3. Limita funkce
[Pages:5]3. Limita funkce
1. Spoctte:
lim
x0
sin
5x - sin sin x
3x
,
lim
x0
1
- cos x2
x
,
lim
x0
1 1
+ -
sin sin
x x
- -
cos cos
x x
.
2. Spoctte:
lim
x3
x + 13 - 2 x2 - 9
x + 1,
lim n 1 + x - 1 (n N),
x0
x
lim 1 + x - 1 - x ,
x0 3 1 + x -
31 - x
lim
x0
1 + tg x - x3
1 + sin x .
3. Spoctte n?sleduj?c? limity:
lim x 3 x3 + 7x - x ,
x
lim
x0
log cos x2
x
,
lim (ex - 1) (t3g xx)22 .
x0+
4. Spoctte n?sleduj?c? limity:
lim
x+2
x2
,
x 2x - 1
lim (tg x)tg 2x,
x
4
1
lim
x0
1 + x 2x 1 + x 3x
x2
,
lim log(1 + 2x) log
x
1
+
3 x
.
5. Spoctte limity n?sleduj?c?ch funkc?
lim log(1 + sin x) , x0 2x + 1 - x + 1
lim
x0
esin 2x
- tg
earcsin x
x
,
lim
x0
ex
-
2
sin(
6
tg x
+
x) .
6. Spoctte limity n?sleduj?c?ch funkc? a posloupnost?
lim 1 +
1
n
,
n
n4 + 2n3 - n4 + 1
lim n( n 2 - 1),
n
lim
x0+
arcsin log(1 +
xx)
.
7. Spoctte:
lim (1 + sin x)cotg x,
x1
lim x
x
4
-
arctg
x
x +
1
,
lim
x2
.
x0 1 + x sin x - cos x
8. Spoctte limitu posloupnosti:
lim sin 2 n2 + 1 .
n
V?sledky a nkter? esen?
1. 2, 1/2, -1
2. -1/16, 3/2, 1/n, 1/4
3. 7/3, -1/2, 1
4. 0, 1/e, 2/3, log 8
5.1. Plat?
lim log(1 + sin x)
x0 2x + 1 - x + 1
=
lim
x0
log(1 + sin sin x
x)
?
sin
x
?
2x
+
1 1-
x
+
1
?
2x 2x
+ +
1 1
+ +
x x
+ +
1 1
=
lim
x0
log(1 + sin sin x
x)
?
sin x x
?
2x + 1 + x + 1
.
V?me, ze
(i)
limy0
log(1+y) y
= 1,
(ii)
limx0
sin x x
= 1,
(iii) limx0 sin x = 0,
(iv) (v)
sinjejespproojistt??
na ve
-/2, /2 , sv?m definicn?m
oboru.
Z
(i),
(iii),
(iv)
a
vty
o
limit
slozen?
funkce
plyne
lim
x0
log(1 + sin x) sin x
=
1.
Posledn? rovnost, spolu s (ii), (v) a vtou o limit soucinu d?v?
lim
x0
log(1 + sin
sin x
x)
sin x
x
2x + 1 + x + 1
= 1 ? 1 ? 2 = 2.
5.2. Pisme
lim
x0
esin
2x
- tg
earcsin x
x
=
lim
x0
esin
2x
- earcsin x
x
?
x tg x
=
lim
x0
esin 2x
- earcsin x x
? lim
x0
x tg x
.
V?me,
ze
limx0
x tg x
=
1.
Zab?vejme
se
te
prvn?
limitou
ve v?se uveden?m
soucinu
limit.
lim
x0
esin 2x
- earcsin x x
=
lim
x0
esin 2x - 1 sin 2x
?
sin 2x x
-
earcsin x - 1 arcsin x
?
arcsin x x
=
1.
Pouzili jsme
(1)
limx0
arcsin x x
= 1,
(2)
limx0
sin x x
= 1,
(3)
limy0
ey -1 y
= 1,
(4) sin je prost? funkce na jist?m okol? 0,
(5) arcsin je prost? funkce,
(6) x 2x je prost? funkce,
(7) vtu o limit slozen? funkce ve verzi s podm?nkou (P1),
(8) vtu o aritmetice limit.
Dohromady tedy m?me
lim
x0
esin 2x - earcsin x tg x
= 1.
5.3. Upravme nejprve v?raz jehoz limitu poc?t?me
ex
- 2 sin(/6 tg x
+ x)
=
x tg x
ex - x
1
+
1
-
2
sin(/6 x
+
x)
=
x tg x
ex - x
1
+
1
- cos x
x
-
3 sin x x
=
x tg x
ex - x
1
+
1
- cos x
x
?
1 1
+ +
cos cos
x x
-
3 sin x x
=
x tg x
ex - x
1
+
sin x x
?
sin x
?
1
1 + cos
x
-
3 sin x x
.
()
V?me, ze
(1)
limx0
tg x x
= 1,
(2)
limx0
sin x x
= 1,
(3)
limx0
ex -1 x
= 1,
(4) limx0 cos x = 1.
Z (), (1)?(4) a z vty o aritmetice limit vypl?v?
lim
x0
ex
-
2 sin(/6 + tg x
x)
=
1 - 3.
6.1. Pokusme se nejprve spoc?tat limitu funkce
lim 1 +
1
x
.
x
x4 + 2x3 - x4 + 1
Upravme nejprve v?raz jehoz limitu poc?t?me:
1+
1
x
x4 + 2x3 - x4 + 1
= exp log
1+
1
x
x4 + 2x3 - x4 + 1
= exp x log 1 +
1
x4 + 2x3 - x4 + 1
log = exp
1 + x4+2x31-x4+1
x4 +2x31-x4 +1
?
x
?
x4
+
1 2x3 -
x4
+
1
log
= exp
1 + x4+2x31-x4+1
x4 +2x31-x4 +1
?x?
x4 + 2x3 + x4 + 1
2x3 - 1
log = exp
1 + x4+2x31-x4+1
x4 +2x31-x4 +1
?
1
+
2 x
2
+ -
1 x3
1
+
1 x4
.
()
D?le plat?:
(1) (2)
limz0
log(1+z) z
=
1,
limx
1
x4 +2x3 -x4 +1
=
limx
x4 +2x3 +x4 +1 2x3 -1
= limx
1+ 2 +
x
1+
1 x4
2x-
1 x2
= 2 = 0,
(3) funkce x x4+2x31-x4+1 je na jist?m okol? rzn? od nuly,
(4) exp je spojit? na R.
Z (1)?(3) a z vty o limit slozen? funkce
log
lim
x
1 + x4+2x31-x4+1
x4 +2x31-x4 +1
= 1.
()
D?le m?me
lim
x
1+ 2 +
x
1
+
1 x4
2
-
1 x3
= 1.
Z (), (), ( ) a (4) plyne
( )
lim 1 +
1
x
= e1 = e
x
x4 + 2x3 - x4 + 1
a tedy podle Heineho vty
lim 1 +
1
n
= e.
n
n4 + 2n3 - n4 + 1
6.2. M?sto limity posloupnosti {n( n 2 - 1)} n=1 poc?tejme limitu funkce f (x) =
x(2
1 x
-
1)
v
.
Pokud totiz uk?zeme, ze limx f (x) = A, pak podle Heineho
vty tak? limn f (n) = A. Plat?
lim
x(2
1 x
x
- 1)
=
lim
x
21 x
-1
1
=
lim
x
e log 2 x
-1
log 2
? log 2
=
log 2.
x
x
Uzili jsme
(1)
limy0
ey -1 y
= 1,
(2)
limx
log 2 x
= 0,
(3) log 2 = 0 pro kazd? x > 0,
x
(4) vtu o limit slozen? funkce ve verzi s podm?nkou (P1).
6.3. Uvdomme si, ze plat?:
(1)
limy0
arcsin y y
= 1,
(2)
limy0
log(1+y)
y
= 1,
(3) (4)
limjxep0r+ost?xn=a
0, sv?m
definicn?m
oboru.
Mzeme ps?t
lim
x0+
arcsin log(1 +
xx)
=
lim
x0+
arcsinx
x
?
log(1
x +
x)
.
Z vty o limit slozen? funkce, (1), (3) a (4) plyne
lim
x0+
arcsinx
x = 1.
Z vty o limit slozen? funkce, (2), (3) a (4) plyne
lim log(1+ x) = 1.
x0+
x
Vta o aritmetice limit pak d?v?
lim
x0+
arcsinx
x
?
log(1
x +
x)
=
1.
7. 1/e, 1/2, 4/3 8. 0
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