Calculo 1 - Limites -

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- C´lculo 1 - Limites a
1. Calcule, se existirem, os seguintes limites:


(a) lim (x − 3);
3

x→1


x4 − 8;
x→2

x3 + 2x + 3
(c) lim
;
x→2
x2 + 5
x2 − 9
(d) lim
;
x→−3 x + 3
3x2 − x
(e) lim
;
1 3x − 1
x→ 3
x3 − 27
(f ) lim
;
x→3 √ − 3
x

x+3− 3
(g) lim
;
x→0
x
(b) lim


 |x|
6
2. Fa¸a o esbo¸o do gr´fico de f (x) =
c
c
a

−4x + 20
entre lim f(x) e f (4)?
x→4

{

3. Seja f a fun¸˜o definida por f (x) =
ca

2x − 1
1

se
se

8t3 − 27
;
4t2 − 9
x→ 2
2x3 − 5x2 − 2x − 3
;
(i) lim 3
x→3 4x − 13x2 + 4x − 3

y2 − 9
(j) lim
;
2 + 7y + 3
y→−3
2y
h
√ ;
(k) lim √
h→5
5
√5+h− √
3 + 3h − 3
(l) lim
;
h→0
h
x4 − 16
(m) lim
;
x→2 x − 2
x−1
(n) lim 2
.
x→1 x − 1
(h) lim
3

se
se
se

x4

x ̸= 2x=2

(a) Encontre lim f (x) e verifique que lim f (x) ̸= f (2).
x→2

x→2

(b) Fa¸a um esbo¸o do gr´fico de f .
c
c
a
{ 2
x − 9 se
4. Seja f a fun¸˜o definida por f (x) =
ca
4
se

x ̸= −3
x = −3

(a) Encontre lim f (x) e verifique que lim f (x) ̸= f (3)
x→−3

x→−3

(b) Fa¸a um esbo¸o do gr´fico de f .
c
c
a
f (x + h) − f (x)
quando
h
2
b) f (x) = x
c) f (x) = x3 .

5.Determine o valor de lim

h→0

a) f (x) = x

6. Nos ´
ıtens a seguir, calcule os limites laterais pedidos e verifique se o limite (bilateral) existe. Caso exista dˆ seu valor.
e
(a) f (x) =

|x|
x ,

lim f (x), lim− f (x), lim f (x).
x→0
x→0+
x→0

se x < 1
 2
−1 se x = 1 ; lim f (x), lim f (x), lim f (x)
(b) f (x) =
+
x→1

x→1−
−3 se x > 1 x→1

 2r + 3 se r < 1
2se r = 1 ; lim f (r), lim f (r), lim f (r)
(c) f (r) =
+
r→1

r→1−
7 − 2r se r > 1 r→1

se x < −2
 2 + x2
0
se x = −2 ; lim f (x), lim f (x), lim f (x)
(d) g(x) =
+
x→−2

x→−2−
11 − x2 se x > −2 x→−2
7. Dada f (x) =
8. Dada f (x) =
a) lim f (x)
x→−1

|x|+x
x .

Existe lim f (x)?
x→0

|x2 +x|
.
x

Verifique se existem os limites abaixo e, caso existam,determine seus valores:
b) lim f (x).
x→0

- Gabarito 1. Calcule, se existirem, os seguintes limites:


(a) lim (x − 3) = −2;
3

x→1



x4 − 8 = 2 2;
x→2


x3 + 2x + 3
5
(c) lim
=
;
x→2
x2 + 5
3
x2 − 9
(d) lim
= −6;
x→−3 x + 3
3x2 − x
1
(e) lim
= ;
1 3x − 1
3
x→ 3
x3 − 27
(f ) lim
= 27;
x→3 √ − 3
x


x+3− 3
3
(g) lim
=
;
x→0
x
6
(b) lim


|x|
6
2. f (x) =

−4x + 20
{
2x − 1 se
3. f (x) =
1
se
{ 2
x − 9 se
4. f (x) =
4
se

se
se
se

x4

x ̸= 2
x=2
x ̸= −3
x = −3

lim f (x) = 4 ̸= f (4) = 6

x→4

lim f (x) = 3 ̸= f (2) = 1.

x→2

lim f (x) = 0 ̸= f (−3) = 4.

x→−3

(a) Figura ex.2

5. a) 1


8t3 − 27
9
(h) lim
=
;
4t2 − 9
2
x→ 3
2
2x3 − 5x2 − 2x − 3
11
=
;
(i) lim 3
x→3 4x −13x2 + 4x − 3

√ 17
y2 − 9
6
(j) lim
=
;
2 + 7y + 3
y→−3
2y
5


h
√ = 10 + 5;
(k) lim √
h→5
5

√5+h− √
3 + 3h − 3
3
(l) lim
=
;
h→0
h
2
x4 − 16
(m) lim
= 32;
x→2 x − 2
x−1
1
(n) lim 2
= .
x→1 x − 1
2

(b) Figura ex.3

c) 3x2 .

b) 2x

6. (a) lim f (x) = 1, lim f (x) = −1,
+


x→0

(b) lim+ f (x) = −3, lim− f (x) = 2,

x→1

x→0

x→0x→1

x→1

lim f (x).
lim f (x)

(c) lim+ f (r) = lim− f (r) = 5, lim f (r) = 5
r→1

(d)
7.

r→1

r→1

lim f (x) = 5,

x→−2+

lim f (x) = 6,

x→−2−

lim f (x)

x→−2

lim f (x), pois lim f (x) = 2 e lim f (x) = 0.

x→0

8. a) lim f (x) = 0
x→−1

x→0−

x→0+

b) lim+ f (x) = 1, lim− f (x) = −1,
x→0

x→0

lim f (x).

x→0

(c) Figura ex.4

- C´lculo 1- Limites - Lista 2
a
1. Determine, caso existam, os seguintes limites:
a) lim+ (3 −



x)

b) lim+

x→0

x→2

1
e) lim− √
x→2
2−x

x−3
i) lim √
x→9
x2 − 9x

x→−2
1
y

j) lim

1
2−x
−1
5

n) lim (−2x − x − 12x + 1)
6

x→−∞

2

2x2 + x − 21
x→+∞ x3 − 2x2 + 9
x4 − 2 4
v) lim
x→+∞ 2 − x

x2 + 4
z) lim
x→+∞ x + 4
2x2 − 5x + 2
δ) lim 2
x→2 5x√...
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