APS 3 - Limites e Derivadas - Resolução

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FUNDAÇÃO EDUCACIONAL MONSENHOR MESSIAS 
UNIFEMM – CENTRO UNIVERSITÁRIO DE SETE LAGOAS 
Unidade Acadêmica de Ensino de Ciências Gerenciais – UEGE 
Curso de Engenharias 
 
 
 
 
 
CAROLINE OLIVEIRA MOURA 027776 
CAROLINE RODRIGUES FONSECA- 033108 
FELIPE DA SILVA GOMES 034762 
MARIANA MARTINS DE CARVALHO – 029960 
SAMUEL PABLO SOUZA RODRIGUES-031107 
 
 
 
 
 
 
 
APS 3 - Limites e derivadas 
 
 
 
 
 
SETE LAGOAS 
2020 
 
 
1. Calcule os seguintes limites: 
 
(a) lim
𝑥→0
(𝑥2 − 7𝑥4 + 9)= (0) 5 − 7 (0) 4 + 9 = 0 − 0 + 9 = 9 
 
(b) lim
𝑥→2 
(−3𝑥2 + 2𝑥 + 10) = −3(2)2 + 2(2) + 10 = −3 . 4 + 4 + 10 = −12 + 14 =
2 
 
(c) lim
𝑥→−5
(
𝑥2−1
𝑥²+1
)= 
(−5)2−1
(−5)²+1
 = 1 
 
(d) lim
𝑥→2
(
𝑥2−4
𝑥²−2
) = 
(2)2−4
(2)2−2
= 
0
2
= 0 
 
(e) lim
𝑥→2
(
𝑥2−4
𝑥−2
) = 
(2)2−4
2−2
 = 0 
 
(f) lim
𝑥→−2
(𝑥 − 𝑥3 + 2 . |𝑥|) = (−2) − (−2)3 = −2 + 8 = 6 
 
(g) lim
𝑥→2 
(𝑥2 + 3𝑥 − 2)² = (2)2 + 3(2) − 2)² = (4 + 6 − 2)2 = 64 
 
(h) lim
𝑥→2 
 
𝑥3−2𝑥+1
𝑥²−1
= 
(2)3−2.2+1
(2)²−1
= 
8−4+1
3
= 
5
3
 
 
(i) lim
𝑥→1 
𝑥3−2𝑥+1
𝑥²−1
= 
(1)3−2.1+1
(1)²−1
= 
0
0
 
 
(j) lim
𝑥→9 
9−𝑡
3− √𝑡
= 
9−9
3− √9
= 
0
3− 3
= 0 
 
(k) lim
𝑥→9 
√6−𝑥 − 2
√3−𝑥 − 1
=
√6−(9) − 2
√3−(9) − 1
= 
√3 − 2
√3 − 1
= −2 
 
(l) lim
𝑥→3 
√𝑥 − √3 
𝑥−3
= 
√3 − √3 
3−3
= 
0
0
 
lim
𝑥→3 
√𝑥 − √3 . √𝑥 + √3 
(𝑥 − 3). √𝑥 + √3 
= 
1
(√𝑥 + √3 )
=
1
2√3 
 
 
(m) lim
𝑥→1 
√𝑥 – 1
√2𝑥+3 – √5 
= 
√1 – 1
√2.1+3 – √5 
= 
0
0
 
 lim
𝑥→1 
√𝑥 – 1
√2𝑥+3 – √5 
.
√𝑥 + 1
√2𝑥+3 + √5 
.
√2𝑥+3 + √5 
√𝑥 + 1
= 
 lim
𝑥→1 
(𝑥−1)
(2𝑥+3−5)
.
√2𝑥+3 + √5
√𝑥 + 1
= 
 lim
𝑥→1 
(𝑥−1).(𝑥+1)
2.(𝑥−1)
.
(√2𝑥+3 + √5) 
(√𝑥 + 1)
= 
lim
𝑥→1 
 
1 
2
.
(√2𝑥 + 3 + √5) 
(√𝑥 + 1)
=
1 
2
.
(√2𝑥 + 3 + √5) 
(√1 + 1)
= 
1 
2
.
2. √5 
2
= 
√5 
2
 
(n) lim
𝑥→2 
√𝑥
3
− √2
3
𝑥−2
= 
0
0
 
lim
𝑥→2 
(√𝑥
3
− √2
3
)
(√𝑥
3
− √2
3
). (√𝑥²
3
+ √2
3
𝑥 + √4
3
)
= 
1
(√𝑥²
3
+ √2
3
𝑥 + √4
3
)
 
lim
𝑥→2 
1
( √4
3
+ √4
3
+ √4
3
)
= 
1
(3. √4
3
)
=
√2
3
(3. √4
3
. √2
3
)
=
√2
3
3. √8
3 = 
√2
3
6
 
 
(o) lim
𝑥→2 
(
𝑥2+𝑥−6
𝑥−2
) =
(2)2+2−6
2−2
=
0
0
 
 
(p) lim
𝑥→2 
(
𝑥2+5𝑥+4
𝑥2+3𝑥−4
) =
(2)2+5.2+4
(2)2+3.2−4
= 
4+10+4
4+6−4
=
18
6
 ÷ 3 = 3 
 
(q) lim
𝑥→2 
(
𝑥2−4𝑥
𝑥2−3𝑥−4
) = (
(2)2−4.2
(2)2−3.2−4
) =
4−8
4−2
= −2 
 
 
 
2. Limites no infinito. 
 
 
(a) lim 3x + 2 = lim 3(+∞ )+2 = +∞ 
 x→+∞ x→+∞ 
 
 
(b) lim 3x + 2 = lim 3(+∞ )+2 = +∞ 
 x→+∞ x→+∞ 
 
 -∞ 0 0 
 
(c) lim
x→−∞
2𝑥3−3𝑥+5
4𝑥5−2
 = lim
x→−∞
 
𝑥3
𝑥2
+
3𝑥
𝑥2
−
1
𝑥2
 = lim 
x→−∞ 
 
𝑥
+
3𝑥
𝑥
−
1
𝑥2
 = 
−∞
2
 = −∞ 
 
2𝑥2
𝑥2
+
𝑥
𝑥2
+
1
𝑥2
 2 + 
1
𝑥
+
1
𝑥2
 
 0 0 
 ∞ ∞ 
(d) lim
x→+∞
x − √𝑥 + 3 = lim
x→+∞
 𝑥(1 −
√𝑥+3
𝑥
) = lim
x→+∞
𝑥(1 −
√𝑥+3
𝑥
) = +∞ 
 
 ∞ -∞ ∞ 
(e) lim
x→+∞
3𝑥3 − 5𝑥2 + 𝑥 − 7 = lim
x→+∞
3𝑥3 − 5𝑥2 + 𝑥 − 7 = ∞ − ∞ + ∞ − 7 =∞ 
 
 
 0 0 0 
 0 0 
0 
0 
 0 0 
(f) lim
x→+∞
𝑥5+𝑥4+1
2𝑥5+x+1
= lim
x→+∞
 
𝑥5
𝑥5
+
𝑥4
𝑥5
+
1
𝑥5
 = lim
x→+∞
 1 +
1
𝑥
+
1
𝑥5 
 = 
1+0+0
2+0+0
 = 
1
2
 
 
2𝑥5
𝑥5
+
𝑥
𝑥5
+
1
𝑥5
 2 +
1
𝑥4
+
1
𝑥5
 
 0 0 
 
 
(g) lim
x→+∞
√𝑥2 + 1 − 𝑥 = lim
x→+∞
√∞2 + ∞ − ∞ = √∞ = ∞ 
 
 
(h) lim
x→+∞
sen(x) = é 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑡𝑒 𝑞𝑢𝑎𝑛𝑑𝑜 𝑥 𝑡𝑒𝑛𝑑𝑒 𝑎 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑜 
 
 
(i) lim
x→+∞
7𝑥7−5𝑥4+3𝑥2−x+8
4𝑥7−3𝑥3+5𝑥−1
= lim
x→+∞
 
7𝑥7
𝑥7
−
5𝑥4
𝑥7
+
3𝑥2
𝑥7
−
𝑥
𝑥7
+
8
𝑥7
 = 
 0 0 0 0 
4𝑥7
𝑥7
−
3𝑥3
𝑥7
+
5𝑥
𝑥7
−
1
𝑥7
 
lim
x→+∞
 7 −
5𝑥4
𝑥3
+
3
𝑥5
−
1
𝑥6
+
8
𝑥7
 = 
7−0+0−0+0
4−0+0−0
 = 
7
4
 
 4 −
3
𝑥4
+
5
𝑥6
−
1
𝑥7
 
(j) lim
x→−∞
 2 +
3
𝑥
−
1
𝑥2
 = lim
x→−∞
 2 +
3
𝑥
−
1
𝑥2
= 2 + 0 − 0 = 2 
 
 
 0 
(k) lim
x→−∞
𝑥5+5x
4𝑥5−50𝑥3
= lim
x→−∞
𝑥5
𝑥5
4𝑥5
𝑥5
+
5𝑥
𝑥5
50𝑥3
𝑥5
= lim
x→−∞
 
1+
5
𝑥4
4−
50
𝑥2
= 
1+0
4−0
= 
1
4
 
 0 
 
 0 0 
(l) lim
x→∞
2𝑥7+500x
𝑥8+1
= lim
x→∞
2𝑥7
𝑥8
𝑥8
𝑥8
+
500𝑥
𝑥8
1
𝑥8
= lim
x→∞
2
𝑥
1+
+
500
𝑥7
1
𝑥8
= 
0+0
1+0
=
0
1
= 0 
 0 
 
 0 
(m) lim
x→∞
2𝑥7+500x
𝑥6−900𝑥3
= lim
x→∞
 
2𝑥6
𝑥6
𝑥6
𝑥6
 
+
500𝑥
𝑥
900𝑥3
𝑥6
6 = lim
x→∞
 
2𝑥+
500
𝑥5
1−
900
𝑥3
= 
∞+0
1−0
= 
∞
1
= ∞ 
 0 
 
(n) lim
x→∞
√
𝑥2+2
2𝑥2+1
= lim
x→∞
√
𝑥2
𝑥2
+
2
𝑥2
2𝑥2
𝑥2
+
1
𝑥2
= lim
x→∞
 √
1+
2
𝑥2
2+
1
𝑥2
= √
1+0
2+0
= √
1
2
 
 
 
 ∞ ∞ 
(o) lim
x→0+
3𝑥2+2
𝑥3
= lim
x→∞
3𝑥2
𝑥3
+
2
𝑥3
= ∞ + ∞ = ∞ 
 
 2 ∞ 
(p) lim
x→4+
√
𝑥
𝑥−4
= lim
x→4+
√𝑥 .
1
𝑥−4
= 2. ∞ = ∞ 
 
 2 -∞ 
(q) lim
x→4−
√
𝑥
𝑥−4
= lim
x→4−
√𝑥 .
1
𝑥−4
= 2. −∞ = −∞ 
 
 √14 ∞ 
(r) lim
x→3+
√
𝑥2+𝑥+2
𝑥2−2𝑥−3
= lim
x→3+
√𝑥2 + 𝑥 + 2 .
1
𝑥2−2𝑥−3
= √14 . ∞ = ∞ 
 
 
 
 
3. Limites infinitos. 
(a) lim
x→1−
 
x−1
𝑥2−1
= lim
x→1−
 
x−1
(x+1)(𝑥−1)
= lim
x→1−
 
1
(x+1)
= 
1
1+1
=
1
2
 
 
 
 -2 -∞ 
(b) lim
x→−1−
 
x−1
x+1
= lim
x→−1−
(𝑥 − 1).
1
𝑥+1
= −2. −∞ = ∞ 
 
 
 2 ∞ 
(c) lim
x→−1−
 
2𝑥2−𝑥−1
𝑥2−1
= lim
x→−1−
 (2𝑥2 − 𝑥 − 1).
1
𝑥2−1
= 2. ∞ = ∞ 
 
 
 
 
4- Limite Trigonométrico Fundamental 
 
a) lim
𝑥→0
𝑠𝑒𝑛(7𝑥)
𝑥
= 
0
0
 𝑖𝑛𝑑. 
 
 
 
 
 
 
 
lim
𝑥→0
𝑠𝑒𝑛(7𝑥)
𝑥
 .
7
7
= lim
𝑥→0
7. 𝑠𝑒𝑛(7𝑥)
7𝑥
=
lim
𝑥→0
7 . lim
𝑥→0
𝑠𝑒𝑛(7𝑥)
7𝑥
= 7 . 1 = 7 
 
 
b) lim
𝑥→0
𝑠𝑒𝑛(4𝑥)
5𝑥
=
0
0
 𝑖𝑛𝑑. 
 
 
 
 
 
 
 
 
 
c) lim
𝑥→0
𝑠𝑒𝑛(10𝑥)
9𝑥
=
0
0
 𝑖𝑛𝑑. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
d) lim
𝑥→0
𝑠𝑒𝑛(2𝑥)
𝑥
=
0
0
 𝑖𝑛𝑑. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
lim
𝑥→0
𝑠𝑒𝑛(4𝑥)
5𝑥
 .
4
5
4
5
= lim
𝑥→0
𝑠𝑒𝑛(4𝑥)
4𝑥
 .
4
5
lim
𝑥→0
 
𝑠𝑒𝑛(4𝑥)
4𝑥
 . lim
𝑥→0
4
5
= 1 .
4
5
=
4
5
 
 
lim
𝑥→0
𝑠𝑒𝑛(10𝑥)
9𝑥
 .
10
9
10
9
= lim
𝑥→0
𝑠𝑒𝑛(10𝑥)
10𝑥
 .
10
9
=
lim
𝑥→0
𝑠𝑒𝑛(10𝑥)
10𝑥
 . lim
𝑥→0
10
9
= 1.
10
9
=
10
9
 
 
 
lim
𝑥→0
𝑠𝑒𝑛(2𝑥)𝑥
 .
2
2
= lim
𝑥→0
2 . 𝑠𝑒𝑛(2𝑥)
2𝑥
= 
lim
𝑥→0
2 . lim
𝑥→0
𝑠𝑒𝑛(2𝑥)
2𝑥
= 2 . 1 = 2
 
 
e) lim
𝑥→0
𝑠𝑒𝑛(3𝑥)
5𝑥
=
0
0
 𝑖𝑛𝑑. 
 
 
 
 
 
 
f) lim
𝑥→0
𝑡𝑔𝑥
𝑥
=
0
0
 𝑖𝑛𝑑. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
g) lim
𝑥→0
1−𝑐𝑜𝑠𝑥
𝑥
=
0
0
 𝑖𝑛𝑑. 
 
 
 
 
 
 
 
 
 
lim
𝑥→0
𝑠𝑒𝑛(3𝑥)
5𝑥
 .
3
5
3
5
= lim
𝑥→0
𝑠𝑒𝑛(3𝑥)
3𝑥
.
3
5
= 
lim
𝑥→0
𝑠𝑒𝑛(3𝑥)
3𝑥
 . lim
𝑥→0
3
5
= 1 .
3
5
=
3
5
 
 
lim
𝑥→0
𝑠𝑒𝑛𝑥
𝑐𝑜𝑠𝑥
𝑥
= lim
𝑥→0
𝑠𝑒𝑛𝑥
𝑐𝑜𝑠𝑥 . 𝑥
= 
lim
𝑥→0
1
𝑐𝑜𝑠𝑥
 . lim
𝑥→0
𝑠𝑒𝑛𝑥
𝑥
=
1
1
 . 1 = 1
 
 
lim
𝑥→0
1 − 𝑐𝑜𝑠𝑥
𝑥
 .
1 + 𝑐𝑜𝑠𝑥
1 + 𝑐𝑜𝑠𝑥
= lim
𝑥→0
12 − 𝑐𝑜𝑠2𝑥
𝑥(1 + 𝑐𝑜𝑠𝑥)
= lim
𝑥→0
𝑠𝑒𝑛2𝑥
𝑥(1 + 𝑐𝑜𝑠𝑥)
 
lim
𝑥→0
𝑠𝑒𝑛𝑥 . 𝑠𝑒𝑛𝑥
𝑥(1 + cosx)
= lim
𝑥→0
𝑠𝑒𝑛𝑥
𝑥
. lim
𝑥→0
𝑠𝑒𝑛𝑥
1 + 𝑐𝑜𝑠𝑥
= 1 .
0
2
= 0
 
 
h) lim
𝑥→0
1−𝑐𝑜𝑠𝑥
𝑥2
=
0
0
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
5- 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
lim
𝑥→0
1 − 𝑐𝑜𝑠𝑥
𝑥2
 . 
1 + 𝑐𝑜𝑠𝑥
1 + 𝑐𝑜𝑠𝑥
= lim
𝑥→0
12 − 𝑐𝑜𝑠2𝑥
𝑥2(1 + 𝑐𝑜𝑠𝑥)
= 
lim
𝑥→0
𝑠𝑒𝑛2𝑥
𝑥2(1 + 𝑐𝑜𝑠𝑥)
= lim
𝑥→0
𝑠𝑒𝑛𝑥 . 𝑠𝑒𝑛𝑥
𝑥 . 𝑥(1 + cosx)
lim
𝑥→0
𝑠𝑒𝑛𝑥
𝑥
. lim
𝑥→0
𝑠𝑒𝑛𝑥
𝑥
 . lim
𝑥→0
1
1 + 𝑐𝑜𝑠𝑥
= 1 . 1.
1
2
= 
1
2
 
 
 
9- 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
B- 
lim x-> -1 = Não existe 
lim x-> 0 = 2 
lim x-> 4 = 4 
 
10- Considere a função 𝑓(𝑥) =
1
𝑥2
 e faça o que se pede: 
 
a) Faça o gráfico da função 𝑓, usando o Geogebra. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
b) Analise o valor de 𝑓(𝑥) quando 𝑥 se aproxima de zero pela direita. O que 
podemos afirmar sobre o valor do limite lim
𝑥→0+
𝑓(𝑥)? 
 
 
Ao se aproximar de zero pela DIREITA, o limite irá tender ao infinito positivo. 
 
 
 
 
lim
𝑥→0+
1
𝑥2
=
1
02
= +∞ 
c) Analise agora o valor de 𝑓(𝑥) quando 𝑥 se aproxima de zero pela esquerda. O 
que podemos afirmar sobre o valor do lim
𝑥→0−
𝑓(𝑥)? 
 
 
Ao se aproximar de zero pela ESQUERDA, o limite irá tender ao infinito positivo. 
 
 
 
 
 
 
d) O limite de lim
𝑥→0
𝑓(𝑥) ? 
 
 
 
 
 
 
12- Seja 𝑓(𝑥) = 𝑥2 + 3. Determine a equação da reta tangente ao gráfico de 𝑓 nos 
pontos; 
 
a) (1, 𝑓 (1)) 
 
 
 
 
 
 
 
 
 
 
 
 
b) (−2, 𝑓 (−2)) 
 
 
 
 
lim
𝑥→0+
1
𝑥2
=
1
(−0)2
= +∞ 
lim
𝑥→0+
1
𝑥2
=
1
02
= +∞ 
𝑓(1) = 12 + 3 = 4
𝑃(1,4)
𝑓′(𝑥) = 2𝑥 → 𝑎 = 𝑓′(1) = 2
(𝑦 − 𝑦0) = 𝑎(𝑥 − 𝑥0)
(𝑦 − 4) = 2(𝑥 − 1)
𝑦 = 2𝑥 − 2 + 4
𝑦 = 2𝑥 + 2
 
𝑓(−2) = (−2)2 + 3 = 7
𝑃(−2,7)
𝑓′(𝑥) = 2𝑥 → 𝑎 = 𝑓′(−2) = 4
(𝑦 − 𝑦0) = 𝑎(𝑥 − 𝑥0)
(𝑦 − 7) = −4(𝑥 − (−2))
−4𝑥 − 8 + 7
𝑦 = −4𝑥 − 1