Analisi Matematica
Esercizi sui Limiti Notevoli — Soluzioni
Liceo Scientifico — Classi V — Prof. Roberto Squellati
1. lim
x→π/4
cos 2x
cos x − cos (π/4) = lim
x→π/4
2 cos2x − 1 cos x −√
2/2 = lim
x→π/4
2¡√
2 cos x − 1¢ ¡√
2 cos x + 1¢
√2¡√
2 cos x − 1¢ =
= lim
x→π/4
√2¡√
2 cos x + 1¢
= 2√ 2
2. lim
x→α
sen (x − α)
cos2x − cos2α = lim
t→0
sen t
cos2(t + α) − cos2α = lim
t→0
sen t
[cos (t + α) − cos α] · [cos (t + α) + cos α] =
= lim
t→0
2 sen t 2cos t
2
−2 · sen
µt + 2α 2
¶
· sent
2· [cos (t + α) + cos α]
= lim
t→0
cost 2
− sen µt
2+ α
¶
· [cos (t + α) + cos α]
=
= − 1
2 sen α cos α= − 1 sen 2a
3. 0 ≤ ln¡ 2 + 1x¢
x ≤ ln 3
x ⇒ limx
→+∞
ln¡ 2 + 1x¢
x = 0
4. lim
x→+∞
ln¡
2x2+ 3¢
ln (x3− 1) = lim
x→+∞
ln
∙ x2
µ 2 + 3
x2
¶¸
ln
∙ x3
µ 1 − 1
x3
¶¸ = lim
x→+∞
ln x2+ ln µ
2 + 3 x2
¶
ln x3+ ln µ
1 − 1 x3
¶ =
= lim
x→+∞
2 ln x + ln µ
2 + 3 x2
¶
3 ln x + ln µ
1 − 1 x3
¶ = lim
x→+∞
2 + ln¡
2 + 3/x2¢ ln x 3 + ln¡
1 − 1/x3¢ ln x
= 2 3
5. lim
x→+∞[ln (1 + ex) − x] = limx
→+∞[ln (1 + ex) − x ln e] = limx
→+∞[ln (1 + ex) − ln ex] =
= lim
x→+∞ln
µ1 + ex ex
¶
= lim
x→+∞ln µ
1 + 1 ex
¶
= 0
6. lim
x→+∞
µ2x + 1 2x + 3
¶x−1
= lim
x→+∞
⎛
⎜⎝ 1 + 1
2x 1 + 3
2x
⎞
⎟⎠
x−1
= lim
x→+∞
⎡
⎢⎢
⎢⎢
⎢⎢
⎣
⎛
⎜⎝ 1 + 1
2x 1 + 3
2x
⎞
⎟⎠
x
·
⎛
⎜⎝ 1 + 1
2x 1 + 3
2x
⎞
⎟⎠
−1
| {z }
1
⎤
⎥⎥
⎥⎥
⎥⎥
⎦
=
= lim
x→+∞
µ 1 + 1
2x
¶x
µ 1 + 1
2x/3
¶x = lim
x→+∞
"µ 1 + 1
2x
¶2x#1/2
"µ 1 + 1
2x/3
¶2x/3#3/2 =e1/2 e3/2 = e−1
1
7. lim
x→+∞(x + 1)−1/ ln x= lim
x→+∞exp
∙
−ln (x + 1) ln x
¸
= lim
x→+∞exp
½
−ln [x (1 + 1/x)]
ln x
¾
=
= lim
x→+∞exp
½
−ln x + ln (1 + 1/x) ln x
¾
= lim
x→+∞exp
½
−1 −ln (1 + 1/x) ln x
¾
= e−1
8. lim
x→+∞x1/ ln2x= lim
x→+∞exp µln x
ln2x
¶
= lim
x→+∞exp µ 1
ln x
¶
= 1
9. lim
x→0
(1 + 2x)4− 1
x = lim
x→0
"
2 ·(1 + 2x)4− 1 2x
#
= lim
t→0
"
2 ·(1 + t)4− 1 t
#
= 8
10. lim
x→1+
ex−1− 1
1 − cos (1 − x) = lim
x→1+
ex−1− 1
x − 1 · (x − 1) 1 − cos (1 − x)
(1 − x)2 · (1 − x)2
= lim
x→1+
2 (x − 1) (x − 1)2 = lim
x→1+
2
x − 1 = +∞
11. lim
x→0
1 − cos x ln¡
1 + tg2x¢ = lim
x→0
1 − cos x x2 · x2 ln¡
1 + tg2x¢ tg2x · tg2x
= lim
x→0
x2/2 sen2x cos2x
= lim
x→0
∙cos2x
2 ·³sen x x
´−2¸
= 1 2
12. lim
x→0
3sen x− 1 x = lim
x→0
µesen x ln 3− 1 sen x ln 3 · sen x
x · ln 3
¶
= ln 3
13. lim
x→−1
1 − cos¡ x2− 1¢ ex+1− 1 = lim
x→−1
1 − cos¡ x2− 1¢ (x2− 1)2 ·¡
x2− 1¢2
ex+1− 1
x + 1 · (x + 1)
= lim
x→−1
(x − 1)2(x + 1)2 2 (x + 1) =
= lim
x→+∞
(x − 1)2(x + 1)
2 = 0
14. lim
x→+∞
arctg x −π 2
x − sen x = lim
x→+∞
arctg x − π 2 x ·³
1 −sen x x
´ = 0
15. lim
x→−∞
¡x + 1 +√
3x2− 5x − 1¢
= lim
x→−∞
Ã
x + 1 + |x|
r 3 − 5
x− 1 x2
!
=
= lim
x→−∞
Ã
x + 1 − x r
3 − 5 x− 1
x2
!
= lim
x→−∞x Ã
1 + 1 x−
r 3 − 5
x− 1 x2
!
= +∞
16. lim
x→+∞
µx + 1 x − 1
¶x
= lim
x→+∞
⎛
⎜⎝ 1 + 1
x 1 − 1 x
⎞
⎟⎠
x
= lim
x→+∞
µ 1 + 1
x
¶x
"µ 1 + 1
−x
¶−x#−1 = e2
17. lim
x→π/2
3 sen2x + sen x − 4
cos x = lim
x→π/2
(3 sen x + 4) (sen x − 1)
cos x = lim
x→π/2
− (3 sen x + 4)¡
1 − sen2x¢ (1 + sen x) cos x =
= lim
x→π/2−(3 sen x + 4) cos2x
(1 + sen x) cos x = lim
x→π/2−(3 sen x + 4) cos x 1 + sen x = 0 18. lim
x→π
cos x + cos 2x (x − π)2 = lim
t→0
cos (π + t) + cos (2π + 2t)
t2 = lim
t→0
− cos t + cos 2t t2 = lim
t→0
2 cos2t − cos t − 1
t2 =
= lim
t→0
(2 cos t + 1) (cos t − 1)
t2 = lim
t→0
∙
− (2 cos t + 1) ·1 − cos t t2
¸
= −3 2 2
19. lim
x→∞
µ x x + 1
¶2x+1
= lim
x→∞
µx + 1 x
¶−2x−1
= lim
x→∞
"µ 1 + 1
x
¶−2x
· µ
1 + 1 x
¶−1#
=
= lim
x→∞
(∙µ 1 + 1
x
¶x¸−2
· µ
1 + 1 x
¶−1)
= e−2
20. lim
x→0
(1 − cos x) sen 3x x2sen kx = lim
x→0
"
3x ·sen 3x
3x ·1 − cos x x2 · 1
kx·
µsen kx kx
¶−1#
= lim
x→0
3x 2kx = 3
2k
21. lim
x→0+
sen¡ x2+ x¢
x2 = lim
x→0+
"
sen¡ x2+ x¢
x2+ x ·x2+ x x2
#
= lim
x→0+
µ 1 + 1
x
¶
= +∞
22. lim
x→0
¡1 + x2− x¢√2
− 1
x = lim
x→0
⎡
⎣
¡1 + x2− x¢√2
− 1
x2− x ·x2− x x
⎤
⎦ = lim
x→0
√2 (x − 1) = −√ 2
23. lim
x→∞
µ3x − 4 3x + 2
¶x+13
= lim
x→∞
µ3x + 2 − 6 3x + 2
¶x3+1 3
= lim
x→∞
⎡
⎢⎢
⎣ µ
1 − 6 3x + 2
¶x3
· µ
1 − 6 3x + 2
¶13
| {z }
1
⎤
⎥⎥
⎦ =
= lim
x→∞
µ
1 + 1
−3x+26
¶x3
= lim
t→∞
µ 1 +1
t
¶13 ·−3t+26
= lim
t→∞
µ 1 + 1
t
¶−23t−2 9 =
= lim
t→∞
⎧⎨
⎩
"µ 1 + 1
t
¶t#−2/3
· µ
1 +1 t
¶−2/9⎫
⎬
⎭= e−2/3
24. lim
x→1
ln (7x − 6) ln (3x − 2) = lim
x→1
ln [7 (x − 1) + 1]
ln [3 (x − 1) + 1] = lim
t→0
ln (1 + 7t) ln (1 + 3t) = lim
t→0
∙ln (1 + 7t)
7t · 7t · 3t
ln (1 + 3t)· 1 3t
¸
=
= lim
t→0
7 3
∙ln (1 + 3t) 3t
¸−1
= 7 3
25. lim
x→0
e2x− ex
ln (1 + 2x) = lim
x→0
∙
ex·ex− 1
x · x 2x
ln (1 + 2x)· 1 2x
¸
= lim
x→0
(ex 2
∙ln (1 + 2x) 2x
¸−1)
= 1 2
26. lim
x→4
4x−1− 64
2 (x2− 3x − 4)= lim
x→4
4x−1− 43
2 (x − 4) (x + 1) = lim
x→4
∙ 43
2 (x + 1)·4x−1− 1 x − 4
¸
= lim
t→0
µ 32
t + 5· 4t− 1 t
¶
=
= 32 ln 4 5 = 64
5 ln 2
27. lim
x→1
x3− 3x + 2 x4− 4x + 3 = lim
x→1
(x + 2) (x − 1)2
(x2+ 2x + 3) (x − 1)2 = lim
x→1
x + 2
x2+ 2x + 3 = 1 2 28. lim
x→+∞
√x q
x +p x +√
x
= lim
x→+∞
√x vu
utx Ã
1 +
px +√ x x
! = limx→+∞
√x
√x · s
1 +
rx +√ x x2
=
= lim
x→+∞
s 1 1 +
r1 x+ 1
x√ x
= 1
29. lim
x→0
sen 5x sen 2x = lim
x→0
µ
5x ·sen 5x 5x · 2x
sen 2x· 1 2x
¶
= lim
x→0
"
5 2
µsen 2x 2x
¶−1#
= 5 2
3
30. lim
x→1
sen πx sen 3πx = lim
t→0
sen [π (t + 1)]
sen [3π (t + 1)] = lim
t→0
sen (πt + π)
sen [(3πt + π) + 2π] = lim
t→0
− sen πt
sen (3πt + π) = lim
t→0
− sen πt
− sen 3πt =
= lim
t→0
µsen πt
πt · πt · 3πt sen 3πt · 1
3πt
¶
= lim
t→0
"
1 3
µsen 3πt 3πt
¶−1#
= 1 3 31. lim
x→0
µ x sen1
x
¶
= lim
x→0
∙
x ·sen (1/x) x
¸
= 0
32. lim
x→0+(ln x − ln sen 2x) = lim
x→0+ln x
sen 2x = lim
x→0+ln µ1
2· 2x sen 2x
¶
= lim
x→0+ln
"
1 2
µsen 2x 2x
¶−1#
= − ln 2
33. lim
x→0
µsen 2x x
¶x+1
= lim
x→0
µ
2 ·sen 2x 2x
¶x+1
= 2 34. 0 ≤√
x |sen ln x| ≤√
x ⇒ limx
→0(√
x sen ln x) = 0
35.
¯¯
¯¯1 x
³
2 + senπ x
´¯¯¯¯ ≥
¯¯
¯¯1 x
¯¯
¯¯ ⇒ limx→0
1 x
³
2 + senπ x
´
= ∞
36. lim
x→α
sen x − sen α x − α = lim
t→0
sen (t + α) − sen α
t = lim
t→0
2 cos
µt + 2α 2
¶ sen
µt 2
¶
t =
= lim
t→0
⎡
⎢⎢
⎣ sen
µt 2
¶ µt
2
¶ · cos
µt + 2α 2
¶⎤
⎥⎥
⎦ = cos α
37. lim
x→0
x + sen 3x x − sen 2x = lim
x→0
x µ
1 + sen 3x x
¶
x µ
1 − sen 2x x
¶ = lim
x→0
1 + 3 ·sen 3x 3x 1 − 2 ·sen 2x
2x
= −4
38. lim
x→∞
x + sen x x + cos x = lim
x→∞
x³
1 + sen x x
´
x³
1 +cos x x
´ = lim
x→∞
1 + sen x x 1 +cos x
x
= 1
39. lim
x→+∞
log2x +√3
log x − 4
3 log x − 1 = lim
x→+∞
µ 1 +
√3
log x log2x − 4
log2x
¶ log2x µ
3 − 1 log x
¶ log x
=
= lim
x→+∞
µ 1 + 3
r 1
log5x− 4 log2x
¶ log x 3 − 1
log x
= +∞
4
