1 29 "Because the choice of delta (" }{XPPEDIT 18 0 "del
ta" "6#%&deltaG" }{TEXT -1 41 ") will most certainly depend on epsilon
 (" }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 72 "), we will s
eek the largest such delta that wins one round of the game. " }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "(1)   I give yo
u  " }{XPPEDIT 18 0 "epsilon" "6#%(epsilonG" }{TEXT -1 6 " = 0.1" }}
{PARA 0 "" 0 "" {TEXT -1 38 "(2)  You estimate the largest postive " }
{XPPEDIT 18 0 "delta;" "6#%&deltaG" }{TEXT -1 14 " such that if " }
{TEXT 262 1 "x" }{TEXT -1 11 " is within " }{XPPEDIT 18 0 "delta;" "6#
%&deltaG" }{TEXT -1 16 " of 3 (ie. 0 <  " }{XPPEDIT 18 0 "abs(x-3);" "
6#-%$absG6#,&%\"xG\"\"\"\"\"$!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "
delta;" "6#%&deltaG" }{TEXT -1 37 ") then f(x) is within 0.1 of  9 (ie
. " }{XPPEDIT 18 0 "abs(f(x)-9);" "6#-%$absG6#,&-%\"fG6#%\"xG\"\"\"\"
\"*!\"\"" }{TEXT -1 98 " < 0.1).  You can illustrate whether f(x) is w
ithin 0.1 of 9 by plotting the horizontal lines 9 - " }{XPPEDIT 18 0 "
epsilon" "6#%(epsilonG" }{TEXT -1 9 " and 9 + " }{XPPEDIT 18 0 "epsilo
n" "6#%(epsilonG" }{TEXT -1 72 " and seeing if f(x) lies between these
 lines over the interval x =  3 - " }{XPPEDIT 18 0 "delta" "6#%&deltaG
" }{TEXT -1 13 " to x =  3 + " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }
{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "First, type in \+
the function and value of " }{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }
{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f := x -> \+
x^2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "eps := 0.1:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Now we take a first guess at " }
{XPPEDIT 18 0 "delta;" "6#%&deltaG" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "delta := 0.1:  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
34 "and see if f(x) lies between  9 - " }{XPPEDIT 18 0 "epsilon" "6#%(
epsilonG" }{TEXT -1 9 " and 9 + " }{XPPEDIT 18 0 "epsilon" "6#%(epsilo
nG" }{TEXT -1 25 " over the interval ( 3 - " }{XPPEDIT 18 0 "delta" "6
#%&deltaG" }{TEXT -1 7 " , 3 + " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }
{TEXT -1 4 ").  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "plot([9
-eps,f(x),9+eps],x=3-delta .. 3 + delta,color = [blue,red,blue]); " }}
{PARA 13 "" 1 "" {GLPLOT2D 373 373 373 {PLOTDATA 2 "6'-%'CURVESG6$7S7$
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[D:3HF*F+7$$\"3.LLe0$=C\"HF*F+7$$\"3QLL3RBr;HF*F+7$$\"3Um;zjf)4#HF*F+7
$$\"3VL$e4;[\\#HF*F+7$$\"3'***\\i'y]!HHF*F+7$$\"35L$ezs$HLHF*F+7$$\"3#
***\\7iI_PHF*F+7$$\"3^mm;_M(=%HF*F+7$$\"3[LL3y_qXHF*F+7$$\"3#)****\\1!
>+&HF*F+7$$\"3)*****\\Z/NaHF*F+7$$\"3'*****\\$fC&eHF*F+7$$\"39L$ez6:B'
HF*F+7$$\"3qmm;=C#o'HF*F+7$$\"3kmmm#pS1(HF*F+7$$\"37+]i`A3vHF*F+7$$\"3
qmmm(y8!zHF*F+7$$\"3))**\\i.tK$)HF*F+7$$\"33+](3zMu)HF*F+7$$\"3!om;H_?
<*HF*F+7$$\"3om;zihl&*HF*F+7$$\"3]LL3#G,***HF*F+7$$\"3=L$ezw5V+$F*F+7$
$\"3:+]PQ#\\\"3IF*F+7$$\"3>LLe\"*[H7IF*F+7$$\"3')*****pvxl,$F*F+7$$\"3
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3!pm\"H!o-*\\IF*F+7$$\"3/+]7k.6aIF*F+7$$\"3mmm;WTAeIF*F+7$$\"3;+]i!*3`
iIF*F+7$$\"3KLLL*zym1$F*F+7$$\"3eLL3N1#42$F*F+7$$\"3im;HYt7vIF*F+7$$\"
37+++xG**yIF*F+7$$\"3kmmT6KU$3$F*F+7$$\"3[LLLbdQ(3$F*F+7$$\"33+]i`1h\"
4$F*F+7$$\"3E+]P?Wl&4$F*F+7$$\"33+++++++JF*F+-%'COLOURG6&%$RGBG$\"\"!F
\\uF[u$\"*++++\"!\")-F$6$7S7$F($\"39++++++5%)F*7$F.$\"3$Q_qMq.`V)F*7$F
1$\"3!QJNTC^tX)F*7$F4$\"3y#*op'Q!=#[)F*7$F7$\"3cp4'*o3@2&)F*7$F:$\"3wH
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Jv()F*7$Fen$\"3G2j$4EN?!))F*7$Fhn$\"3]$4F\\71Z#))F*7$F[o$\"3]4W$eT96&)
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F`p$\"3,!Rq/>$)e-*F*7$Fcp$\"3I%4O.%='*[!*F*7$Ffp$\"3IX!fQ^?R2*F*7$Fip$
\"3g\\r-k8u*4*F*7$F\\q$\"38]83Hv.D\"*F*7$F_q$\"3;+PK!HY&\\\"*F*7$Fbq$
\"3wYr_+wzw\"*F*7$Feq$\"3T(GP#z'=8?*F*7$Fhq$\"3N!Hh,GOvA*F*7$F[r$\"3y:
S-NbK^#*F*7$F^r$\"3!*fw^<&otF*F*7$Far$\"3>%H]#fj!>I*F*7$Fdr$\"3(*RdD;,
fF$*F*7$Fgr$\"3-TR'f\"\\t_$*F*7$Fjr$\"3GcB.ca4z$*F*7$F]s$\"38M0Ge)=XS*
F*7$F`s$\"34C1)pa`0V*F*7$Fcs$\"3gG3\"f>3kX*F*7$Ffs$\"3#zt;m8(>![*F*7$F
is$\"3\"*oIq+()\\2&*F*7$F\\t$\"3*z<NA!3&>`*F*7$F_t$\"3;+wgLk0e&*F*7$Fb
t$\"3#)ysQ!HwIe*F*7$Fet$\"3@,+++++5'*F*-Fht6&FjtF]uF[uF[u-F$6$7S7$F($
\"3k*************4*F*7$F.F\\_l7$F1F\\_l7$F4F\\_l7$F7F\\_l7$F:F\\_l7$F=
F\\_l7$F@F\\_l7$FCF\\_l7$FFF\\_l7$FIF\\_l7$FLF\\_l7$FOF\\_l7$FRF\\_l7$
FUF\\_l7$FXF\\_l7$FenF\\_l7$FhnF\\_l7$F[oF\\_l7$F^oF\\_l7$FaoF\\_l7$Fd
oF\\_l7$FgoF\\_l7$FjoF\\_l7$F]pF\\_l7$F`pF\\_l7$FcpF\\_l7$FfpF\\_l7$Fi
pF\\_l7$F\\qF\\_l7$F_qF\\_l7$FbqF\\_l7$FeqF\\_l7$FhqF\\_l7$F[rF\\_l7$F
^rF\\_l7$FarF\\_l7$FdrF\\_l7$FgrF\\_l7$FjrF\\_l7$F]sF\\_l7$F`sF\\_l7$F
csF\\_l7$FfsF\\_l7$FisF\\_l7$F\\tF\\_l7$F_tF\\_l7$FbtF\\_l7$FetF\\_lFg
t-%+AXESLABELSG6$Q\"x6\"Q!Fbbl-%%VIEWG6$;$\"#H!\"\"$\"#JFjbl%(DEFAULTG
" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "C
urve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "The above co
mmand plots 3 functions (y = 9-" }{XPPEDIT 18 0 "epsilon;" "6#%(epsilo
nG" }{TEXT -1 18 ", y = f(x), y = 9+" }{XPPEDIT 18 0 "epsilon;" "6#%(e
psilonG" }{TEXT -1 80 ") where the constant functions are in blue and \+
f(x) is in red. If we had chosen " }{XPPEDIT 18 0 "delta;" "6#%&deltaG
" }{TEXT -1 111 " small enough, the graph would have fallen entirely w
ithin the blue lines.  Judging from the above graph I try " }{XPPEDIT 
18 0 "delta;" "6#%&deltaG" }{TEXT -1 124 " = 0.02. You can go back, ch
ange delta and then execute the same plot command to save on typing or
 you can do the following:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
14 "delta := 0.02:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "plot(
[9-eps,f(x),9+eps],x=3-delta .. 3 + delta,color = [blue,red,blue]);" }
}{PARA 13 "" 1 "" {GLPLOT2D 373 373 373 {PLOTDATA 2 "6'-%'CURVESG6$7S7
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$\"3)***\\U7Y]()HF*F+7$$\"3[LLV!pu$))HF*F+7$$\"3)om;c0T\"*)HF*F+7$$\"3
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/TM)*HF*F+7$$\"3CL$eDBJ\"**HF*F+7$$\"3)om;kD!)***HF*F+7$$\"3km;f`@'3+$
F*F+7$$\"3&)**\\nZ)H;+$F*F+7$$\"3kmmJy*eC+$F*F+7$$\"31++S^bJ.IF*F+7$$
\"3*****\\5a`T+$F*F+7$$\"3$)**\\7RV'\\+$F*F+7$$\"3,++:#fke+$F*F+7$$\"3
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$\"36+]sI@K=IF*F+7$$\"3(***\\2%)38>IF*F+7$$\"3-++++++?IF*F+-%'COLOURG6
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$\"3_KkF!ey.4*F*7$Ffs$\"3yoYQZ5/&4*F*7$Fis$\"3;B\"))4B'Q+\"*F*7$F\\t$
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1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "We are not quite there.  Judging f
rom the graph let" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "delta \+
:= 3 - 2.9835;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&deltaG$\"$l\"!\"%
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "plot([9-eps,f(x),9+eps]
,x=3-delta .. 3 + delta,color = [blue,red,blue]);" }}{PARA 13 "" 1 "" 
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7$FhnF\\_l7$F[oF\\_l7$F^oF\\_l7$FaoF\\_l7$FdoF\\_l7$FgoF\\_l7$FjoF\\_l
7$F]pF\\_l7$F`pF\\_l7$FcpF\\_l7$FfpF\\_l7$FipF\\_l7$F\\qF\\_l7$F_qF\\_
l7$FbqF\\_l7$FeqF\\_l7$FhqF\\_l7$F[rF\\_l7$F^rF\\_l7$FarF\\_l7$FdrF\\_
l7$FgrF\\_l7$FjrF\\_l7$F]sF\\_l7$F`sF\\_l7$FcsF\\_l7$FfsF\\_l7$FisF\\_
l7$F\\tF\\_l7$F_tF\\_l7$FbtF\\_l7$FetF\\_lFgt-%+AXESLABELSG6$Q\"x6\"Q!
Fbbl-%%VIEWG6$;$\"&N)H!\"%$\"&l,$Fjbl%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 
2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 208 "Did it !  We can now say that if \+
x is within 0.0165 of 3 then f(x) is within 0.1 of 9.  This is the gen
eral game in proving that a limit exists.  Do you get the game?   Of c
ourse, showing that you can find a " }{XPPEDIT 18 0 "delta" "6#%&delta
G" }{TEXT -1 18 " for an arbitrary " }{XPPEDIT 18 0 "epsilon" "6#%(eps
ilonG" }{TEXT -1 70 " is considerably trickier but the strategy is the
 same. Specifically: " }{XPPEDIT 18 0 "limit(f(x),x = a);" "6#-%&limit
G6$-%\"fG6#%\"xG/F)%\"aG" }{TEXT -1 20 " = L   if , for any " }
{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }{TEXT -1 18 " > 0,  I can fin
d " }{XPPEDIT 18 0 "delta;" "6#%&deltaG" }{TEXT -1 26 " > 0, such that
 if   0 <  " }{XPPEDIT 18 0 "abs(x-a);" "6#-%$absG6#,&%\"xG\"\"\"%\"aG
!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "delta;" "6#%&deltaG" }{TEXT 
-1 8 "  then  " }{XPPEDIT 18 0 "abs(L-f(x));" "6#-%$absG6#,&%\"LG\"\"
\"-%\"fG6#%\"xG!\"\"" }{TEXT -1 3 " < " }{XPPEDIT 18 0 "epsilon;" "6#%
(epsilonG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "# end of section" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Assignm
ent " }}{PARA 0 "" 0 "" {TEXT -1 145 "Put your name, course, and secti
on number in the upper left corner. You will hand in two graphs (one p
age) with a little information about each. " }}{EXCHG {PARA 256 "" 0 "
" {TEXT -1 0 "" }{TEXT 257 11 "Graph # 1: " }}{PARA 258 "" 0 "" {TEXT 
-1 23 "Consider  the function " }{XPPEDIT 18 0 "(1+2*abs(x))/x;" "6#*&
,&\"\"\"F%*&\"\"#F%-%$absG6#%\"xGF%F%F%F+!\"\"" }{TEXT -1 63 "  .  Not
e: the absolute value function is called  by \"abs(x)\". " }}{PARA 0 "
" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 188 "(a) Display the \+
2 limits at infinity, and the 2 infinite limits of the function by plo
tting it over an appropriate interval.  Display only one graph which c
aptures all of these features.  " }{TEXT 265 5 "Hint:" }{TEXT -1 66 " \+
 See the graphing commands for the final graph in the section on " }
{TEXT 266 38 "Infinite Limits and Limits at Infinity" }{TEXT -1 3 ".  \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 124 "(b) \+
Describe each of the limits using mathematical terms.  You can get the
 editor to type limit notation by clicking on the " }{XPPEDIT 18 0 "Si
gma;" "6#%&SigmaG" }{TEXT -1 209 " button.  A small box appears in the
 context bar.  You can then type in the command using Maple notation s
uch as \"limit(f(x),x=0,left)\", \"enter\",  and the mathematical expr
ession described by this command:  \" " }{XPPEDIT 18 0 "limit(f(x),x =
 0,left);" "6#-%&limitG6%-%\"fG6#%\"xG/F)\"\"!%%leftG" }{TEXT -1 300 "
\"   is printed at the cursor position. If you can't get this to work,
 you can always express the limit as a statement like: \"the limit of \+
f(x) as x goes to (negative) infinity is ...\" and \"the limit of f(x)
 as x go to zero from the right (left) is ...\" .  Or you can write th
ese on the page by hand. " }}{PARA 0 "" 0 "" {TEXT -1 2 "  " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 258 0 "" }{TEXT 259 0 "" }
{TEXT 260 0 "" }{TEXT 261 11 "Problem #2:" }}{PARA 0 "" 0 "" {TEXT -1 
19 "(a)  Display that  " }{XPPEDIT 18 0 "limit(x^2,x = 2);" "6#-%&limi
tG6$*$%\"xG\"\"#/F'F(" }{TEXT -1 35 "  =  4   by estimating the larges
t " }{XPPEDIT 18 0 "delta;" "6#%&deltaG" }{TEXT -1 20 " > 0, such that
: if " }{TEXT 256 1 "x" }{TEXT -1 11 " is within " }{XPPEDIT 18 0 "del
ta;" "6#%&deltaG" }{TEXT -1 12 " of 2, then " }{XPPEDIT 18 0 "x^2;" "6
#*$%\"xG\"\"#" }{TEXT -1 65 " is               within 0.1 of 4.  State
 the specific value of  " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT 
-1 129 " that  satisfies this requirement and print only the final gra
ph that displays this (similar to the last graph in the section on " }
{TEXT 267 14 "Finite Limits)" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "(b) Explain why this valu
e of " }{XPPEDIT 18 0 "delta;" "6#%&deltaG" }{TEXT -1 82 " is larger t
han the one discovered in the example for the same function at x = 3. \+
" }{TEXT 264 5 "Hint:" }{TEXT -1 65 " It has something to do with the \+
steepness of the function.      " }}}}}{MARK "3" 0 }{VIEWOPTS 1 1 0 1 
1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }