En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Calcul de limitesFonctions trigonométriques/Exercices/Calcul de limites », n'a pu être restituée correctement ci-dessus.
Calculer
lim
x
→
0
sin
5
x
7
x
{\displaystyle \lim _{x\to 0}{\frac {\sin 5x}{7x}}}
et
lim
x
→
0
sin
5
x
sin
7
x
{\displaystyle \lim _{x\to 0}{\frac {\sin 5x}{\sin 7x}}}
.
Solution
Ces deux limites valent
5
7
{\displaystyle {\frac {5}{7}}}
, puisque
lim
X
→
0
sin
X
X
=
1
{\displaystyle \lim _{X\to 0}{\frac {\sin X}{X}}=1}
.
Calculer
lim
x
→
0
tan
5
x
7
x
{\displaystyle \lim _{x\to 0}{\frac {\tan 5x}{7x}}}
et
lim
x
→
0
sin
5
x
tan
7
x
{\displaystyle \lim _{x\to 0}{\frac {\sin 5x}{\tan 7x}}}
.
Solution
Ces deux limites valent
5
7
{\displaystyle {\frac {5}{7}}}
, puisque
lim
X
→
0
tan
X
X
=
lim
X
→
0
sin
X
X
=
1
{\displaystyle \lim _{X\to 0}{\frac {\tan X}{X}}=\lim _{X\to 0}{\frac {\sin X}{X}}=1}
.
Calculer :
1°
lim
x
→
0
sin
2
x
tan
2
x
{\displaystyle \lim _{x\to 0}{\frac {\sin ^{2}x}{\tan ^{2}x}}}
;
2°
lim
x
→
0
sin
2
x
1
−
cos
x
{\displaystyle \lim _{x\to 0}{\frac {\sin ^{2}x}{1-\cos x}}}
.
Calculer
lim
x
→
0
1
−
cos
x
x
2
{\displaystyle \lim _{x\to 0}{\frac {1-\cos x}{x^{2}}}}
, puis
lim
x
→
0
sin
x
−
cos
x
+
1
x
{\displaystyle \lim _{x\to 0}{\frac {\sin x-\cos x+1}{x}}}
.
Solution
lim
x
→
0
1
−
cos
x
x
=
0
{\displaystyle \lim _{x\to 0}{\frac {1-\cos x}{x}}=0}
, puisque
lim
x
→
0
1
−
cos
x
x
2
=
1
2
{\displaystyle \lim _{x\to 0}{\frac {1-\cos x}{x^{2}}}={\frac {1}{2}}}
. Par conséquent,
lim
x
→
0
sin
x
−
cos
x
+
1
x
=
lim
x
→
0
sin
x
x
=
1
{\displaystyle \lim _{x\to 0}{\frac {\sin x-\cos x+1}{x}}=\lim _{x\to 0}{\frac {\sin x}{x}}=1}
.
Calculer :
1°
lim
x
→
0
+
sin
2
x
1
−
cos
x
{\displaystyle \lim _{x\to 0^{+}}{\frac {\sin 2x}{\sqrt {1-\cos x}}}}
;
2°
lim
x
→
0
tan
x
−
sin
x
x
3
{\displaystyle \lim _{x\to 0}{\frac {\tan x-\sin x}{x^{3}}}}
.
Solution
1°
lim
x
→
0
1
−
cos
x
|
x
|
=
lim
x
→
0
1
−
cos
x
x
2
=
1
2
{\displaystyle \lim _{x\to 0}{\frac {\sqrt {1-\cos x}}{|x|}}={\sqrt {\lim _{x\to 0}{\frac {1-\cos x}{x^{2}}}}}={\frac {1}{\sqrt {2}}}}
donc
lim
x
→
0
+
sin
2
x
1
−
cos
x
=
2
2
{\displaystyle \lim _{x\to 0^{+}}{\frac {\sin 2x}{\sqrt {1-\cos x}}}=2{\sqrt {2}}}
.
2°
lim
x
→
0
tan
x
−
sin
x
x
3
=
lim
x
→
0
tan
x
x
1
−
cos
x
x
2
=
1
2
{\displaystyle \lim _{x\to 0}{\frac {\tan x-\sin x}{x^{3}}}=\lim _{x\to 0}{\frac {\tan x}{x}}{\frac {1-\cos x}{x^{2}}}={\frac {1}{2}}}
.
Calculer :
1°
lim
x
→
π
2
(
sin
x
−
1
)
tan
2
x
{\displaystyle \lim _{x\to {\frac {\pi }{2}}}\left(\sin x-1\right)\tan ^{2}x}
;
2°
lim
x
→
π
(
1
+
cos
x
)
tan
x
2
{\displaystyle \lim _{x\to \pi }\left(1+\cos x\right)\tan {\frac {x}{2}}}
.
Solution
1°
lim
h
→
0
(
sin
(
π
2
+
h
)
−
1
)
tan
2
(
π
2
+
h
)
=
lim
h
→
0
cos
h
−
1
tan
2
h
=
−
1
2
{\displaystyle \lim _{h\to 0}\left(\sin \left({\frac {\pi }{2}}+h\right)-1\right)\tan ^{2}\left({\frac {\pi }{2}}+h\right)=\lim _{h\to 0}{\frac {\cos h-1}{\tan ^{2}h}}=-{\frac {1}{2}}}
.
2°
lim
h
→
0
(
1
+
cos
(
π
+
h
)
)
tan
π
+
h
2
=
lim
h
→
0
1
−
cos
h
−
tan
h
2
=
0
{\displaystyle \lim _{h\to 0}\left(1+\cos \left(\pi +h\right)\right)\tan {\frac {\pi +h}{2}}=\lim _{h\to 0}{\frac {1-\cos h}{-\tan {\frac {h}{2}}}}=0}
.
Calculer :
1°
lim
x
→
a
cos
x
−
cos
a
sin
x
−
sin
a
{\displaystyle \lim _{x\to a}{\frac {\cos x-\cos a}{\sin x-\sin a}}}
;
2°
lim
x
→
a
tan
x
−
tan
a
cos
x
−
cos
a
{\displaystyle \lim _{x\to a}{\frac {\tan x-\tan a}{\cos x-\cos a}}}
;
3°
lim
x
→
π
3
sin
3
x
1
−
2
cos
x
{\displaystyle \lim _{x\to {\frac {\pi }{3}}}{\frac {\sin 3x}{1-2\cos x}}}
.
Solution
1°
cos
′
a
sin
′
a
=
−
tan
a
{\displaystyle {\frac {\cos 'a}{\sin 'a}}=-\tan a}
.
2°
tan
′
a
cos
′
a
=
−
1
cos
2
a
sin
a
{\displaystyle {\frac {\tan 'a}{\cos 'a}}=-{\frac {1}{\cos ^{2}a\sin a}}}
.
3°
3
cos
3
π
3
2
sin
π
3
=
−
3
3
=
−
3
{\displaystyle {\frac {3\cos 3{\frac {\pi }{3}}}{2\sin {\frac {\pi }{3}}}}={\frac {-3}{\sqrt {3}}}=-{\sqrt {3}}}
.