---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------
The compute of limits of a function (and order of infinite/infinitesimal)
g(x) -> L as x -> p
#
# sin(sin(x))/(7*sin(x)+sin(x)^3) as x -> 0
g = function(x) sin(sin(x))/(7*sin(x)+sin(x)^3)
BF=4; HF=3; graph2F(g, -pi,pi, "brown")
g(0)
# NaN
point0(0,g(1e-10),"black")
# x -> 0+ a 0 da destra
for(x in 10^-(2:8) ) more(g(x))
# 0.142852721310892
# 0.142857098639478
# 0.142857142414966
# 0.142857142852721
# 0.142857142857099
# 0.142857142857142
# 0.142857142857143
# A more sophisticated way (also to evaluate variations)
A=1;B=10; L=0;for(n in A:B) {x=10^-n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 1 0.1424172 [1] 0.142417182596157
# 2 0.0004355387 [1] 0.142852721310892
# 3 4.377329e-06 [1] 0.142857098639478
# 4 4.377549e-08 [1] 0.142857142414966
# 5 4.377551e-10 [1] 0.142857142852721
# 6 4.377554e-12 [1] 0.142857142857099
# 7 4.377054e-14 [1] 0.142857142857142
# 8 4.440892e-16 [1] 0.142857142857143
# 9 0 [1] 0.142857142857143
# 10 0 [1] 0.142857142857143
# n g(x)-L L
# L is about 0.142857142857143. It is rational?
fraction( g(10^-8) )
# 1/7
# x -> 0- a 0 da sinistra
fraction( g(-(10^-8)) )
# 1/7
#
# With which order g tends to 1/7? x is divided by 10 -> g(x)-L is divided by 100
# g(x)-L is an infinitesimal of order 2
# --------------------
# If I want, I can approximate g near 0 (I compute the ratio between ...)
L = 1/7
A=1;B=8; L0=0;for(n in A:B) {x=10^-n; U=(L-g(x))/x^2; cat(n,'\t',U-L0,'\t'); more(U); L0=U}
# 1 0.04399603 [1] 0.0439960260985861
# 2 0.0002194364 [1] 0.0442154625035496
# 3 2.202332e-06 [1] 0.0442176648352 (/100)
# 4 2.023381e-08 [1] 0.0442176850690146 there is no longer a regular approach
# 5 -6.938894e-18 [1] 0.0442176850690146
# 6 -3.053113e-06 [1] 0.0442146319556969
# 7 0.000194289 [1] 0.0444089209850063
# 8 -0.04440892 [1] 0
# n variation (L-g(x))/x^2
h = function(x) 1/7 - 0.0442177*x^2 # as x -> 0 g(x) ~ h(x)
graph1(h, -pi,pi, "blue")
# If I want a more accurate valutation I can use 2^-n instead of 10^-n
A=9;B=15; L0=0;for(n in A:B) {x=2^-n; U=(L-g(x))/x^2; cat(n,'\t',U-L0,'\t'); more(U); L0=U}
# 9 0.0442176 [1] 0.0442176022042986
# 10 6.365008e-08 [1] 0.0442176658543758
# 11 1.589069e-08 [1] 0.0442176817450672 (/4)
# 12 3.958121e-09 [1] 0.0442176857031882
# 13 1.396984e-09 [1] 0.044217687100172 there is no longer a regular approach
# 14 -3.72529e-09 [1] 0.0442176833748817
# 15 -7.450581e-09 [1] 0.0442176759243011
# 0.044217687
# --------------------
#
g(x) -> L as x -> ∞
#
# (1+1/x)^x as x -> ∞
g = function(x) (1+1/x)^x
BF=4; HF=3; Plane(0,50, 0,3); graph2(g, 0,50, "brown")
# I prove with 10^n:
A=1;B=12; L=0;for(n in A:B) {x=10^n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 1 2.593742 [1] 2.5937424601
# 2 0.1110714 [1] 2.70481382942153
# 3 0.0121101 [1] 2.71692393223559
# 4 0.001221995 [1] 2.71814592682493 (/10)
# 5 0.0001223104 [1] 2.7182682371923
# 6 1.22319e-05 [1] 2.71828046909573
# 7 1.225037e-06 [1] 2.71828169413255
# 8 1.042066e-07 [1] 2.71828179833913 there is no longer a regular approach
# 9 2.536574e-07 [1] 2.71828205199649
# 10 1.238298e-09 [1] 2.71828205323479
# 11 1.223226e-10 [1] 2.71828205335711
# 12 0.0002414427 [1] 2.71852349603724
#
# If I want a more accurate valutation I can use 2^-n instead of 10^-n
A=30;B=32; L=0;for(n in A:B) {x=2^n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 30 2.718282 [1] 2.71828182719316
# 31 6.329839e-10 [1] 2.71828182782615
# 32 3.164495e-10 [1] 2.7182818281426 (/2)
A=40;B=42; L=0;for(n in A:B) {x=2^n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 40 2.718282 [1] 2.71828182845781
# 41 6.181722e-13 [1] 2.71828182845843
# 42 3.08642e-13 [1] 2.71828182845874
A=50;B=52; L=0;for(n in A:B) {x=2^n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 50 2.718282 [1] 2.71828182845904
# 51 4.440892e-16 [1] 2.71828182845904 there is no longer a regular approach
# 52 4.440892e-16 [1] 2.71828182845905
# I deepen the situation
A=48;B=52; L=0;for(n in A:B) {x=2^n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 48 2.718282 [1] 2.71828182845904
# 49 2.664535e-15 [1] 2.71828182845904
# 50 1.332268e-15 [1] 2.71828182845904
# 51 4.440892e-16 [1] 2.71828182845904
# 52 4.440892e-16 [1] 2.71828182845905
# 2.71828182845905
# I can find that this nummber is Napier's constant (or Euler's number) e, exp(1)
more(exp(1))
# 2.71828182845905
#
L = exp(1)
# With which order g tends to L? x is multiplied by 2 -> g(x)-L is divided by 2
# g(x)-L is an infinitesimal of order 1 (respect to 1/x)
#
# --------------------
# If I want, I can approximate g(x) as x -> ∞ (I compute the ratio between ...)
A=20;B=23; L0=0;for(n in A:B) {x=2^n; U=(L-g(x))/(1/x); cat(n,'\t',U-L0,'\t'); more(U); L0=U}
# 20 1.35914 [1] 1.3591397292912
# 21 6.016344e-07 [1] 1.35914033092558
# 22 -3.613532e-07 [1] 1.35913996957242
# 23 4.824251e-07 [1] 1.35914045199752
A=18;B=23; L0=0;for(n in A:B) {x=2^n; U=(L-g(x))/(1/x); cat(n,'\t',U-L0,'\t'); more(U); L0=U}
# 18 1.359136 [1] 1.35913616209291
# 19 2.376735e-06 [1] 1.35913853882812
# 20 1.190463e-06 [1] 1.3591397292912 (/2)
# 21 6.016344e-07 [1] 1.35914033092558
# 22 -3.613532e-07 [1] 1.35913996957242 there is no longer a regular approach
# 23 4.824251e-07 [1] 1.35914045199752
#
h = function(x) exp(1) - 1.35914/x # as x -> ∞ g(x) ~ h(x)
graph1(h, 0,50, "red")
# --------------------
#
g(x) -> ∞ as x -> ∞
#
# sqrt(x^5)*sin(5/x^2) as x -> ∞
g = function(x) sqrt(x^5)*sin(5/x^2)
graph2F(g, 0,3, "brown")
graph2F(g, 0,15, "brown")
# It's clear the limit of g as x -> ∞ is ∞. We find the order of infinite.
for(x in 10^(4:10) ) more(g(x))
# 500
# 1581.13883008419
# 5000
# 15811.3883008419
# 50000
# 158113.883008419
# 5e+05
# If I multiply the input by 100 the output multiplies by 10:
# as x -> ∞ g(x) is an infinite of order 1/2
#
# --------------------
# If I want, I can approximate g(x) as x ->∞ (I compute the ratio between ...)
A=5;B=15; L0=0;for(n in A:B) {x=2^n; U=g(x)/sqrt(x); cat(n,'\t',U-L0,'\t'); more(U); L0=U }
# 5 4.99998 [1] 4.99998013180876
# 6 1.862643e-05 [1] 4.99999875823666
# 7 1.164153e-06 [1] 4.99999992238979
# 8 7.275958e-08 [1] 4.99999999514936
# 9 4.547474e-09 [1] 4.99999999969684
# 10 2.842171e-10 [1] 4.99999999998105
# 11 1.776357e-11 [1] 4.99999999999882
# 12 1.110223e-12 [1] 4.99999999999993
# 13 6.927792e-14 [1] 5
# 14 4.440892e-15 [1] 5
# 15 0 [1] 5
# The ratio is 5. Then: as x -> ∞ g(x) ~ 5*sqrt(x)
h = function(x) 5*sqrt(x)
graph2F(g, 0,15, "brown")
graph1(h, 0,15, "seagreen")
# --------------------
#
Other examples
#
# (cos(x)-1)/x^2 as x -> 0
g = function(x) (cos(x)-1)/x^2
graph2F(g, -4*pi,4*pi, "brown")
pointO(0,-1/2, "blue")
# x -> 0+
for(x in 10^-(2:8) ) more(g(x))
# -0.499995833347366
# -0.499999958325503
# -0.499999996961265
# -0.500000041370185
# -0.500044450291171
# -0.49960036108132
# 0
# It is best to approach 0 more slowly (and to evaluate variations)
A=7;B=14; L=0;for(n in A:B) {x=2^-n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 7 -0.4999975 [1] -0.499997456874553
# 8 -1.907345e-06 [1] -0.499999364219548
# 9 -4.768444e-07 [1] -0.499999841063982
# 10 -1.192384e-07 [1] -0.499999960302375 (/4)
# 11 -2.991874e-08 [1] -0.499999990221113
# 12 -7.916242e-09 [1] -0.499999998137355
# 13 -1.862645e-09 [1] -0.5
# 14 0 [1] -0.5
# -1/2
# x -> 0-
g(2^-13)
# -1/2
L = -1/2
# With which order g tends to -1/2?
# As I divide by 2 the distance from 0 the distance of g(x) from the limit L is
# divided by 4: g(x)-L is an infinitesimal of order 2
#
# --------------------
#
# If I want, I can approximate g near 0 (I compute the ratio between ...)
A=4;B=12; L0=0;for(n in A:B) {x=2^-n; U=(g(x)-L)/x^2; cat(n,'\t',U-L0,'\t'); more(U); L0=U }
# 4 0.04166124 [1] 0.0416612416956923
# 5 4.068694e-06 [1] 0.0416653103893623
# 6 1.016655e-06 [1] 0.0416663270443678 (/4)
# 7 2.402812e-07 [1] 0.041666567325592
# 8 -5.960464e-08 [1] 0.0416665077209473 there is no longer a regular approach
# 9 -2.384186e-06 [1] 0.0416641235351562
# 10 -3.814697e-05 [1] 0.0416259765625
# 11 -0.0006103516 [1] 0.041015625
# 12 -0.009765625 [1] 0.03125
fraction(0.0416666)
# 1/24
h = function(x) 1/24*x^2-1/2; graph1(h, -4*pi,4*pi, "seagreen")
# as x -> 0 g(x) ~ h(x)
# --------------------
#
# (x-1)^2+(1-x)^3+sqrt(x^2-1) as x -> 1
g = function(x) (x-1)^2+(1-x)^3+sqrt(x^2-1)
graph2F(g, -2,4, "brown")
# It's clear the limit as x -> 1+ is 0. With which order g tends to 0?
A=2;B=8; L=0;for(n in A:B) {x=(1+10^-n); cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 2 0.1418735 [1] 0.141873468787578
# 3 -0.09713993 [1] 0.0447335374926867
# 4 -0.03059104 [1] 0.0141424991716993
# 5 -0.009670352 [1] 0.00447214723534007
# 6 -0.003057933 [1] 0.0014142139168997
# 7 -0.0009670003 [1] 0.000447213606811918
# 8 -0.0003057923 [1] 0.000141421355807667
# As I divide by 100 the distance from 1 the distance of g(x) from 0 is divided by 10:
# as x -> 1+ g(x) is an infinitesimal of order 1/2
#
# --------------------
# If I want, I can approximate g near 0 (I compute the ratio between ...)
A=5;B=13; L0=0;for(n in A:B) {x=1+10^-n; U=g(x)/(x-1)^(1/2); cat(n,'\t',U-L0,'\t'); more(U); L0=U }
# 5 1.414217 [1] 1.41421712952533
# 6 -3.212567e-06 [1] 1.41421391695787
# 7 -3.192261e-07 [1] 1.4142135977318
# 8 -3.53577e-08 [1] 1.4142135623741
# 9 -9.681145e-13 [1] 1.41421356237313
# 10 -3.08642e-14 [1] 1.4142135623731
# 11 -1.110223e-15 [1] 1.41421356237309
# 12 2.220446e-16 [1] 1.4142135623731
# 13 -2.220446e-16 [1] 1.41421356237309
more(sqrt(2))
# 1.4142135623731
# The ratio is sqrt(2). Then: as x -> 1+ g(x) ~ sqrt(2)*sqrt(x-1)
h = function(x) sqrt(2)*sqrt(x-1)
graph2F(g, 1,3, "brown"); graph1(h, 1,4, "seagreen")
# --------------------
#
# (1+1/x)^sqrt(x) as x -> ∞
g = function(x) (1+1/x)^sqrt(x)
BF=4; HF=3; Plane(0,50, 0,3); graph2(g, 0,50, "brown")
A=1;B=12; L=0;for(n in A:B) {x=10^n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
1 1.351746 [1] 1.35174621929824
2 -0.2471241 [1] 1.1046221254112
3 -0.07251034 [1] 1.03211178061832
4 -0.02206212 [1] 1.01004966209288
5 -0.006882395 [1] 1.00316726707346
6 -0.002166767 [1] 1.00100049966613
7 -0.0006842219 [1] 1.00031627775566
8 -0.0002162728 [1] 1.00010000499906
9 -6.838172e-05 [1] 1.00003162327921
10 -2.162323e-05 [1] 1.00001000005083
11 -6.837768e-06 [1] 1.00000316228292
12 -2.162194e-06 [1] 1.0000010000894
#
# If I want a more accurate valutation I can use 2^-n instead of 10^-n
A=45;B=55; L=0;for(n in A:B) {x=2^n; cat(n,'\t'); cat(g(x)-L,'\t'); L=g(x); more(L) }
# 45 1 [1] 1.00000016858741
# 46 -4.937811e-08 [1] 1.0000001192093
# 47 -3.49156e-08 [1] 1.0000000842937
# 48 -2.468905e-08 [1] 1.00000005960465
# 49 -1.74578e-08 [1] 1.00000004214685
# 50 -1.234453e-08 [1] 1.00000002980232
# 51 -8.728898e-09 [1] 1.00000002107342
# 52 -6.172263e-09 [1] 1.00000001490116
# 53 -1.490116e-08 [1] 1
# 54 0 [1] 1
# 55 0 [1] 1
L = 1
# With which order g tends to L? x is multiplied by 2 -> g(x)-L is divided by √2:
4.937811/3.49156
# 1.414213
3.49156/2.468905
# 1.414214
sqrt(2)
# 1.414214
# As x -> ∞ g(x)-1 ~ K*1/sqrt(x).
#
# --------------------
K = ?
A=40;B=43; L0=0;for(n in A:B) {x=2^n; U=(1-g(x))/(1/sqrt(x)); cat(n,'\t',U-L0,'\t'); more(U); L0=U}
# 40 -1 [1] -1.00000047660433
# 41 1.394217e-07 [1] -1.00000033718267
# 42 9.87641e-08 [1] -1.00000023841858
# 43 6.98233e-08 [1] -1.00000016859528
# K = 1
h = function(x) 1+1/sqrt(x)
graph2(h, 0,50, "seagreen")
# --------------------
Other examples of use