{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Lab 5 Maple Activities for Sept 12: (Lesson 3 - Limits)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 32 "1. In lecture, we claimed that " }{XPPEDIT 18 0 " Limit(3*x^2-x+4,x = -2);" "6#-%&LimitG6$,(*&\"\"$\"\"\"*$%\"xG\"\"#F)F )F+!\"\"\"\"%F)/F+,$\"\"#F-" }{TEXT -1 50 " = 18. Let's see this usi ng Maple. Recall that " }{XPPEDIT 18 0 "Limit(3*x^2-x+4,x = -2);" "6# -%&LimitG6$,(*&\"\"$\"\"\"*$%\"xG\"\"#F)F)F+!\"\"\"\"%F)/F+,$\"\"#F-" }{TEXT -1 21 " = 18 means that as " }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 17 " approaches - 2, " }{XPPEDIT 18 0 "3*x^2-x+4;" "6#,(*&\" \"$\"\"\"*$%\"xG\"\"#F&F&F(!\"\"\"\"%F&" }{TEXT -1 29 " approaches 18. Let's let f(" }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 4 ") = " } {XPPEDIT 18 0 "3*x^2-x+4;" "6#,(*&\"\"$\"\"\"*$%\"xG\"\"#F&F&F(!\"\"\" \"%F&" }{TEXT -1 13 " and show as " }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 19 " approaches - 2, f(" }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 16 ") approaches 18." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "f := x -> 3*x^2 - x + 4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 137 "Ten significant digits is the defaul t for Maple. Let's change the number of significant digits to twenty \+ by the following Maple command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Digits := 20;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 20 "Let's first look at " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 53 " approaching - 2 from values that are larger t han -2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f(-1.9); f(-1.99 9); f(-1.99999); f(-1.9999999); f(-1.9999999999);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "Now, let's look at " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 54 " approaching - 2 from va lues that are smaller than -2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f(-2.1); f(-2.001); f(-2.00001); f(-2.0000001); f(-2.000000000 1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "It does appear that as " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 23 " is approaching - 2, f(" }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 34 ") is approaching 18. That is, as " }{XPPEDIT 18 0 "x;" "6#%\"x G" }{TEXT -1 21 " is approaching - 2, " }{XPPEDIT 18 0 "3*x^2-x+4;" "6 #,(*&\"\"$\"\"\"*$%\"xG\"\"#F&F&F(!\"\"\"\"%F&" }{TEXT -1 19 " is appr oaching 18." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 67 "Now, let's see the Maple command, that will find this l imit for us:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Limit(f(x), x=-2); value(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Digits := 25:" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "2. In lecture, we claimed that " }{XPPEDIT 18 0 "Limit((x^2+2*x-15)/(2* x^2-5*x-3),x = 3);" "6#-%&LimitG6$*&,(*$%\"xG\"\"#\"\"\"*&\"\"#F+F)F+F +\"#:!\"\"F+,(*&\"\"#F+*$F)\"\"#F+F+*&\"\"&F+F)F+F/\"\"$F/F//F)\"\"$" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "8/7;" "6#*&\"\")\"\"\"\"\"(!\"\"" } {TEXT -1 57 " . Let's see this using Maple. We want to show that as \+ " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 15 " approaches 3, " } {XPPEDIT 18 0 "(x^2+2*x-15)/(2*x^2-5*x-3);" "6#*&,(*$%\"xG\"\"#\"\"\"* &\"\"#F(F&F(F(\"#:!\"\"F(,(*&\"\"#F(*$F&\"\"#F(F(*&\"\"&F(F&F(F,\"\"$F ,F," }{TEXT -1 13 " approaches " }{XPPEDIT 18 0 "8/7;" "6#*&\"\")\"\" \"\"\"(!\"\"" }{TEXT -1 22 " . Let's again let f(" }{XPPEDIT 18 0 "x; " "6#%\"xG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "(x^2+2*x-15)/(2*x^2-5*x -3)" "6#*&,(*$%\"xG\"\"#\"\"\"*&\"\"#F(F&F(F(\"#:!\"\"F(,(*&\"\"#F(*$F &\"\"#F(F(*&\"\"&F(F&F(F,\"\"$F,F," }{TEXT -1 14 " and show as " } {XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 17 " approaches 3, f(" } {XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 13 ") approaches " }{XPPEDIT 18 0 "8/7;" "6#*&\"\")\"\"\"\"\"(!\"\"" }{TEXT -1 49 " . Also, let's \+ find a decimal approximation for " }{XPPEDIT 18 0 "8/7;" "6#*&\"\")\" \"\"\"\"(!\"\"" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "A decimal approximation for " } {XPPEDIT 18 0 "8/7;" "6#*&\"\")\"\"\"\"\"(!\"\"" }{TEXT -1 74 " to fi fteen significant decimal can be found using the following command:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(8/7, 15);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "f := x -> (x^2 + 2*x - 15)/( 2*x^2 - 5*x - 3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 20 "Let's first look at " }{XPPEDIT 18 0 "x" "6#%\"xG " }{TEXT -1 50 " approaching 3 from values that are larger than 3." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "f(3.1); f(3.001); f(3.00001 ); f(3.0000001); f(3.0000000001);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "No w, let's look at " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 51 " approac hing 3 from values that are smaller than 3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "f(2.9); f(2.999); f(2.99999); f(2.9999999); f(2.99 99999999);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "It does appear that as " }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 21 " is approaching 3, f(" }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 17 ") is approaching " }{XPPEDIT 18 0 "8/7" "6#*&\"\")\"\"\" \"\"(!\"\"" }{TEXT -1 16 " . That is, as " }{XPPEDIT 18 0 "x;" "6#%\" xG" }{TEXT -1 19 " is approaching 3, " }{XPPEDIT 18 0 "(x^2+2*x-15)/(2 *x^2-5*x-3)" "6#*&,(*$%\"xG\"\"#\"\"\"*&\"\"#F(F&F(F(\"#:!\"\"F(,(*&\" \"#F(*$F&\"\"#F(F(*&\"\"&F(F&F(F,\"\"$F,F," }{TEXT -1 17 " is approac hing " }{XPPEDIT 18 0 "1.14285714285714;" "6#$\"09dG9dG9\"!#9" }{TEXT -1 32 ", the decimal approximation for " }{XPPEDIT 18 0 "8/7" "6#*&\" \")\"\"\"\"\"(!\"\"" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Now, let's find the limit using Maple:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Limit(f(x), x=3) ; value(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Digits := 25:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "3. In le cture, we claimed that " }{XPPEDIT 18 0 "Limit((w^2+3*w)/(w^2+6*w+9),w = -3);" "6#-%&LimitG6$*&,&*$%\"wG\"\"#\"\"\"*&\"\"$F+F)F+F+F+,(*$F)\" \"#F+*&\"\"'F+F)F+F+\"\"*F+!\"\"/F),$\"\"$F4" }{TEXT -1 73 " does not exist . Let's see this using Maple. We want to show that as " } {XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 18 " approaches - 3, " } {XPPEDIT 18 0 "(w^2+3*w)/(w^2+6*w+9);" "6#*&,&*$%\"wG\"\"#\"\"\"*&\"\" $F(F&F(F(F(,(*$F&\"\"#F(*&\"\"'F(F&F(F(\"\"*F(!\"\"" }{TEXT -1 50 " d oes not approach any value. Let's again let f(" }{XPPEDIT 18 0 "w;" " 6#%\"wG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "(w^2+3*w)/(w^2+6*w+9);" "6 #*&,&*$%\"wG\"\"#\"\"\"*&\"\"$F(F&F(F(F(,(*$F&\"\"#F(*&\"\"'F(F&F(F(\" \"*F(!\"\"" }{TEXT -1 14 " and show as " }{XPPEDIT 18 0 "x;" "6#%\"xG " }{TEXT -1 19 " approaches - 3, f(" }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 30 ") does not approach any value." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "f := w -> (w^2 + 3*w)/(w^2 + 6*w + 9);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Let's first look at " }{XPPEDIT 18 0 "w" "6#%\" wG" }{TEXT -1 54 " approaching - 3 from values that are larger than - \+ 3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f(-2.9); f(-2.999); f (-2.99999); f(-2.9999999); f(-2.9999999999);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Let's first look at " }{XPPEDIT 18 0 "w" "6#%\"wG" } {TEXT -1 55 " approaching - 3 from values that are smaller than - 3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f(-3.1); f(-3.001); f(-3. 00001); f(-3.0000001); f(-3.0000000001);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Notice as " }{XPPEDIT 18 0 "w" "6#%\"wG" }{TEXT -1 19 " approaches - 3, f(" }{XPPEDIT 18 0 "w" "6#%\"wG" }{TEXT -1 45 ") is not approaching any value. That is, as \+ " }{XPPEDIT 18 0 "w" "6#%\"wG" }{TEXT -1 17 " approaches - 3, " } {XPPEDIT 18 0 "(w^2+3*w)/(w^2+6*w+9)" "6#*&,&*$%\"wG\"\"#\"\"\"*&\"\"$ F(F&F(F(F(,(*$F&\"\"#F(*&\"\"'F(F&F(F(\"\"*F(!\"\"" }{TEXT -1 38 " is \+ not approaching any value. Thus, " }{XPPEDIT 18 0 "Limit((w^2+3*w)/(w ^2+6*w+9),w = -3);" "6#-%&LimitG6$*&,&*$%\"wG\"\"#\"\"\"*&\"\"$F+F)F+F +F+,(*$F)\"\"#F+*&\"\"'F+F)F+F+\"\"*F+!\"\"/F),$\"\"$F4" }{TEXT -1 19 " does not exist . " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Now, let's find the limit using Maple:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Limit(f(w),w=-3); value(%); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Unfortunately, this answer of " }{TEXT 256 9 "undefined" }{TEXT -1 4 " is " }{TEXT 257 3 "not" }{TEXT -1 33 " correct. The correct an swer is " }{TEXT 258 20 "does not exist (DNE)" }{TEXT -1 1 "." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 41 "4. In lecture notes, it is claimed that " }{XPPEDIT 18 0 "Limit(((x+h)^2-x^2)/h,h = 0);" "6#-%&LimitG6$*&,&*$,&%\"xG\"\"\" %\"hGF+\"\"#F+*$F*\"\"#!\"\"F+F,F0/F,\"\"!" }{TEXT -1 4 " = " } {XPPEDIT 18 0 "2*x;" "6#*&\"\"#\"\"\"%\"xGF%" }{TEXT -1 57 " . Let's \+ see this using Maple. We want to show that as " }{XPPEDIT 18 0 "h;" " 6#%\"hG" }{TEXT -1 15 " approaches 0, " }{XPPEDIT 18 0 "((x+h)^2-x^2)/ h;" "6#*&,&*$,&%\"xG\"\"\"%\"hGF(\"\"#F(*$F'\"\"#!\"\"F(F)F-" }{TEXT -1 12 " approaches " }{XPPEDIT 18 0 "2*x;" "6#*&\"\"#\"\"\"%\"xGF%" } {TEXT -1 21 ". Let's again let f(" }{XPPEDIT 18 0 "h;" "6#%\"hG" } {TEXT -1 4 ") = " }{XPPEDIT 18 0 "((x+h)^2-x^2)/h;" "6#*&,&*$,&%\"xG\" \"\"%\"hGF(\"\"#F(*$F'\"\"#!\"\"F(F)F-" }{TEXT -1 14 " and show as " }{XPPEDIT 18 0 "h;" "6#%\"hG" }{TEXT -1 17 " approaches 0, f(" } {XPPEDIT 18 0 "h;" "6#%\"hG" }{TEXT -1 13 ") approaches " }{XPPEDIT 18 0 "2*x" "6#*&\"\"#\"\"\"%\"xGF%" }{TEXT -1 29 ". Note: f is a fun ction of " }{XPPEDIT 18 0 "h;" "6#%\"hG" }{TEXT -1 50 " since this is \+ limiting variable. We are letting " }{XPPEDIT 18 0 "h;" "6#%\"hG" } {TEXT -1 12 " approach 0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f := h -> ( (x + h)^2 - x^2 )/h;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Let's first look at " }{XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 100 " ap proaching 0 from values that are larger than 0. Note: We will need t o simplify each expression." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "f(0.1); simplify(%); f(0.001); simplify(%); f(0.00001); simplify( %); f(0.0000001); simplify(%); f(0.0000000001); simplify(%);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 20 "Let's first look at " }{XPPEDIT 18 0 "h " "6#%\"hG" }{TEXT -1 108 " approaching 0 from values that are smaller than 0. Note: Again, we will need to simplify each expression." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "f(-0.1); simplify(%); f(-0. 001); simplify(%); f(-0.00001); simplify(%); f(-0.0000001); simplify(% ); f(-0.0000000001); simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "It does appear that as " } {XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 21 " is approaching 0, f(" } {XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 17 ") is approaching " } {XPPEDIT 18 0 "2*x" "6#*&\"\"#\"\"\"%\"xGF%" }{TEXT -1 15 ". That is, as " }{XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 19 " is approaching 0, " }{XPPEDIT 18 0 "((x+h)^2-x^2)/h" "6#*&,&*$,&%\"xG\"\"\"%\"hGF(\"\"#F(* $F'\"\"#!\"\"F(F)F-" }{TEXT -1 16 " is approaching " }{XPPEDIT 18 0 "2 *x" "6#*&\"\"#\"\"\"%\"xGF%" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Now, let's find the lim it using Maple:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Limit(f( h), h=0); value(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "Use Maple to find the following limits from the lecture notes. However, only use the " }{TEXT 259 6 "Limit " }{TEXT -1 180 "command to check your answer. You might want to work with twe nty significant digits instead of the default value of ten. To get tw enty significant digits, type the Maple command " }{TEXT 260 15 "Digit s := 20: " }{TEXT -1 59 "You will need to retype this command after e ach use of the " }{TEXT 261 7 "restart" }{TEXT -1 9 " command." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "1. " } {XPPEDIT 18 0 "Limit(3*x+5,x = 1);" "6#-%&LimitG6$,&*&\"\"$\"\"\"%\"xG F)F)\"\"&F)/F*\"\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 2 " \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "2. " }{XPPEDIT 18 0 "Limit((x^2-4)/(x-2),x = 2);" "6 #-%&LimitG6$*&,&*$%\"xG\"\"#\"\"\"\"\"%!\"\"F+,&F)F+\"\"#F-F-/F)\"\"# " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "3. " }{XPPEDIT 18 0 "Limit((r^2-r-2)/(3*r^2+11*r+6), r = -1);" "6#-%&LimitG6$*&,(*$%\"rG\"\"#\"\"\"F)!\"\"\"\"#F,F+,(*&\"\" $F+*$F)\"\"#F+F+*&\"#6F+F)F+F+\"\"'F+F,/F),$\"\"\"F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "4. " }{XPPEDIT 18 0 "Limit((sqrt(t+1)-2)/(t-3),t = 3);" "6#-%&LimitG6$* &,&-%%sqrtG6#,&%\"tG\"\"\"\"\"\"F-F-\"\"#!\"\"F-,&F,F-\"\"$F0F0/F,\"\" $" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "Find the following l imits, which we will be discussing later." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "1. " }{XPPEDIT 18 0 "Limit(((x+h)^ 3-x^3)/h,h = 0);" "6#-%&LimitG6$*&,&*$,&%\"xG\"\"\"%\"hGF+\"\"$F+*$F* \"\"$!\"\"F+F,F0/F,\"\"!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "2. " }{XPPEDIT 18 0 "Limit(((x+h)^4-x^4)/h,h = 0);" "6#-% &LimitG6$*&,&*$,&%\"xG\"\"\"%\"hGF+\"\"%F+*$F*\"\"%!\"\"F+F,F0/F,\"\"! " }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "1 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }