{VERSION 6 0 "IBM INTEL NT" "6.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0
1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0
0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }
{CSTYLE "" -1 256 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal
" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1
1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1
-1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2
0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1
2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }}
{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 192 "Chapter 2 selected end pr
oblems meant as an example of how to expand into a nice report from wh
at might be an innocently posed problem, doing more if possible with e
ach part of the discussion." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 48 "2
.5.3 continuity of a piecewise defined function" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
34 "For what value(s) of the constant " }{XPPEDIT 18 0 "a;" "6#%\"aG"
}{TEXT -1 16 " is the function" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 69 "f:=x->piecewise(0<=x and x<1, ln(1-a^2+a^2*x)/(x-1),x>=1, 1/(1
-a*x));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "continuous at each poi
nt of the interval " }{XPPEDIT 18 0 "[0, infinity];" "6#7$\"\"!%)infin
ityG" }{TEXT -1 2 "?\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "The fi
rst thing we must check is how we can establish continuity at the poin
t where the piecewise formula changes at " }{XPPEDIT 18 0 "x = 1;" "6#
/%\"xG\"\"\"" }{TEXT -1 108 " and we also have to make sure that the l
n function does not have a nonpositive value on the interval (0,1)." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "L_R:=limit(f(x),x=1,right)
;\nL_L:=limit(f(x),x=1,left);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "S
o " }{XPPEDIT 18 0 "a = 1;" "6#/%\"aG\"\"\"" }{TEXT -1 173 " is not al
lowed or this is not finite. And for the left limit to be finite, sinc
e the denominator vanishes the numerator must vanish, in which case L'
Hopital's rule applies:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "
limit(a^2/(1-a^2+a^2*x),x=1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 26 "L_L:=limit(f(x),x=1,left);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
17 "But of course at " }{XPPEDIT 18 0 "x = 1;" "6#/%\"xG\"\"\"" }
{TEXT -1 102 " the argument of the log is automatically 1 and so the l
og vanishes as required, no matter what value " }{XPPEDIT 18 0 "a^%?;
" "6#)%\"aG%#%?G" }{TEXT -1 106 " has as long as it yields a positive \+
input on the interval.\n\nSetting the two limits equal and solving for
" }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 27 " leads to a cubic equat
ion:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "L_R=L_L;\nsimplify(
%*(1-a));\nCubicC:=simplify(lhs(%)-rhs(%))=0;" }}}{EXCHG {PARA 0 "" 0
"" {TEXT -1 28 "which can be solved exactly:" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 36 "solve(CubicC,a);\nA:=%[1];\nevalf(%%);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Just to show that we can check thi
s exactly:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "a:='a':\nsubs
(a=A,CubicC);\nsimplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "On
ly the first root is real, so:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 6 "a:=A:\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Now we can plot
the piecewise function:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42
"plot(f(x),x=0..10);\nplot(f(x),x=0.9..1.1);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 51 "Notice that this function is not differentiable at " }
{XPPEDIT 18 0 "x = 1;" "6#/%\"xG\"\"\"" }{TEXT -1 92 " but the derivat
ive has a finite jump discontinuityu as the slope decreases suddenly t
here.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 213 "Of course we can't pl
ot all the way out to infinity, so we should make sure that the second
formula is well behaved for all values greater than 1 if we are to gu
arantee that this function is continuous everywhere." }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 17 "1/(1-evalf(A)*x);" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 49 "Since this only vanishes for a negative value of " }
{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 36 ", this is continous out to \+
infinity:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(f(x),x=0.
.infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "and of course goes
to zero at infinity. This is a suggestive graph of its behavior at in
finity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 126 "And since the linear function argument of the log is non
zero, there can be no problem inside that interval from 0 to 1 either.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "evalf(1-A^2+A^2*x);\npl
ot(1-A^2+A^2*x,x=0..1,y=0..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
120 "Thus our piecewise function is continuous on the positive real ax
is and differentiable everywhere but at the join point." }}}}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 23 "2.5.7 limit of sequence" }}{EXCHG {PARA
0 "" 0 "" {TEXT -1 53 "It is not very difficult to check that the sequ
ence (" }{XPPEDIT 18 0 "x[n];" "6#&%\"xG6#%\"nG" }{TEXT -1 78 "), gene
rated by repeated application of the sine function to an initial value
" }{XPPEDIT 18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 18 " in the inte
rval (" }{XPPEDIT 18 0 "0,Pi;" "6$\"\"!%#PiG" }{TEXT -1 55 "), is stri
ctly monotonically decreasing to 0. Thus for " }{XPPEDIT 18 0 "x[0];"
"6#&%\"xG6#\"\"!" }{TEXT -1 5 " in (" }{XPPEDIT 18 0 "0,Pi;" "6$\"\"!%
#PiG" }{TEXT -1 7 ") and \n" }{XPPEDIT 18 0 "x[n+1] = sin(x[n]);" "6#/
&%\"xG6#,&%\"nG\"\"\"F)F)-%$sinG6#&F%6#F(" }{TEXT -1 4 " , " }
{XPPEDIT 18 0 "n = 0,1;" "6$/%\"nG\"\"!\"\"\"" }{TEXT -1 17 ",...\nwe \+
have 0 < " }{XPPEDIT 18 0 "x[n+1];" "6#&%\"xG6#,&%\"nG\"\"\"F(F(" }
{TEXT -1 3 " < " }{XPPEDIT 18 0 "x[n];" "6#&%\"xG6#%\"nG" }{TEXT -1 5
" and " }{XPPEDIT 18 0 "Lim(x[n],n = infinity) = 0;" "6#/-%$LimG6$&%\"
xG6#%\"nG/F*%)infinityG\"\"!" }{TEXT -1 38 ". Prove this. Also calcula
te the limit" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Limit(1/x[n+1]^2-1/x[n]^2,n
=infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "Just to get starte
d, notice that the sine is smaller than its argument for this interval
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "plot([x,sin(x)],x=0..P
i/2,color=[red,blue]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "so " }
{XPPEDIT 18 0 "x[n+1] < x[n];" "6#2&%\"xG6#,&%\"nG\"\"\"F)F)&F%6#F(" }
{TEXT -1 165 ". Since this is a sequence of positive terms bounded bel
ow by zero, and it is monotonically decreasing, we know that it has to
have a limit. If it does have a limit " }{XPPEDIT 18 0 "L;" "6#%\"LG
" }{TEXT -1 23 ", then it will satisfy:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 23 "L=sin(L);\nL=solve(%,L);" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 27 "so it has to converge to 0." }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 38 "Let's look at how fast this converges:" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 75 "x:=proc(n::integer) option remember;\nif \+
n=0 then X\nelse sin(x(n-1))\nfi end;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 94 "X:=0.9;\nseq(x(n),n=0..10);\nseq(x(n),n=90..100);\nse
q(x(n),n=990..1000);\nseq(x(n),n=4990..5000);" }}}{EXCHG {PARA 0 "" 0
"" {TEXT -1 147 "It is converging to zero very slowly. Why? Because th
e closer you get to 0 the smaller an effect the sin has on its making \+
its input a bit smaller." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 60 "Now let's define a new sequence whose lim
it we want to test:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "y:=p
roc(n::integer)\ny(n):=1/x(n+1)^2-1/x(n)^2\nend;" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 46 "X:=0.9;\nseq(y(n),n=0..10);\nseq(y(n),n=10..2
0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "X:=0.1;\nseq(y(n),n=
0..10);\nseq(y(n),n=10..20);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 51 "X:=0.00001;\nseq(y(n),n=0..10);\nseq(y(n),n=90..100);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 167 "It looks like it is converging pe
rhaps. The closer to zero we start it, the fewer iterations we have to
do to get closer to the limit. Now let us examine the formula: " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "`y(n)`=1/`x(n+1)`^2-1/`x(n)`
^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "Notice that since " }
{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 222 " is a decreasing positive \+
fractional sequence its reciprocal sequence is greater than 1 and is i
ncreasing, and squaring that makes it larger, but this difference is o
nly guaranteed to be positive. Suppose this had a limit " }{XPPEDIT
18 0 "M;" "6#%\"MG" }{TEXT -1 17 ", then for small " }{XPPEDIT 18 0 "x
(n) = Delta;" "6#/-%\"xG6#%\"nG%&DeltaG" }{TEXT -1 16 ", we would have
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "1/sin(Delta)^2-1/Delta
^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "is approximately " }
{XPPEDIT 18 0 "M;" "6#%\"MG" }{TEXT -1 25 ". Can we take this limit?"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Limit(1/sin(Delta)^2-1/De
lta^2,Delta=0);\nvalue(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 173 "On
e way of understanding this limit would be to combine the fractions an
d look at the limit of the Taylor expansion of the numerator and denom
inator (the latter is obvious):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 113 "(Delta^2-sin(Delta)^2)/sin(Delta)^2/Delta^2;\nN:=numer(%); De
n:=denom(%%);\ntaylor(N,Delta,6);\ntaylor(Den,Delta,6);" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 75 "So the 1/3 comes from the fourth order ta
ylor coefficient of the numerator." }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 26 "So what have we done here?" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 23 "g:=x->1/sin(x)^2-1/x^2;" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 35 "We have looked at this function of " }{XPPEDIT 18 0 "x;"
"6#%\"xG" }{TEXT -1 146 " on the interval from 0 to 1 where it is cont
inuous and examined its limit at 0 where it is undefined by evaluating
it on a sequence of values of " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT
-1 64 " which slowly converge to 0. The plot will not show any problem
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot(g(x),x=-1..1,y=0.
.g(1));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "Notice that this is a
n even function which would also be differentiable at 0 if we defined \+
its value by its limit." }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 26 "2.5.
10 limit of a sequence" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "res
tart:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "It is a well-known fact \+
that the sequence with general term" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 16 "a:=n->(1-1/n)^n;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
187 "is strictly monotonically increasing. Use Maple to prove this. Cl
early, the sequence is bounded and hence convergent. Compute its limit
. Apply the technique used in Worksheet FuncW2b.mws." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "First let's plo
t the beginning of the sequence:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 42 "plot([seq([n,a(n)],n=1..10)],style=point);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "1) Consider the remark: \"Clearly
the sequence is bounded.\" Well, what is the upper bound then? We sho
uld think about this." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "a(n
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 289 "We are raising a proper fr
action to a natural number power, but any positive power of a proper f
raction is smaller than its starting value, which in particular is sma
ller than 1. Yes, now that we have thought about it, the sequence is c
learly bounded above by 1. Had we taken the sequence:" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "b:=n->(1+1/n)^n: b(n); limit(b(n),n
=infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "it would no longe
r be clear that it is bounded above, although looking more and more te
rms might plausibly suggest that it is bounded above." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "2) To be an in
creasing sequence, the ratio of successive terms should be greater tha
n 1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "seq(evalf(a(n+1)/a(
n)),n=2..10);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 " How can we prov
e this?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "a(n+1)/a(n);\nsi
mplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 166 "We have to combine
these by hand since Maple won't cooperate. Use the negative exponent \+
to take the reciprocal of the inside fraction, and combine the like po
wers of " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 66 " and then the su
m and difference to get the difference of squares:" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 24 "(n^2/(n^2-1))^n*n/(n+1);" }}}{EXCHG {PARA
0 "" 0 "" {TEXT -1 90 "The argument of the power is greater than one s
o removing the power gives a smaller value:" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 40 "(n^2/(n^2-1))*n/(n+1);\nexpand(denom(%));" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "Dead end. Okay, I give up. The bac
k of the book says to rewrite this in the form:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "(1+1/(n^2-1))^n*n/(n+1);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 109 "and expanding using the binomial theorem dropping
the terms after the second (decreasing the result) gives us" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "(1+n/(n^2-1))*n/(n+1);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "simplify(%);\nexpand(numer(%
));\nexpand(denom(%%));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 241 "Aha! \+
the denominator is always one less than the numerator, so the ratio is
always bigger than one, and hence the original ratio is always bigger
than 1. Clearly Maple cannot easily do these manipulations without us
doing some steps by hand." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 143 "3) Now we compute the limit of the seque
nce, which we know it must have since it is a monotonically increasing
sequence which is bounded above:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 46 "Limit(a[n],n=infinity)=limit(a(n),n=infinity);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "4) Of course we should have recogn
ized the limit from the first year calculus continuum limit:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "Limit((1+x)^(1/x),x=0)=limit
((1+x)^(1/x),x=0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Changing si
gn and taking the reciprocal variable:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 60 "Limit((1-1/u)^(u),u=infinity)=limit((1-1/u)^(u),u=inf
inity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "This is also true for \+
the discrete limit through positive integers as well." }}}}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 45 "2.5.11 limit of a sequence (l'Hopital's r
ule)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT 256 17 "Problem Statement" }{TEXT -1 8 "\nAssume
" }{XPPEDIT 18 0 "k;" "6#%\"kG" }{TEXT -1 62 " < 0 and constant. Let \+
Maple compute the limit of the sequence" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 22 "a:=n->(2*k^(1/n)-1)^n;" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 169 "The answer may surprise you, it may even make you suspec
t that it could be wrong. To find out, apply the natural logarithm fun
ction to the general term and remember that" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 44 "Limit(ln(1+x)/x,x=0) = limit(ln(1+x)/x,x=0);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "It may also be a good idea to firs
t consider the limit" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Lim
it(n*(k^(1/n)-1),n=infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103
"Show in a few lines of Maple code (with corresponding output of cours
e) what you tried and discovered.\n" }}}{EXCHG {PARA 256 "" 0 ""
{TEXT -1 16 "Problem Response" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "
Here is the possibly surprising limit:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 46 "Limit(a(n),n=infinity)=limit(a(n),n=infinity);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "Why do they say that it might be \+
surprising? It seems that the multiplicative factor of 2 in front of \+
" }{XPPEDIT 18 0 "k;" "6#%\"kG" }{TEXT -1 71 " is squaring the result,
which you might expect instead from replacing " }{XPPEDIT 18 0 "k;" "
6#%\"kG" }{TEXT -1 4 " by " }{XPPEDIT 18 0 "k^2;" "6#*$%\"kG\"\"#" }
{TEXT -1 246 " instead of multiplying its root by 2. So we might consi
der the other sequence in which a 2 is not present as suggested by the
problem statement. \nOr maybe we are supposed to see this as an indet
erminate limit since no matter what positive value " }{XPPEDIT 18 0 "k
;" "6#%\"kG" }{TEXT -1 6 " has, " }{XPPEDIT 18 0 "k^(1/n);" "6#)%\"kG*
&\"\"\"F&%\"nG!\"\"" }{TEXT -1 16 "approaches 1 as " }{XPPEDIT 18 0 "n
;" "6#%\"nG" }{TEXT -1 95 " grows, so the what is inside the parenthes
is approaches 1, so we have the indeterminate limit " }{XPPEDIT 18 0 "
1^infinity;" "6#)\"\"\"%)infinityG" }{TEXT -1 124 ", and l'Hopital's R
ule applies to the corresponding continuous limit.\nGuessing what some
one else had in mind is not so easy." }}{PARA 0 "" 0 "" {TEXT -1 93 "
\nTaking the logarithm is the certainly the first step in the applicat
ion of l'Hopital's rule:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61
"ln(a(n))=n*ln(2*k^(1/n)-1);\nln(a(n))=ln(2*k^(1/n)-1)/`(1/n)`;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "diff(ln(2*k^(1/n)-1),n)/diff
( 1/n,n);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "This is a determinat
e limit:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Limit(%,n=infin
ity)=limit(%,n=infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "whos
e exponential is the above therefore correct result:" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 12 "exp(rhs(%));" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 14 "The factor of " }{XPPEDIT 18 0 "2*ln(k);" "6#*&\"\"#\"\"
\"-%#lnG6#%\"kGF%" }{TEXT -1 60 " just comes from the derivative of th
e exponential function " }{XPPEDIT 18 0 "2*exp(ln(k)*x);" "6#*&\"\"#\"
\"\"-%$expG6#*&-%#lnG6#%\"kGF%%\"xGF%F%" }{TEXT -1 16 "with respect to
" }{XPPEDIT 18 0 "x = 1/n;" "6#/%\"xG*&\"\"\"F&%\"nG!\"\"" }{TEXT -1
15 ".\nEnd of story." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA
0 "" 0 "" {TEXT -1 51 "Now we consider the alternative suggested seque
nce:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "b:= n->(k^(1/n)-1)^
n;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "b := ln(2*k^(1/n)-1);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "This limit is indeterminate o
f type " }{XPPEDIT 18 0 "0^infinity;" "6#)\"\"!%)infinityG" }{TEXT -1
1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ln(b(n))=n*ln(k^(1/
n)-1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 46 "Limit(b(n),n=infinity)=limit(b(n),n=infinity);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "Limit(n*ln(k^(1/n)-1),n
=infinity)=limit(n*ln(k^(1/n)-1),n=infinity);" }}}{EXCHG {PARA 0 "" 0
"" {TEXT -1 36 "This to is a l'Hopital rule problem." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 186 "I have no clue
why he is talking about the following \"standard limit\" and not just
referring to L'Hopital's rule which is also how this is easily evalua
ted as a 0/0 indetermintate limit:" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 42 "Limit(ln(1+x)/x,x=0)=limit(ln(1+x)/x,x=0);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 0 0" 192 }{VIEWOPTS 1 1 0 1 1 1803
1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }