Initiation aux DL
Pour avoir une limite avec Maple, on utilse la fnction "limit". Pour un équivalent et/ou un développement limité, on utilise en général "taylor", ou "series" si "taylor" n’y arrive pas. Si l’affaire se passe au voisinage de l’infini, on peut utiliser "asympt"
> ?limit
• The limit(f,x=a,dir) function attempts to compute the limiting value of f as x approaches a.
• If dir is not specified, the limit is the real bidirectional limit, except in the case where the limit point is infinity or -infinity, in which case the limit is from the left to infinity and from the right to
-infinity. For help with directional limits see limit[dir].
> ?taylor
• The function taylor computes the Taylor series expansion of expr, with respect to the variable x, about the point a, up to order n.
• The taylor function is a restriction of the more general series function. See series for a complete explanation of the parameters.
> ?asympt
• The function asympt computes the asymptotic expansion of f with respect to the variable x (as x approaches infinity).
• The third argument n specifies the truncation order of the series expansion. If no third argument is given, the value of the global variable Order (default Order = 6) is used.
Exercice 3
> limit((1-2/n)^(3*n),n=infinity);
e(-6)
Exercice 7
> limit((1-1/n)^n,n=infinity);
e(-1)
Exercice 13
> plot({x,x^2,sqrt(x)},x=0..1);
0 0.2 0.4 0.6 0.8 1
0.2 0.4 x 0.6 0.8 1
> plot({1/x,1/x^ 2,-ln(x)},x=0..1,0..5);
0 1 2 3 4 5
0.2 0.4 x 0.6 0.8 1
Exercice 14
> plot({x,x^2,sqrt(x),ln(x)},x=1..3);
0 2 4 6 8
1.2 1.6 2x 2.4 2.8
> plot({1/x,1/x^2},x=1..3);
0.2 0.4 0.6 0.8 1
1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 x
Exercice 15
> limit((1-2/n)^(3*n),n=infinity);
e(-6) Incroyable !
Exercice 16
> taylor(1/sin(x)-1/x,x=0);
+ +
1
6x 7
360x3 O x( 4)
C’est un "développement de Taylor" (théorie en cours d’année). Le premier terme nous donne l’équivalent.
Exercice 17
> taylor(1/sin(x)^2-1/x^2,x=0);
+ + + 1
3 1
15x2 2
189x4 O x( 5)
> limit(1/sin(x)^2-1/x^2,x=0);
1 3
Exercice 20
> limit((t^t-1)/(1-t+ln(1+t)),t=1);
0
> limit(ln(t)/(t-1),t=1);
1
> limit((1-1/(n*sqrt(n)))^(n^(5/3)),n=infinity);
0
un peu plus subtil
> limit((t^t-1)/(1-t+ln(t)),t=1);
undefined
> limit((t^t-1)/(1-t+ln(t)),t=1,right);
−∞
> limit((t^t-1)/(1-t+ln(t)),t=1,left);
∞
> taylor((t^t-1)/(1-t+ln(t)),t=1);
Error, does not have a taylor expansion, try series()
> series((t^t-1)/(1-t+ln(t)),t=1);
−2 (t−1)-1−10− − +
3 20
9 (t−1) 173
135(t−1)2 O (( t−1)3)
> limit(x^(sqrt(x))/(sqrt(x))^x,x=infinity);
0
> limit((Pi/2-x)^(sin(x)/ln(cos(x))),x=Pi/2);
e
> limit(tan(x)^(tan(2*x)),x=0);
1
Exercice 21
> limit(1/(1-x^x)-1/(x*ln(x)),x=0,right);
∞
> limit(sqrt(x+sqrt(x))-sqrt(x),x=infinity);
1 2
> limit(cos(x)^(1/cotan(x)^2),x=0);
1
> limit(cos(x)^(1/tan(x)^2),x=0);
e(-1 2/ )
?????
> cotan(Pi/4);
cotan 1 4π
Il n connait pas cette fonction... cela dit, il est "assez contestable" de donner la réponse qu’il donne sur limit(cos(x)^(1/cotan(x)^2),x=0)...
> limit(tan(x)^(tan(2*x)),x=Pi/2,left);
1
> limit(tan(x)^(tan(2*x)),x=Pi/4);
e(-1)
Exercice 22
22.1
> limit(1/cos(x)^2+1/ln(sin(x)^2),x=Pi/2);
1 2
> taylor(ln(sin(x)^2)+cos(x)^2,x=Pi/2);
−1 +
2
− x 1
2π
4
O
− x 1
2π
6
Soit encore :
> taylor(ln(sin(Pi/2+u)^2)+cos(Pi/2+u)^2,u=0);
−1 +
2u4 O u( 6)
22.2
> limit((x^sin(x)-sin(x)^x)/(x^tan(x)-tan(x)^x),x=0);
-1 2
> taylor(x^sin(x)-sin(x)^x,x=0);
Error, does not have a taylor expansion, try series()
> series(x^sin(x)-sin(x)^x,x=0,4);
+
1− 6
1
6ln x( ) x3 O x( 4)
> series(x^tan(x)-tan(x)^x,x=0,4);
+
− +1 3
1
3ln x( ) x3 O x( 4)
22.3
> limit((2^x+3^x-12)^(tan(Pi/4*x)),x=2);
1
2
1 π
16
3
1 π
36
> simplify(%);
9836602018824134393856
−1 π
mouais...
22.4
> limit((exp(1)-(1+1/x)^x)^(sqrt(x^2+1)-sqrt(x^2-1)),x=infinity );
1
> taylor(exp(1)-(1+1/x)^x,x=infinity,2);
1 + 2
e x
O 1 x2
> taylor(sqrt(x^2+1)-sqrt(x^2-1),x=infinity,2);
1 + x
O 1 x3
Exercice 23
> series(ln(tan(x)),x=0,1);
+ ( )
ln x O x( 2)
> taylor(ln(tan(x)),x=Pi/4,2);
+ 2
− x 1
4π
O
− x 1
4π
3
> series(sqrt(x^2+x)-(x^3+2*x^2)^(1/3),x=0,1);
+
x O x( (2 3/ ))
> asympt(sqrt(x^2+x)-(x^3+2*x^2)^(1/3),x,2);
− +1 +
6 23 72 x
O 1 x2
> series(1/x-1/tan(x),x=0,4);
1 +
3x O x( 2)
> asympt(exp(1/x)-x*(x+1)/x^2,x,3);
1 + 2
1 x2
O 1 x3
> asympt(sqrt(ln(2*n+1))-sqrt(ln(2*n)),n,2);
1 + 4
1 + ( )
ln 2 ln n n( )
O 1 n2
> asympt((ln(n+1)/ln(n))^n-1,n,1);
− + e
1 ( )
ln n 1
O 1 n
> simplify(%);
− + e
1 ( )
ln n 1
O 1 n
Etrange... je n’ai pas réussi à lui faire "voir" que exp(1/ln(n))-1 est équivalent à 1/ln(n). Comme quoi...
Exercice 24
> asympt((ln(x)/ln(x-1))^(x^2),x,1);
+ e
− /1 2−ln x( )−1 ( )
ln x2
O 1 x
e
− x
( ) ln x
Arf ! On fait le calcul à la main... et onpeut tout de même vérifier :
> limit(exp(-x/ln(x))*(ln(x)/ln(x-1))^(x^2),x=infinity);
1
> asympt(exp(sqrt(x^2+x+1)),x,1);
+ e(1 2/ )
O 1 x ex
> expand(%);
+ exe(1 2/ ) ex
O 1 x
O(1/x) est uu terme qui tend vers 0, donc le tout est équivalent à exp(x)exp(1/2)
> series(exp(tan(x)^2),x=Pi/2);
Error, (in series/exp) unable to compute series
> series(tan(x)^2,x=Pi/2);
− + + +
− x 1
2π
-2 2 3
1 15
− x 1
2π
2 2
189
− x 1
2π
4
O
− x 1
2π
5
"donc" exp(tan^2(Pi/2+u)) est equivalent à exp(-2/3)exp(1/u^2)
> asympt(ln(n)/n-ln(n)^2/2+ln(n-1)^2/2,n,3);
+
−1 +
2ln n( ) 1 2 n2
O 1 n3
> asympt((ln(x)/ln(x-1))^(x^2),x,2);
+ e
− /1 2−ln x( )−1 ( )
ln x2
O 1 x
e
− x
( ) ln x
> simplify(%);
+ e
/
1 2ln x( )+1 ( )
ln x2
O 1
x e
x ( ) ln x
A la main :
∞
> asympt(ln(x)/ln(x-1),x,3);
+ + +
1 1
( ) ln x x
1+ 2
1 ( ) ln x ( ) ln x x2
O 1 x3
> asympt(x^2*ln(ln(x)/ln(x-1)),x,1);
+ − +
x ( ) ln x
1+ 2
1 ( ) ln x ( ) ln x
1 2
1 ( ) ln x 2
O 1 x Et donc...
> series(x^x-(sin(x))^x-x^3/6,x=0,5);
1 +
6ln x x( ) 4 O x( 5)
