1. lim - Sweb.cz

1.
lim (x2 − 1) = 22 − 1 = 3
x→2
2.
2·2−3
1
2x − 3
= 2
=
2
x→2 x + 1
2 +1
5
lim
3.
√
lim sin x =
x→ Π
3
4.
3
2
x(1 + 2x)
x + 2x2
= lim
= lim (1 + 2x) = 1 + 2 · 0 = 1
x→0
x→0
x→0
x
x
lim
5.
x2 + 3x + 2
−1 + 2
1
(x + 1)(x + 2)
x+2
= lim
= lim 2
=
=
x→−1
x→−1 (x + 1)(x2 − x + 1)
x→−1 x − x + 1
x3 + 1
(−1)2 − (−1) + 1
3
lim
6.
x2 − 9
(x + 3)(x − 3)
= lim
= lim (x − 3) = −3 − 3 = −6
x→−3 x + 3
x→−3
x→−3
x+3
lim
7.
x3 − 1
(x − 1)(x2 + x + 1)
= lim
= lim (x2 + x + 1) = 12 + 1 + 1 = 3
x→1 x − 1
x→1
x→1
x−1
lim
8.
(x − 3)(x − 1)
x2 − 4x + 3
= lim
= lim (x − 1) = 3 − 1 = 2
x→3
x→3
x→3
x−3
x−3
lim
9.
x2 + 5x + 6
1
1
(x + 2)(x + 3)
x+3
= lim
= lim
=
=−
x→−2 x2 + x − 2
x→−2 (x + 2)(x − 1)
x→−2 x − 1
−3
3
lim
10.
x3 − 5x2 + 8x − 4
x→2
x3 − 3x2 + 4
Protože (x3 − 5x2 + 8x − 4) : (x − 2) = x2 − 3x + 2
a (x3 − 3x2 + 4) : (x − 2) = x2 − x − 2, máme
lim
(x − 2)(x2 − 3x + 2)
1
x2 − 3x + 2
(x − 2)(x − 1)
x−1
= lim 2
= lim
= lim
=
2
x→2 (x − 2)(x − x − 2)
x→2 x − x − 2
x→2 (x − 2)(x + 1)
x→2 x + 1
3
lim
11.
1 3 sin 3x
3
sin 3x
3
3
sin 3x
= lim ·
= lim
= ·1=
x→0 5
x→0 5x
3x
5 x→0 3x
5
5
lim
12.
2 cos2 x sin x
sin x
= lim 2 · lim cos2 x · lim
= 2 · cos2 0 · 1 = 2 · 1 · 1 = 2
x→0
x→0
x→0
x→0 x
x
lim
13.
1 − cos x
1 − cos x 1 + cos x
1 − cos2 x
sin2 x
1
= lim
·
= lim 2
= lim
·
2
2
x→0
x→0
x
x
1 + cos x x→0 x (1 + cos x) x→0 x2 1 + cos x
2
1
1
1
sin x
· lim
= 12 · =
= lim
x→0 1 + cos x
x→0 x
2
2
lim
1
14.
1 − cos 2x
1 − cos2 x + sin2 x
2 sin2 x
2 sin x
= lim
= lim
= lim
= 2·1 = 2
x→0
x→0
x→0 x sin x
x→0
x sin x
x sin x
x
lim
15.
lim
x→0
3
x
−
x2 + 1 sin x
= lim
x→0
3
x
3
1
1
− lim
= 2
− lim
= 3− = 2
x2 + 1 x→0 sin x
0 + 1 x→0 sinx x
1
16.
2 tan2 x
2 sin2 x
2
= lim
· lim
= lim 2
2
x→0 3 cos2 x x→0
x→0
x→0 3x cos2 x
3x
lim
sin x
x
2
=
2
2
2
·12 =
=
3 cos2 0
3·1
3
17.
lim
x→0
sin 3x
3 sin 3x x
sin 3x
= lim
·
= 3· lim
· lim
x→0
x→0
tan x
3x
tan x
3x x→0
1
tan x
x
1
= 3·1· lim cos x· sin x = 3·1 = 3
x→0
x
18.
√
√
x+2+1
x+1
x+1
(x + 1)( x + 2 + 1)
= lim √
·√
= lim
x→−1
x→−1
x+2−1
x + 2 − 1 x→−1
x+2−1
x+2+1
√
√
√
(x + 1)( x + 2 + 1)
= lim
= lim ( x + 2 + 1) = −1 + 2 + 1 = 2
x→−1
x→−1
x+1
lim √
19.
√
√
√
2− x−3 2+ x−3
2− x−3
4 − (x − 3)
√
√
= lim
·
lim
= lim 2
x→7
x→7
x→7
x2 − 49
x2 − 49
2+ x−3
(x − 49)(2 + x − 3)
= lim
x→7
7−x
−(x − 7)
√
√
= lim
(x2 − 49)(2 + x − 3) x→7 (x − 7)(x + 7)(2 + x − 3)
(−1)
−1
−1
1
√
√
=
=
=−
x→7 (x + 7)(2 +
14 · 4
56
x − 3)
14 · (2 + 7 − 3)
= lim
20.
√ !
√
x+2+ 2
sin 2x
sin 2x
√ = lim √
√ ·√
√
lim √
x→0
x + 2 − 2 x→0
x+2− 2
x+2+ 2
√
√
√
√
sin 2x · ( x + 2 + 2)
2 · sin 2x · ( x + 2 + 2)
= lim
= lim
x→0
x→0
x+2−2
2x
√
√
√
√
√
sin 2x
= 2 · lim
· lim ( x + 2 + 2) = 2 · 1 · ( 2 + 2) = 4 2
x→0 2x
x→0
21.
lim (5x3 −x2 +2x+1) = lim 5x3 (1−
x→+∞
x→+∞
x2
2x
1
1
2
1
+
+
) = lim 5x3 · lim (1− + 2 + 3 )
x→+∞
x→+∞
5x3 5x3 5x3
5x 5x x
= lim 5x3 · 1 = lim 5x3 = +∞
x→+∞
x→+∞
22.
lim (5x3 −x2 +2x+1) = lim 5x3 (1−
x→−∞
x→−∞
x2
2x
1
1
2
1
+
+
) = lim 5x3 · lim (1− + 2 + 3 )
x→−∞
x→−∞
5x3 5x3 5x3
5x 5x x
= lim 5x3 · 1 = lim 5x3 = −∞
x→−∞
x→−∞
2
23.
lim (−5x3 −x2 +2x+1) = lim −5x3 · lim (1+
1
1
2
− 2 − 3 ) = lim −5x3 ·1 = −∞
x→+∞
5x 5x x
lim (−5x3 −x2 +2x+1) = lim −5x3 · lim (1+
1
2
1
−
− ) = lim −5x3 ·1 = +∞
x→−∞
5x 5x2 x3
x→+∞
x→+∞
x→+∞
24.
x→−∞
x→−∞
x→−∞
25.
lim (2x4 −3x3 +x+1) = lim 2x4 (1−
1
1
3
+
+
) = lim 2x4 = +∞
x→+∞
2x 2x3 2x4
lim (2x4 −3x3 +x+1) = lim 2x4 (1−
1
1
3
+ 3 + 4 ) = lim 2x4 = +∞
x→−∞
2x 2x 2x
x→+∞
x→+∞
26.
x→−∞
x→−∞
27.
lim (−2x4 −3x3 +x+1) = lim −2x4 (1+
3
1
1
−
−
) = lim −2x4 = −∞
x→+∞
2x 2x3 2x4
lim (−2x4 −3x3 +x+1) = lim −2x4 (1+
1
1
3
−
−
) = lim −2x4 = −∞
x→−∞
2x 2x3 2x4
x→+∞
x→+∞
28.
x→−∞
29.
x→−∞
3
1
x2 · ( x3 − x12 )
0
3x − 1
x − x2
=
lim
= =0
=
lim
x→+∞ 1 + 1
x→+∞ x2 · (1 + 1 )
x→+∞ x2 + 1
1
x
x
lim
30.
x2 (5 −
5x2 − 4x + 2
=
lim
x→+∞ 3x2 + x − 3
x→+∞ x2 (3 +
lim
4
x
1
x
+
−
2
x2 )
3
x2 )
5−
x→+∞ 3 +
= lim
4
x
1
x
+
−
2
x2
3
x2
=
5
3
31.
x2 (x2 − x5 )
x2 − x5
x4 − 5x
= lim
=
lim
= lim x2 = +∞
3
1
x→−∞
− 3x + 1 x→−∞ x2 (1 − x + x2 ) x→−∞ 1 − x3 + x12
lim
x→−∞ x2
32.
√
lim
x→+∞
√
x2
x
x2
q
1−
1
x2
q
+
1+
−1+
+1
= lim
x→+∞
x+3
x(1 + x3 )
q
q
√
√
1 − x12 + 1 + x12
1+ 1
=2
= lim
=
x→+∞
1
1 + x3
1
x2
33.
lim x(
p
x→+∞
x2
√
p
x · (x2 + 1 − x2 )
x2 + 1 + x
2
√
+ 1−x) = lim x( x + 1−x)· √
= lim
x→+∞
x2 + 1 + x x→+∞
x2 + 1 + x
x→+∞
x
q
= lim
x·
1+
1
x2
1
= lim q
+1
x→+∞
3
1+
1
x2
=√
+1
1
1
=
2
1+0+1
34.
tan x
z
√
√
[tan x = z, x → 0 ⇔ z → 0] = lim
z→0
1 − 1 + tan x
1− 1+z
√
√
√
√
z
1+ 1+z
z(1 + 1 + z)
√
√
= lim
·
= lim
= − lim (1+ 1 + z) = −(1+ 1 + 0) = −2
z→0 1 −
z→0
z→0
1
−
(1
+
z)
1+z 1+ 1+z
lim
x→0
35.
−x
lim 2
x→−∞
x −∞
1
1
= lim
=
= 2∞ = ∞
x→−∞ 2
2
36.
lim 3−x =
x→+∞
37.
2x 1 −
2x − 1
lim x
= lim x
x→+∞ 2 + 1
x→+∞ 2
1+
38.
1
=0
3∞
1
2x 1
2x
3x+1
1 − 3x
lim
=
lim
x→+∞ 2 + 3x+1
x→+∞ 3x+1
1−
x→+∞ 1 +
= lim
1
3x+1
2
3x+1
1
2x
1
2x
1−0
=1
1+0
=
− 13
0 − 13
1
=
=−
0
+
1
3
+1
39.
lim
x→+∞
x
a· xa
a x
1
1
=
= lim
1+
= lim
1+ a
1+ a
x→+∞
x→+∞
x
x
x
lim
1+
x→+∞
1
xa !a
a
x
40.
lim
x→+∞
x
1+x
= lim
1−
x→+∞
x
= lim
1−
x→+∞
1
x+1
x+1
1
x+1
x
· lim
1−
x→+∞
= lim
x→+∞
1
x+1
1−
1
x+1
x+1−1
−1
= e−1 · (1 − 0)−1 =
1
e
41.
lim
x→+∞
x
x+5
lim
1−
x→+∞
2x−1
= lim
x→+∞
5
x+5
1−
x+5 !2
· lim
x→+∞
5
x+5
1−
2x−1
= lim
x→+∞
5
x+5
1−
5
x+5
2(x+5)−11
=
−11
= (e−5 )2 ·(1−0)−11 =
1
e10
42.
lim
x→+∞
2x + 4
2x + 5
= lim
1−
x→+∞
x+3
= lim
1−
x→+∞
1
2x + 5
2x+5
2
1
2x + 5
· lim
1−
x→+∞
43.
1
3
lim (1 + 3x) x = lim (1 + 3x) 3x =
x→0
x→0
4
x+3
= lim
1−
x→+∞
1
2x + 5
12
2x+6
2
1
1
1
= (e−1 ) 2 · 1 2 = √
e
1
lim (1 + 3x) 3x
x→0
1
2x + 5
3
= e3
= ea
44.
x2 + 2x + 1
x2 + 2x + 1
x2 + 2x + 1
1
4
1
= lim
= lim
· lim
= − · lim
=?
2
x→1 x − 4x + 3
x→1 (x − 3)(x − 1)
x→1
x→1 x − 1
x−3
2 x→1 x − 1
lim
x2 + 2x + 1
1
= −2 · lim
= −2 · (+∞) = −∞
2
x→1+ x − 4x + 3
x→1+ x − 1
lim
x2 + 2x + 1
1
= −2 · lim
= −2 · (−∞) = +∞
x→1− x2 − 4x + 3
x→1− x − 1
Limita neexistuje.
lim
45.
x2 + 2x + 1
x2 + 2x + 1
1
1
= lim
= lim (x2 +2x+1)· lim
= 9· lim
=?
2
x→2 x − 4x + 4
x→2 (x − 2)2
x→2
x→2 (x − 2)2
x→2 (x − 2)2
lim
1
=∞
(x − 2)2
lim
x→2+
1
=∞
x→2− (x − 2)2
lim
x2 + 2x + 1
=9·∞=∞
x→2 x2 − 4x + 4
lim
46.
x
x
1
1
= lim
· lim
= 1 · lim
2
x→0
x→0
x→0
sin
x
sin
x
sin
x
sin x
1
lim
=∞
x→0+ sin x
1
lim
= −∞
x→0− sin x
Limita neexistuje.
lim
x→0
47.
x
x
= lim
=1
x→0+ x
x→0+ |x|
x
x
lim
= lim
= −1
x→0− |x|
x→0+ −x
lim
48.
1
lim 2 x = 2+∞ = ∞
x→0+
1
lim 2 x = 2−∞ = 0
x→0−
49.
3
1
lim
x→0+
2x + 3
1
x
3 +2
= lim
x→0+
1
x
1
2x
1
3x
3
1
x
+
1+
x→0−
2x + 3
1
x
3 +2
2x
1
3x
= lim
2
1
3x
1
lim
1
3
=
x→0+
1
3x
+
1+
3
1
3x
2
1
3x
0+3
3
2−∞ + 3
=
=
−∞
3
+2
0+2
2
5
= lim
x→0+
2
3
x1
1+
3
+
1
2
3x
1
3x
=
0+0
=0
1+0