Propiedades de los limites Cálculo Diferencial e Integral

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INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
95 
 
Calculo de limites 
Fecha: _______________ 
 
 
Propiedades de los limites 
 
En Matemáticas a nivel superior se trabaja con indeterminaciones, que son expresiones tales como: 
∞
∞
, ∞, −∞, 00,
0
0
, 𝑦 
𝑐
0
, con 𝑐 ≠ 0. 
Para calcular limites es necesario considerar los siguientes teoremas: 
1 lim
𝑥→𝑎
𝑐 = 𝑐, si 𝑐 es una constante 
 
 
 5 lim
𝑥→𝑎
[𝑓(𝑥)𝑔(𝑥)] = lim
𝑥→𝑎
𝑓(𝑥) lim
𝑥→𝑎
𝑔(𝑥) 
2 lim
𝑥→𝑎
𝑥 = 𝑎 
 
 
 6 
lim
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
= 
lim
𝑥→𝑎
𝑓(𝑥)
lim
𝑥→𝑎
𝑔(𝑥)
 
3 lim
𝑥→𝑎
𝑐𝑓(𝑥) = 𝑐 lim
𝑥→𝑎
𝑓(𝑥) 
 
 
 
 7 lim
𝑥→𝑎
[𝑓(𝑥)]𝑛 = [lim
𝑥→𝑎
𝑓(𝑥)]
𝑛
 
4 lim
𝑥→𝑎
[𝑓(𝑥) ± 𝑔(𝑥)] = lim
𝑥→𝑎
𝑓(𝑥) ± lim
𝑥→𝑎
𝑔(𝑥) 
 
 
 
 8 lim
𝑥→𝑎
√𝑓(𝑥)
𝑛 = √lim
𝑥→𝑎
𝑓(𝑥)𝑛 , con 𝑛 ∈ 𝑍+ 
 
 
 
Ahora bien, en muchos casos sucede que al calcular directamente el límite de una función real de variable real, resulta una 
indeterminación y para quitarla (o eliminarla) puede emplearse la Regla de L´Hopital. 
La Regla de L´Hopital establece lo siguiente: si lim
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
=
0
0
, o bien lim
𝑥→∞
𝑓(𝑥)
𝑔(𝑥)
=
∞
∞
, donde 𝑓(𝑥) 𝑦 𝑔(𝑥) son funciones 
continuas, entonces lim
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
= lim
𝑥→𝑎
𝑓(𝑛)(𝑥)
𝑔(𝑛)(𝑥)
 , o bien lim
𝑥→∞
𝑓(𝑥)
𝑔(𝑥)
= lim
𝑥→∞
𝑓(𝑛)(𝑥)
𝑔(𝑛)(𝑥)
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
96 
 
Calculo de limites 
Fecha: _______________ 
 
Limites 
 
Ejemplos I 
 
 
1) lim
𝑥→2
 
𝑥2−4
𝑥2−3𝑥+2
= 
22−4
22−3(2)+2
=
4−4
4−6+2
=
0
0
 
 
lim
𝑥→2
 
𝑥2 − 4
𝑥2 − 3𝑥 + 2
= lim
𝑥→2
 
𝑥2 − 22
𝑥2 − 3𝑥 + 2
 
= lim
𝑥→2
(𝑥 − 2)(𝑥 + 2)
(𝑥 − 2)(𝑥 − 1)
 
= lim
𝑥→2
(𝑥 − 2)
1
(𝑥 − 2)
(
𝑥 + 2
𝑥 − 1
) 
= lim
𝑥→2
𝑥 + 2
𝑥 − 1
 
=
2+2
2−1
=
4
1
= 4 
 
 
 
2) lim
𝑥→2 3⁄
9𝑥2−4
3𝑥−2
 =
9(2 3⁄ )
2
−4
3(2 3⁄ )−2
=
9(4 9⁄ )−4
3(2 3⁄ )−2
 =
4−4
2−2
 =
0
0
 
lim
𝑥→2 3⁄
9𝑥2 − 4
3𝑥 − 2
= lim
𝑥→2 3⁄
32𝑥2 − 22
3𝑥 − 2
 
= lim
𝑥→2 3⁄
(3𝑥)2 − 22
3𝑥 − 2
 
= lim
𝑥→2 3⁄
(3𝑥 − 2)(3𝑥 + 2)
(3𝑥 − 2) 
 
= lim
𝑥→2 3⁄
(3𝑥 − 2)
1
(3𝑥 − 2)
(3𝑥 + 2) 
= lim
𝑥→2 3⁄
3𝑥 + 2 = 3 (
2
3
) + 2 = 2 + 2 = 4 
 
 
 
 lim
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
=
0
0
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
97 
 
Calculo de límites 
Fecha: _______________ 
 
 
3) lim
𝑥→3
𝑥2−2𝑥−3
𝑥2−3𝑥
=
32−2(3)−3
32−3(3)
=
9−6−3
9−9
=
0
0
 
 
 
 Lim
𝑥→3
𝑥2−2𝑥−3
𝑥2−3𝑥
= lim
𝑥→3
(𝑥−3)(𝑥+1)
𝑥(𝑥−3)
 
= lim
𝑥→3
(𝑥 − 3) (
1
𝑥 − 3
) (
𝑥 + 1
𝑥
) 
= lim
𝑥→3
𝑥 + 1
𝑥
 
=
3 + 1
3
=
4
3
 
 
lim
𝑥→3
𝑥2 − 2𝑥 − 3
𝑥2 − 3𝑥
= lim
𝑥→3
𝑑
𝑑𝑥
(𝑥2 − 2𝑥 − 3)
𝑑
𝑑𝑥
(𝑥2 − 3𝑥)
= lim
𝑥→3
2𝑥 − 2
2𝑥 − 3
=
2(3) − 2
2(3) − 3
=
6 − 2
6 − 3
=
4
3
 
4) lim
𝑥→1
𝑥3−1
𝑥−1
=
13−1
1−1
=
1−1
1−1
=
0
0
 
 lim
𝑥→1
𝑥3 − 1
𝑥 − 1
= lim
𝑥→1
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
𝑥 − 1
= lim
𝑥→1
(𝑥2 + 𝑥 + 1) = 1 + 1 + 1 = 3 
 
Por división de polinomios. 
 
 
 
 
 
 
 
 
5) lim
𝑥→−1
𝑥3+1
𝑥2+1
=
−1+1
1+1
=
0
2
 = 0 
 
 
 
 
 
 𝑥3 − 1 
 𝑥2 + 𝑥 + 1 
 𝑥 − 1 
 − 𝑥3 + 𝑥2 
 
−𝑥 + 1 
 −𝑥2 + 𝑥 
 
0 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
98 
 
Calculo de Limites. 
Fecha: _______________ 
 
 
 
 
6) lim
𝑥→1
(
1
1−𝑥
−
3
1−𝑥3
) =
1
1−1
− 
3
1−1
=
1
0
−
3
0
 
1
1 − 𝑥
−
3
1 − 𝑥3
 =
1
(−1)(𝑥 − 1)
+
3
(−1)(−𝑥3 + 1)
 
 
=
−1
𝑥 − 1
+
3
𝑥3 − 13
 
 
=
−1
𝑥 − 1
+
3
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
 
=
−1
(𝑥 − 1)
(1) +
3
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
=
−1
(𝑥 − 1)
(𝑥2 + 𝑥 + 1) (
1
𝑥2 + 𝑥 + 1
) +
3
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
=
−1(𝑥2 + 𝑥 + 1)
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
+
3
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
 
=
−(𝑥2 + 𝑥 + 1) + 3
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
 
 
=
−𝑥2 − 𝑥 − 1 + 3
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
 
 
=
−𝑥2 − 𝑥 + 2
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
 
 
=
−(𝑥2 + 𝑥 − 2)
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
 
 
=
−(𝑥 − 1)(𝑥 + 2)
(𝑥 − 1)(𝑥2 + 𝑥 + 1)
 
 
= −(𝑥 − 1) (
1
𝑥 − 1
) (
𝑥 + 2
𝑥2 + 𝑥 + 1
) 
 
 
= − (
𝑥 + 2
𝑥2 + 𝑥 + 1
) 
 
lim
𝑥→1
(
1
𝑥 − 1
− 
1
𝑥3 − 1
) = lim
𝑥→1
 − (
𝑥 + 2
𝑥2 + 𝑥 + 1
) = −(
1 + 2
12 + 1 + 1
) = −
3
3
= −1 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
99 
 
Calculo de Limites. 
Fecha: _______________ 
 
 
Ejemplos II lim
𝑥→∞
(
𝑓(𝑥)
𝑔(𝑥)
) =
∞
∞
 
lim
𝑥→∞
(
1
𝑥
) = 0, es un límite que se utiliza con mucha frecuencia 
 
 
7) lim
𝑥→∞
4𝑥
𝑥2−𝑥
=
∞
∞
 
 
 
 
 
lim
𝑥→∞
4𝑥
𝑥2 − 𝑥
 = lim
𝑥→∞
(
4𝑥
𝑥2 − 𝑥
) (1) 
= lim
𝑥→∞
4𝑥
(𝑥2 − 𝑥)
 
1
𝑥2
1
𝑥2
 
= lim
𝑥→∞
4 (𝑥
1
𝑥
)
1
𝑥
𝑥2
𝑥2
−
𝑥
𝑥2
 
= lim
𝑥→∞
4
1
𝑥
1 −
1
𝑥
 
 
 
=
lim
𝑥→∞
4
1
𝑥
lim
𝑥→∞
(1−
1
𝑥
)
 ….. Por Teorema 6 de Limites 
 
=
4 lim
𝑥→∞
1
𝑥
lim
𝑥→∞
1− lim
𝑥→∞
1
𝑥
 ….. Por Teorema 3 y 4 de Limites 
 
=
4 lim
𝑥→∞
1
𝑥
lim
𝑥→∞
1− lim
𝑥→∞
1
𝑥
 ……Aplicando lim
𝑥→∞
(
1
𝑥
) = 0 
 
=
4(0)
1 − 0
 =
0
1
= 0 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
100 
 
Calculo de Limites. 
Fecha: _______________ 
 
 
 
8) lim
𝑥→∞
2𝑥2−𝑥+3
𝑥3−8𝑥+5
=
∞
∞
 
 
lim
𝑥→∞
2𝑥2 − 𝑥 + 3
𝑥3 − 8𝑥 + 5
= lim
𝑥→∞
2𝑥2 − 𝑥 + 3
𝑥3
𝑥3 − 8𝑥 + 5
𝑥3
 
 
= lim
𝑥→∞
2𝑥2
𝑥3
−
𝑥
𝑥3
+
3
𝑥3
𝑥3
𝑥3
−
8𝑥
𝑥3
+
5
𝑥3
 
 
= lim
𝑥→∞
2
𝑥
−
1
𝑥2
+
3
𝑥3
1 −
8
𝑥2
+
5
𝑥3
 
 
= lim
𝑥→∞
2
1
𝑥
− (
1
𝑥
)
2
+ 3(
1
𝑥
)
3
1 − 8 (
1
𝑥
)
2
+ 5(
1
𝑥
)
3 
 
Utilizando las propiedades de límites 
lim
𝑥→∞
2
1
𝑥
− (
1
𝑥
)
2
+ 3(
1
𝑥
)
3
1 − 8 (
1
𝑥
)
2
+ 5(
1
𝑥
)
3 =
lim
𝑥→∞
(2
1
𝑥
− (
1
𝑥
)
2
+ 3(
1
𝑥
)
3
)
lim
𝑥→∞
(1 − 8 (
1
𝑥
)
2
+ 5(
1
𝑥
)
3
)
 
 
 
=
lim
𝑥→∞
2
1
𝑥
− lim
𝑥→∞
(
1
𝑥
)
2
+ lim
𝑥→∞
3 (
1
𝑥
)
3
lim
𝑥→∞
1 − lim
𝑥→∞
8 (
1
𝑥
)
2
+ lim
𝑥→∞
5 (
1
𝑥
)
3 
 
 
=
2 lim
𝑥→∞
1
𝑥
− lim
𝑥→∞
(
1
𝑥
)
2
+ 3 lim
𝑥→∞
(
1
𝑥
)
3
lim
𝑥→∞
1 − 8 lim
𝑥→∞
(
1
𝑥
)
2
+ 5 lim
𝑥→∞
(
1
𝑥
)
3 
 
 
=
2 lim
𝑥→∞
1
𝑥
− ( lim
𝑥→∞
1
𝑥
)
2
+ 3( lim
𝑥→∞
1
𝑥
)
3
lim
𝑥→∞
1 − 8 ( lim
𝑥→∞
1
𝑥
)
2
+ 5( lim
𝑥→∞
1
𝑥
)
3 
 
 
=
2(0) − 0 + 3(0)
1 − 8(0) + 5(0)
=
0
1
= 0 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
101 
 
Calculo de Limites. 
Fecha: _______________ 
 
9) lim
𝑥→∞
(−𝑥+1)(2𝑥+1)(4𝑥−6)
20𝑥3+2
=
∞
∞
 
 
lim
𝑥→∞
(−𝑥 + 1)(2𝑥 + 1)(4𝑥 − 6)
20𝑥3 + 2
 = lim
𝑥→∞
(−𝑥 + 1)(2𝑥 + 1)(4𝑥 − 6)
𝑥3
20𝑥3 + 2
𝑥3
 
 
= lim
𝑥→∞
(
−𝑥 + 1
𝑥
) (
2𝑥 + 1
𝑥
) (
4𝑥 − 6
𝑥
)
 
20𝑥3 + 2
𝑥3
 
 
 
= lim
𝑥→∞
(−
𝑥
𝑥
+
1
𝑥
) (2
𝑥
𝑥
+
1
𝑥
) (4
𝑥
𝑥
−
6
𝑥
)
 20
𝑥3
𝑥3
+ 2 
1
𝑥3= lim
𝑥→∞
(−1 +
1
𝑥
) (2 +
1
𝑥
) (4 − 6
1
𝑥
)
 20 + 2 (
1
𝑥
)
3 
 
 
=
lim
𝑥→∞
((−1 +
1
𝑥
) (2 +
1
𝑥
) (4 − 6
1
𝑥
))
lim
𝑥→∞
(20 + 2 (
1
𝑥
)
3
)
 
 
 
=
lim
𝑥→∞
(−1 +
1
𝑥
) lim
𝑥→∞
(2 +
1
𝑥
) lim
𝑥→∞
(4 − 6
1
𝑥
)
lim
𝑥→∞
20 + lim 2 (
1
𝑥
)
3
𝑥→∞
 
 
 
=
( lim
𝑥→∞
− 1 + lim
𝑥→∞
1
𝑥
) ( lim
𝑥→∞
2 + lim
𝑥→∞
1
𝑥
) ( lim
𝑥→∞
4 − lim
𝑥→∞
6
1
𝑥
)
lim
𝑥→∞
20 + 2 lim
𝑥→∞
(
1
𝑥
)
3 
 
=
( lim
𝑥→∞
− 1 + lim
𝑥→∞
1
𝑥
) ( lim
𝑥→∞
2 + lim
𝑥→∞
1
𝑥
) ( lim
𝑥→∞
4 − 6 lim
𝑥→∞
1
𝑥
)
( lim
𝑥→∞
20 + 2 ( lim
𝑥→∞
1
𝑥
)
3
)
 
 
 
=
(−1 + 0)(2 + 0)(4 − 0)
(20 + 2(0)3)
 
 
=
(−1)(2)(4)
(20)
 
=
−8
20
= −
2
5
 
 
 
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
102 
 
Calculo de Limites. 
Fecha: _______________ 
 
 
 
 
10) lim
𝑥→∞
(2𝑥−3)2
3𝑥2+𝑥−1
=
∞
∞
 
Utilizando regla.. lim
𝑥→∞
(2𝑥−3)2
3𝑥2+𝑥−1
= lim
𝑥→∞
2(2𝑥−3)1(2)
6𝑥+1
= lim
𝑥→∞
4(2𝑥−3)1
6𝑥+1
= lim
𝑥→∞
8𝑥−12
6𝑥+1
= lim
𝑥→∞
8
6
 = lim
𝑥→∞
4
3
= 
4
3
 
lim
𝑥→∞
(2𝑥 − 3)2
3𝑥2 + 𝑥 − 1
 = lim
𝑥→∞
(2𝑥 − 3)2
(𝑥)2
3𝑥2 + 𝑥 − 1
𝑥2
 
 
 
= lim
𝑥→∞
(
2𝑥 − 3
𝑥
)
2
3𝑥2 + 𝑥 − 1
𝑥2
 
 
= lim
𝑥→∞
(2
𝑥
𝑥
− 3
1
𝑥
)
2
3
𝑥2
𝑥2
+
𝑥
𝑥2
−
1
𝑥2
 
 
= lim
𝑥→∞
(2 − 3
1
𝑥
)
2
3 +
1
𝑥
− (
1
𝑥
)
2 
 
 
=
lim
𝑥→∞
(2 − 3
1
𝑥
)
2
lim
𝑥→∞
(3 +
1
𝑥
− (
1
𝑥
)
2
)
 
 
 
=
(lim2
𝑥→∞
− lim
𝑥→∞
3
1
𝑥
)
2
lim
𝑥→∞
3 + lim
𝑥→∞
1
𝑥
− lim
𝑥→∞
(
1
𝑥
)
2 
 
 
=
( lim
𝑥→∞
2 − 3 lim
𝑥→∞
1
𝑥
)
2
lim
𝑥→∞
3 + lim
𝑥→∞
1
𝑥
− ( lim
𝑥→∞
1
𝑥
)
2 
 
 
=
(2 − 0)2
3 + 0 − 0
 
 
 
=
(2)2
3
 
 
 
=
4
3
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
103 
 
Calculo de Limites. 
Fecha: _______________ 
 
 
 Ejemplos III. 𝐥𝐢𝐦
𝒙→∞
 (Funciones Irracionales) 
 
 
11) lim
𝑥→∞
2𝑥2−3𝑥−4
√𝑥4+1 
=
∞
∞
 
lim
𝑥→∞
 
2𝑥2 − 3𝑥 − 4
√𝑥4 + 1 
 = lim
𝑥→∞
2𝑥2 − 3𝑥 − 4
𝑥2
√𝑥4 + 1
𝑥2
 
 
 
= lim
𝑥→∞
2𝑥2 − 3𝑥 − 4
𝑥2
√𝑥4 + 1
√(𝑥2)2
 
 
 
= lim
𝑥→∞
2𝑥2 − 3𝑥 − 4
𝑥2
√𝑥
4 + 1
𝑥4
 
 
= lim
𝑥→∞
2
𝑥2
𝑥2
−
3𝑥
𝑥2
−
4
𝑥2
√𝑥
4 
𝑥4
+
1 
𝑥4
 
 
= lim
𝑥→∞
2 − 3
1
𝑥
 − 4 
1
𝑥2
√1 +
1
𝑥4
 
 
= lim
𝑥→∞
2 − 3
1
𝑥
 − 4 (
1
𝑥
)
2
√1 + (
1
𝑥
)
4
 
 
=
lim
𝑥→∞
(2 − 3
1
𝑥
 − 4 (
1
𝑥
)
2
)
lim
𝑥→∞
√1 + (
1
𝑥
)
4
 
 
 
= 
lim
𝑥→∞
2 − 3 lim
𝑥→∞
1
𝑥
 − 4 ( lim
𝑥→∞
 
1
 𝑥
)
2
√lim
𝑥→∞
1 + ( lim
𝑥→∞
1
𝑥
)
4
 
 
= 
2 − 3 (0) − 4(0)2
√1 + (0)4
=
2
1
= 2 
 
 
 Conocimiento Previo: 𝑥
𝑛
𝑚 = √𝑥𝑛
𝑚
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
104 
 
Calculo de Limites 
Fecha: _______________ 
 
 
12) lim
𝑥→∞
4𝑥3−𝑥−1
√𝑥6−3 
=
∞
∞
 
 
Lim
𝑥→∞
4𝑥3 − 𝑥 − 1
√𝑥6 − 3 
 = lim
𝑥→∞
4𝑥3 − 𝑥 − 1
𝑥3
√𝑥6 − 3 
𝑥3
 
 
 
= lim
𝑥→∞
4𝑥3 − 𝑥 − 1
𝑥3
√𝑥6 − 3 
√(𝑥3)2
 
 
 
= lim
𝑥→∞
4𝑥3
𝑥3
 − 
𝑥
𝑥3
 − 
1
𝑥3
√𝑥
6 − 3
𝑥6
 
 
 
= lim
𝑥→∞
4
𝑥3
𝑥3
 − 
𝑥
𝑥3
 − 
1
𝑥3
√𝑥
6
𝑥6
−
3
𝑥6
 
 
 
= lim
𝑥→∞
4 − 
1
𝑥2
 − 
1
𝑥3
√1 − 3
1
𝑥6
 
 
 
= 
lim
𝑥→∞
(4 − (
1
𝑥
)
2
 − (
1
𝑥
)
3
)
lim
𝑥→∞
√1 − 3 (
1
𝑥
)
6
 
 
 
= 
lim
𝑥→∞
4 − lim
𝑥→∞
 (
1
𝑥
)
2
 − lim
𝑥→∞
(
1
𝑥
)
3
√lim
𝑥→∞
1 − 3 lim
𝑥→∞
 (
1
𝑥
)
6
 
 
 
=
lim
𝑥→∞
4 − ( lim
𝑥→∞
1
𝑥
)
2
− ( lim
𝑥→∞
1
𝑥
)
3
√lim
𝑥→∞
1 − 3 ( lim
𝑥→∞
1
𝑥
)
6
 
 
 
= 
4 − 0 − 0
√1 − 3(0)
= 
4
1
= 4 
 INSTITUTO POLITÉCNICO NACIONAL 
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RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
105 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
 
13) lim
𝑥→∞
𝑥−4
𝑥+ √𝑥
3 = 
∞
∞
 
 
 
 
 
lim
𝑥→∞
𝑥 − 4
𝑥 + √𝑥
3 = lim𝑥→∞
𝑥 − 4
𝑥
𝑥 + √𝑥
3
𝑥
 
 
= lim
𝑥→∞
𝑥
𝑥
 − 
4
𝑥
𝑥
𝑥
 + 
√𝑥
3
√𝑥3
3
 
 
= lim
𝑥→∞
1 − 4 
1
𝑥
1 + √
𝑥
𝑥3
3
 
 
= lim
𝑥→∞
1 − 4
1
𝑥
1 + √
1
𝑥2
3
 
= 
lim
𝑥→∞
(1 − 4
1
𝑥
)
lim
𝑥→∞
(1 + √
1
𝑥2
3
)
 
 
= 
lim
𝑥→∞
1 − 4 lim
𝑥→∞
 
1
𝑥
lim
𝑥→∞
1 + √ lim
𝑥→∞
 (
1
𝑥
)
23
 
=
lim
𝑥→∞
1 − 4 lim
𝑥→∞
 
1
𝑥
lim
𝑥→∞
1 + √( lim
𝑥→∞
 
1
𝑥
)
23
 
 
=
1 − 4(0)
1 + 0
=
1
1
= 1 
 INSTITUTO POLITÉCNICO NACIONAL 
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RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
106 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
14) lim
𝑥→∞
√𝑥3−3
5
𝑥4−2
 =
∞
∞
 
 
 
 
lim
𝑥→∞
√𝑥3 − 3
5
𝑥4 − 2
 
= lim
𝑥→∞
√𝑥3 − 3
5
𝑥4
𝑥4 − 2
𝑥4
 
 
 
= lim
𝑥→∞
√𝑥3 − 3
5
√(𝑥4)5
5
𝑥4 − 2
𝑥4
 = lim
𝑥→∞
√𝑥
3 − 3
𝑥20
5
𝑥4
𝑥4
− 
2
𝑥4
 
 
 
 
 
= lim
𝑥→∞
√ 𝑥
3
𝑥20
− 
3
𝑥20
5
1 − 
2
𝑥4
= lim
𝑥→∞
√
1
𝑥17
− 
3
𝑥20
5
1 − 2 
1
𝑥4
 
 
=
lim
𝑥→∞
√
1
𝑥17
− 3
1
𝑥20
5
lim
𝑥→∞
(1−2 
1
𝑥4
)
 
 
 
 
=
√lim 
𝑥→∞
(
1
𝑥
)
17
− lim
𝑥→∞
 3 (
1
𝑥
)
205
lim
𝑥→∞
1 − 2 lim
𝑥→∞
(
1
𝑥
)
4 
 
 
 
=
√( lim
𝑥→∞
 
1
𝑥
)
17
− 3( lim
𝑥→∞
 
1
𝑥
)
205
lim
𝑥→∞
1 − 2 ( lim
𝑥→∞
 
1
𝑥
)
4 
 
 
=
√0 − 0
5
1 − 2(0)
= 0 
 
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
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RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
107 
 
Calculo de Limites 
Fecha: _______________ 
 
 
15) lim
𝑥→∞
√7𝑥+4
3
2𝑥−3
 =
∞
∞
 
 
 
 
 
 
 
 
lim
𝑥→∞
√7𝑥 + 4
3
2𝑥 − 3
 = lim
𝑥→∞
 
√7𝑥 + 4
3
𝑥
2𝑥 − 3
𝑥
 
 
= lim
𝑥→∞
 
√7𝑥 + 4
3
√𝑥3
3
2𝑥 − 3
𝑥
 
 
= lim
𝑥→∞
√
7𝑥 + 4
𝑥3
3
2𝑥 − 3
𝑥
 
 
= lim
𝑥→∞
√
7𝑥
𝑥3
+
4
𝑥3
 
3
2
𝑥
𝑥
− 3 
1
𝑥
 
 
= lim
𝑥→∞
√7
1
𝑥2
+ 4
1
𝑥3
 
3
2 − 3 
1
𝑥
 
 
= lim
𝑥→∞
√7 (
1
𝑥
)
2
+ 4(
1
𝑥
)
3
 
3
2 − 3
1
𝑥
 
 
=
√7 lim
𝑥→∞
(
1
𝑥
)
2
+ 4 lim
𝑥→∞
(
1
𝑥
)
3
 
3
lim
𝑥→∞
2 − 3 lim
𝑥→∞
1
𝑥
 
 
=
√7 ( lim
𝑥→∞
 
1
𝑥
)
2
+ 4( lim
𝑥→∞
 
1
𝑥
)
3
 
3
lim
𝑥→∞
2 − 3 lim
𝑥→∞
1
𝑥
 
 
= 
√7(0) + 4(0) 
3
2 − 3(0)
=
0
2
= 0 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
108 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
Ejemplos IV racionalización 
 
16) lim
𝑥→1
 
√𝑥−1
𝑥−1
= 
√1−1
1−1
= 
1−1
1−1
= 
0
0
 𝑥2 − 𝑦2 = (𝑥 − 𝑦)(𝑥 + 𝑦) 
 
lim
𝑥→1
 
√𝑥 − 1
𝑥 − 1
 = lim
𝑥→1
 
√𝑥 − 1
𝑥 − 1
∗ 
√𝑥 + 1
√𝑥 + 1
 
 
= lim
𝑥→1
 [
 (√𝑥)
2
− 12 
(𝑥 − 1)(√𝑥 + 1)
] 
 
= lim
𝑥→1
 [(𝑥 − 1) 
1
(𝑥 − 1)
(
1
√𝑥 + 1
)] 
 
= lim
𝑥→1
 
1
√𝑥 + 1
 
=
1
√1 + 1
 =
1
1 + 1
 
 
=
1
2 
 
 
17) lim
𝑥→4
√𝑥−2
𝑥−4
=
√4−2
4−4
=
2−2
4−4
= 
0
0
 
 
lim
𝑥→4
√𝑥 − 2
𝑥 − 4
 = lim
𝑥→4
 
√𝑥 − 2
𝑥 − 4
∗
√𝑥 + 2
√𝑥 + 2
 
 
= lim
𝑥→4
 
 (√𝑥)
2
− 22 
(𝑥 − 4)(√𝑥 + 2)
 
 
= lim
𝑥→4
(𝑥 − 4) 
 1 
(𝑥 − 4)
(
 1 
√𝑥 + 2
) 
 
= lim
𝑥→4
1
√𝑥 + 2
 
 
=
1
√4 + 2
 
 
=
1
4 
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
109 
 
Calculo de Limites 
Fecha: _______________ 
 
 
18) lim
𝑥→4
 
3−√5+𝑥
1−√5−𝑥
=
3−√5+4
1−√5−4=
3−√9
1−√1
=
3−3
1−1
=
0
0
 
lim
𝑥→4
 
3 − √5 + 𝑥
1 − √5 − 𝑥
 = lim
𝑥→4
[
3 − √5 + 𝑥
1 − √5 − 𝑥
∗
3 + √5 + 𝑥
3 + √5 + 𝑥
] 
 
 
= lim
𝑥→4
[
9 − (5 + 𝑥)
(1 − √5 − 𝑥)(3 + √5 + 𝑥)
] = lim
𝑥→4
[
9 − 5 − 𝑥
(1 − √5 − 𝑥)(3 + √5 + 𝑥)
] 
 
 
 
= lim
𝑥→4
 [
4 − 𝑥
(3 + √5 + 𝑥)(1 − √5 − 𝑥)
] [
1 + √5 − 𝑥
1 + √5 − 𝑥
] 
 
 
= lim 
𝑥→4
(4 − 𝑥)(1 + √5 − 𝑥)
(1 − √5 − 𝑥)(1 + √5 − 𝑥)(3 + √5 + 𝑥)
 
 
 
= lim [
𝑥→4
(4 − 𝑥)(1 + √5 − 𝑥)
(1 − (5 − 𝑥))(3 + √5 + 𝑥)
] = lim 
𝑥→4
(4 − 𝑥)(1 + √5 − 𝑥)
(1 − 5 + 𝑥)(3 + √5 + 𝑥)
 
 
 
= lim
𝑥→4
 [
(4 − 𝑥)(1 + √5 − 𝑥)
(−4 + 𝑥)(3 + √5 + 𝑥)
] = lim
𝑥→4
 [
(4 − 𝑥)(1 + √5 − 𝑥)
−1(4 − 𝑥)(3 + √5 + 𝑥)
] 
 
 
= lim
𝑥→4
− 1 [
1 + √5 − 𝑥
3 + √5 + 𝑥
] 
 
 
= −lim
𝑥→4
[
1 + √5 − 𝑥
3 + √5 + 𝑥
] 
 
 
= −
1 + √5 − 4
3 + √5 + 4
 
 
 
= −
1 + √1
3 + √9
 = −
1 + 1
3 + 3
 
 
 
= −
2
6
 
= −
1
3
 
 
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
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RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
110 
 
Calculo de Limites 
Fecha: _______________ 
 
19) lim
𝑥→∞
(√𝑥 + 𝑎 − √𝑥) = ∞ 
lim
𝑥→∞
(√𝑥 + 𝑎 − √𝑥) = lim
𝑥→∞
(√𝑥 + 𝑎 − √𝑥) ∗
√𝑥 + 𝑎 + √𝑥
√𝑥 + 𝑎 + √𝑥
 
 
 
= lim
𝑥→∞
[
𝑥 + 𝑎 − 𝑥
√𝑥 + 𝑎 + √𝑥
] 
 
 = lim
𝑥→∞
[
𝑎
√𝑥 + 𝑎 + √𝑥
] 
 
 
= lim
𝑥→∞
[
 
 
 
 
𝑎
√𝑥
√𝑥 + 𝑎 + √𝑥
√𝑥 ]
 
 
 
 
 
 
 
= lim
𝑥→∞
[
 
 
 
 √𝑎
2
√𝑥
√𝑥 + 𝑎 + √𝑥
√𝑥 ]
 
 
 
 
 = lim
𝑥→∞
[
 
 
 
 √𝑎
2
√𝑥
√𝑥 + 𝑎
√𝑥
+
√𝑥
√𝑥 ]
 
 
 
 
 
 
 
= lim
𝑥→∞
[
 
 
 √
𝑎2
𝑥
√
𝑥 + 𝑎
𝑥
+ 1]
 
 
 
 = lim
𝑥→∞
[
 
 
 √
𝑎2
𝑥
√
𝑥
𝑥
+
𝑎
𝑥
+ 1
]
 
 
 
 
 
= lim
𝑥→∞
[
 
 
 √
𝑎2
𝑥
√1 +
𝑎
𝑥
+ 1
]
 
 
 
= 
lim
𝑥→∞
√𝑎
2
𝑥
lim
𝑥→∞
(√1 +
𝑎
𝑥
+ 1)
 
 
 
=
 √ lim
𝑥→∞
𝑎2
𝑥
√ lim
𝑥→∞
(1 + 𝑎
1
𝑥
) + lim
𝑥→∞
1
 
 
=
 √𝑎2 lim
𝑥→∞
1
𝑥
√ lim
𝑥→∞
1 + 𝑎 lim
𝑥→∞
(
1
𝑥
) + lim
𝑥→∞
1
 
 
=
 √𝑎20
√1 + 𝑎(0) + 1
 
 
=
 0
√1 + 1
 
=
 0
1 + 1
=
0
2
 
 
= 0 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
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RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
111 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
20) lim
𝑥→∞
 𝑥(√𝑥2 + 1 − 𝑥) = ∞ 
lim
𝑥→∞
 𝑥 (√𝑥2 + 1 − 𝑥) = lim
𝑥→∞
 𝑥 [(√𝑥2 + 1 − 𝑥)] 1 
 
 
= lim
𝑥→∞
 𝑥 [(√𝑥2 + 1 − 𝑥) (
√𝑥2 + 1 + 𝑥
√𝑥2 + 1 + 𝑥
)] 
 
 
= lim
𝑥→∞
 𝑥 [
𝑥2 + 1 − 𝑥2
√𝑥2 + 1 + 𝑥
] 
 
 = lim
𝑥→∞
 
𝑥
√𝑥2 + 1 + 𝑥
 
 
 
= lim
𝑥→∞
 
𝑥
𝑥
√𝑥2 + 1 + 𝑥
𝑥
 = lim
𝑥→∞
 
𝑥
𝑥
√𝑥2 + 1
√𝑥2
+
𝑥
𝑥
 
 
 
= lim
𝑥→∞
 
1
√𝑥
2 + 1
𝑥2
+ 1
 
 
 
= lim
𝑥→∞
 
1
√𝑥
2
𝑥2
+
1
𝑥2
+ 1
 
 
= lim
𝑥→∞
 
1
√1 + (
1
𝑥
)
2
+ 1
 
 
 
=
lim
𝑥→∞
1
lim
𝑥→∞
(√1 + (
1
𝑥
)
2
+ 1)
 
 
 
= 
lim
𝑥→∞
1
√lim
𝑥→∞
1 + lim
𝑥→∞
(
1
𝑥
)
2
+ lim
𝑥→∞
1
 
 
 
= 
lim
𝑥→∞
1
√lim
𝑥→∞
1 + ( lim
𝑥→∞
 
1
𝑥
)
2
+ lim
𝑥→∞
1
 
 
=
1
√1 + 0 + 1
=
1
√1 + 1
=
1
1 + 1
=
1
2
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
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RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
112 
 
Ejemplos V 
Calculo de Limites 
Fecha: _______________ 
 
 
 
 
 lim
𝑥→0
(
𝑆𝑒𝑛 𝑥
𝑥
) = 1, es un límite que se utiliza con mucha frecuencia 
 
21) lim
𝑥→0
𝑆𝑒𝑛 5𝑥
𝑥
=
𝑆𝑒𝑛 0
0
=
0
0
 
 
Utilizando regla de H… lim
𝑥→0
𝑆𝑒𝑛 5𝑥
𝑥
= lim
𝑥→0
5𝐶𝑜𝑠(5𝑥)
1
= lim
𝑥→0
5𝐶𝑜𝑠(5𝑥) = 5 𝐶𝑜𝑠 (0) = 5(1) = 5 
 
lim
𝑥→0
𝑆𝑒𝑛 5𝑥
𝑥
 = lim
𝑥→0
(5 
1
5
)
(𝑆𝑒𝑛 5𝑥)
𝑥
 
 
= lim
𝑥→0
5 (
𝑆𝑒𝑛 5𝑥
5𝑥
) 
= 5 lim
𝑥→0
(
𝑆𝑒𝑛 5𝑥
5𝑥
) 
 
= 5(lim
𝑥→0
𝑆𝑒𝑛 5𝑥
5𝑥
) 
 = 5 
 
 
22. lim
𝑥→0
𝑆𝑒𝑛 7𝑥
2𝑥
=
𝑆𝑒𝑛 0
2(0)
=
0
0
 
lim
𝑥→0
𝑆𝑒𝑛 7𝑥
2𝑥
 = lim
𝑥→0
({
𝑆𝑒𝑛 7𝑥
𝑥
} (
1
7
)) (
1
2
) (7) 
 
= lim
𝑥→0
(
𝑆𝑒𝑛 7𝑥
7𝑥
) (
7
2
) 
 
=
7
2
lim
𝑥→0
(
𝑆𝑒𝑛 7𝑥
7𝑥
) 
 
=
7
2
 
 
 
22. lim
𝑛→0
(
𝑆𝑒𝑛 9 𝑥
𝑥
) =
0
0
 
lim
𝑥→0
(
𝑆𝑒𝑛 9 𝑥
𝑥
) = lim
𝑥→0
9 (
𝑆𝑒𝑛 9 𝑥
9 𝑥
) 
 
= 9 (lim
𝑥→0
𝑆𝑒𝑛 9𝑥
9 𝑥
) 
 = 9 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
113 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
 
23. lim
𝑥→0
𝑆𝑒𝑛 𝜋 𝑥
𝑆𝑒𝑛 3 𝜋 𝑥
=
𝑆𝑒𝑛 𝜋 (0)
𝑆𝑒𝑛 3 𝜋 (0)
=
𝑆𝑒𝑛 0
𝑆𝑒𝑛 0
=
0
0
 
 
lim
𝑥→0
𝑆𝑒𝑛 𝜋 𝑥
𝑆𝑒𝑛 3 𝜋 𝑥
= lim
𝑥→0
(𝑆𝑒𝑛 𝜋 𝑥)
(𝑆𝑒𝑛 3 𝜋 𝑥)
 
(𝜋 𝑥
1
𝜋 𝑥
)
(3𝜋 𝑥
1
3𝜋 𝑥
)
 
 
 
 
= lim
𝑥→0
(𝑆𝑒𝑛 𝜋 𝑥) (
1
𝜋 𝑥
) (𝜋 𝑥)
(𝑆𝑒𝑛 3 𝜋 𝑥) (
1
3𝜋 𝑥
) (3𝜋 𝑥)
 
 
= lim
𝑥→0
(
𝑆𝑒𝑛 𝜋 𝑥
𝜋 𝑥
)
(
𝑆𝑒𝑛 3 𝜋 𝑥
3𝜋 𝑥
)
 (
𝜋 𝑥
𝜋 𝑥
 
1
3 
 ) 
 
 
= lim
𝑥→0
(
𝑆𝑒𝑛 𝜋 𝑥
𝜋 𝑥
𝑆𝑒𝑛 3𝜋𝑥
3𝜋𝑥
)(
1
3 
) 
 
 
=
1
3 
lim
𝑥→0
(
𝑆𝑒𝑛 𝜋 𝑥
𝜋 𝑥
𝑆𝑒𝑛 3𝜋𝑥
3𝜋𝑥
) 
 
 
=
1
3 
lim
𝑥→0
(
𝑆𝑒𝑛 𝜋 𝑥
 𝜋 𝑥
)
lim
𝑥→0
(
𝑆𝑒𝑛 3𝜋𝑥
 3𝜋𝑥
)
 
 
 
=
1
3 
1
1
 
 
 
= (
1
3 
) (1) 
 
= 
1
3 
 
 
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
114 
 
 
 
24. lim 
𝑥→0
 𝑥𝑆𝑒𝑛
1
𝑥
= 0𝑆𝑒𝑛
1
0
 
 
lim
𝑥→0
𝑥 𝑆𝑒𝑛
1
𝑥
 = lim 
𝑥→0
𝑥 (𝑆𝑒𝑛
1
𝑥
) (1) 
 
= lim 
𝑥→0
𝑥 (𝑆𝑒𝑛
1
𝑥
)(
1
𝑥
 
1
1
𝑥
) 
 
= lim 
𝑥→0
𝑥 (
𝑆𝑒𝑛 
1
𝑥
1
𝑥
)(
1
𝑥
) 
 
= lim 
𝑥→0
(𝑥
1
𝑥
)(
𝑆𝑒𝑛 
1
𝑥
1
𝑥
) 
 
= lim 
𝑥→0
(1) (
𝑆𝑒𝑛 
1
𝑥
1
𝑥
) 
 
= lim 
𝑥→0
(
𝑆𝑒𝑛 
1
𝑥
1
𝑥
) 
 = 1 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
115 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
 
25. lim
𝑥→0
1−𝐶𝑜𝑠𝑥
𝑥
=
1−𝑐𝑜𝑠0
0
=
1−1
0
=
0
0
 
lim
𝑥→0
1 − 𝐶𝑜𝑠𝑥
𝑥
 = lim
𝑥→0
[
1 − 𝐶𝑜𝑠𝑥
𝑥
] [1] = lim
𝑥→0
[
1 − 𝐶𝑜𝑠𝑥
𝑥
] [(1 + 𝐶𝑜𝑠𝑥) (
1
1 + 𝐶𝑜𝑠𝑥
)] 
 
 
 = lim
𝑥→0
 (
12−(𝐶𝑜𝑠𝑥)2
𝑥
) (
1
1+𝐶𝑜𝑠𝑥
) 
 
= lim
𝑥→0
 
1 − 𝐶𝑜𝑠2𝑥
𝑥(1 + 𝐶𝑜𝑠𝑥)
 
 
 = lim
𝑥→0
 
𝑆𝑒𝑛2𝑥
𝑥(1+𝑐𝑜𝑠𝑥)
 porque 𝑆𝑒𝑛2𝑥 = 1 − 𝐶𝑜𝑠2𝑥 
 
 
= lim
𝑥→0
[[
(𝑆𝑒𝑛 𝑥)2
𝑥
] (
1
1 + 𝑐𝑜𝑠𝑥
)] 
 
 
= lim
𝑥→0
 
𝑥
𝑥
 (
(𝑆𝑒𝑛 𝑥)2
𝑥
) [
1
1 + 𝐶𝑜𝑠𝑥
] 
 
 
= lim
𝑥→0
𝑥 (
(𝑆𝑒𝑛 𝑥)2
(𝑥)2
) [
1
1 + 𝐶𝑜𝑠𝑥
] 
 
 
= lim
𝑥→0
 𝑥 (
𝑆𝑒𝑛 𝑥
𝑥
)
2
[
1
1 + 𝐶𝑜𝑠𝑥
] 
 
 
= (lim
𝑥→0
 𝑥) (lim
𝑥→0
 (
𝑆𝑒𝑛 𝑥
𝑥
)
2
) (lim
𝑥→0
1
1 + 𝐶𝑜𝑠𝑥
) 
 
 
= (lim
𝑥→0
 𝑥) (lim 
𝑥→0
𝑠𝑒𝑛 𝑥
𝑥
)
2
(lim
𝑥→0
1
1 + 𝐶𝑜𝑠𝑥
) 
 
 
= (lim
𝑥→0
 𝑥) (lim 
𝑥→0
𝑠𝑒𝑛 𝑥
𝑥
)
2
(
1
1 + 𝐶𝑜𝑠 (0)
) 
 
 
= (0)(1) (
1
2
) 
 
= 0 
 
 
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
116 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
27. lim
ℎ→0
𝑠𝑒𝑛(𝑥+ℎ)−𝑠𝑒𝑛 𝑥
ℎ
=
𝑠𝑒𝑛(𝑥+0)−𝑠𝑒𝑛 𝑥
0
=
𝑠𝑒𝑛(𝑥)−𝑠𝑒𝑛 𝑥
0
= 
0
0
 
 
 
 
 
 
lim
ℎ→0
𝑆𝑒𝑛(𝑥 + ℎ) − 𝑆𝑒𝑛 𝑥
ℎ
 = lim
ℎ→0
[
𝑆𝑒𝑛 𝑥 Cos ℎ + 𝑆𝑒𝑛 ℎ 𝐶𝑜𝑠𝑥 − 𝑆𝑒𝑛 𝑥
ℎ
] 
 
 
= lim
ℎ→0
[
𝑆𝑒𝑛 𝑥 Cos ℎ − 𝑆𝑒𝑛 𝑥 + 𝑆𝑒𝑛 ℎ 𝐶𝑜𝑠𝑥
ℎ
] 
 
 
= lim
ℎ→0
[
(𝐶𝑜𝑠(ℎ) − 1)𝑆𝑒𝑛𝑥 + 𝑆𝑒𝑛(ℎ)𝐶𝑜𝑠(𝑥)
ℎ
] 
 
 
= lim
ℎ→0
[
(𝐶𝑜𝑠(ℎ) − 1)𝑆𝑒𝑛𝑥
ℎ
+
𝑆𝑒𝑛(ℎ)𝐶𝑜𝑠(𝑥)
ℎ
] 
 
 
= lim
ℎ→0
[
(𝐶𝑜𝑠ℎ − 1)
ℎ
𝑆𝑒𝑛𝑥 +
𝑆𝑒𝑛ℎ
ℎ
𝐶𝑜𝑠𝑥] 
 
Sabemos que:lim
𝑥→0
1 − 𝐶𝑜𝑠𝑥
𝑥
= 0 
lim
𝑥→0
(
(−1)(𝐶𝑜𝑠 𝑥 − 1)
𝑥
) = 0 
lim
𝑥→0
(
𝐶𝑜𝑠 𝑥 − 1
𝑥
) = −0 = 0 
lim
ℎ→0
(
𝐶𝑜𝑠 ℎ − 1
ℎ
) = 0 
 
 
 
= 𝑆𝑒𝑛 𝑥 lim
ℎ→0
[
𝐶𝑜𝑠 ℎ − 1
ℎ
] + 𝐶𝑜𝑠 𝑥 lim
ℎ→0
𝑆𝑒𝑛 ℎ
ℎ
 
 
= 𝑆𝑒𝑛𝑥 lim
ℎ→0
[
𝐶𝑜𝑠 ℎ − 1
ℎ
] + 𝐶𝑜𝑠𝑥 lim
ℎ→0
𝑆𝑒𝑛 ℎ
ℎ
 
 
 = (𝑆𝑒𝑛𝑥)(0) + (𝐶𝑜𝑠𝑥)(1) 
 
 = 𝐶𝑜𝑠𝑥 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
117 
 
Calculo de Limites 
Fecha: _______________ 
 
 
 
28. lim
𝑥→0
1−√𝑐𝑜𝑠𝑥
𝑥2
=
1−√𝑐𝑜𝑠0
02
=
1−√1
0
= 
1−1
0
=
0
0
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
lim
𝑥→0
1 − √𝐶𝑜𝑠𝑥
𝑥2
 = lim
𝑥→0
[
1 − √𝐶𝑜𝑠𝑥
𝑥2
 ] (1) 
 
 
= lim
𝑥→0
[
1 − √𝐶𝑜𝑠𝑥
𝑥2
 ] (1 + √𝐶𝑜𝑠𝑥 
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
= lim
𝑥→0
[
12 − (√𝐶𝑜𝑠𝑥)
2
𝑥2
 ] (
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
 
 
= lim
𝑥→0
(
1 − 𝑐𝑜𝑠𝑥
𝑥2
)(
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
= lim
𝑥→0
(
1 − 𝑐𝑜𝑠𝑥
𝑥2
) (1) (
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
= lim
𝑥→0
 (
1 − 𝐶𝑜𝑠𝑥
𝑥2
) (1 + 𝐶𝑜𝑠𝑥 
1
1 + 𝐶𝑜𝑠𝑥
) (
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
= lim
𝑥→0
 (
12 − (𝐶𝑜𝑠𝑥)2
𝑥2
)(
1
1 + 𝐶𝑜𝑠𝑥
) (
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
= lim
𝑥→0
(
1 − 𝐶𝑜𝑠2 𝑥
𝑥2
)(
1
1 + 𝐶𝑜𝑠 𝑥
) (
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
= lim
𝑥→0
(
𝑆𝑒𝑛2 𝑥
𝑥2
)(
1
1 + 𝐶𝑜𝑠𝑥
) (
1
1 + √𝐶𝑜𝑠𝑥
) 
 
 
= lim
𝑥→0
(
𝑆𝑒𝑛 𝑥
𝑥
)
2
(
1
1 + 𝐶𝑜𝑠𝑥
) (
1
1 + √𝐶𝑜𝑠𝑥
) 
 
= lim
𝑥→0
(
𝑆𝑒𝑛 𝑥
𝑥
)
2
 lim
𝑥→0
(
1
1 + 𝐶𝑜𝑠𝑥
) lim
𝑥→0
(
1
1 + √𝐶𝑜𝑠𝑥
) 
 
= (lim
𝑥→0
𝑆𝑒𝑛 𝑥
𝑥
)
2
 lim
𝑥→0
(
1
1 + 𝐶𝑜𝑠𝑥
) lim
𝑥→0
(
1
1 + √𝐶𝑜𝑠𝑥
) 
= (1) (
1
1 + 𝐶𝑜𝑠 0
) (
1
1 + √𝐶𝑜𝑠0
) 
= 
 
= (1)
1
2
(
1
2
) =
1
4
 
 INSTITUTO POLITÉCNICO NACIONAL 
ESIME CULHUACAN 
Academia de Matemáticas 
 
RAMIREZ ORTIZ MARÍA VERÓNICA 
RAMIREZ CASTELLANOS ERNESTINA CONCEPCIÓN 
118 
 
Calculo de Limites. 
Fecha: _______________ 
 
 
 
 
29. lim
𝑥→0
√1+𝑆𝑒𝑛𝑥−√1−𝑆𝑒𝑛𝑥
𝑥
= 
√1+𝑆𝑒𝑛0 −√1−𝑆𝑒𝑛0
0
=
√1−√1
0
=
0
0
 
 
 
 
lim
𝑥→0
√1 + 𝑆𝑒𝑛𝑥 − √1 − 𝑆𝑒𝑛𝑥
𝑥
 = lim
𝑥→0
[
√1 + 𝑆𝑒𝑛𝑥 − √1 − 𝑆𝑒𝑛𝑥
𝑥2
 ] (1) 
 
 
= lim
𝑥→0
[
√1 + 𝑆𝑒𝑛𝑥 − √1 − 𝑆𝑒𝑛𝑥
𝑥2
 ] (
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
) 
 
 
= lim
𝑥→0
[
(√1 + 𝑆𝑒𝑛𝑥)
2
− (√1 − 𝑆𝑒𝑛𝑥)
2
𝑥2
 ] (
1
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
) 
 
 
= lim
𝑥→0
[
(1 + 𝑆𝑒𝑛𝑥) − (1 − 𝑆𝑒𝑛𝑥)
𝑥
] (
1
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
) 
 
 
= lim
𝑥→0
[
1 + 𝑆𝑒𝑛 𝑥 − 1 + 𝑆𝑒𝑛 𝑥
𝑥
] (
1
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
) 
 
 
= lim
𝑥→0
[
2𝑠𝑒𝑛𝑥
𝑥
] (
1
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
) 
 
 
= lim
𝑥→0
(
𝑠𝑒𝑛𝑥
𝑥
)(
2
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
) 
 
 
= lim
𝑥→0
[
𝑆𝑒𝑛𝑥
𝑥
] . lim
𝑥→0
(
2
√1 + 𝑆𝑒𝑛𝑥 + √1 − 𝑆𝑒𝑛𝑥
) 
 
 
= (1) (
2
√1 + 𝑠𝑒𝑛0 + √1 − 𝑠𝑒𝑛0
) 
 
 
=
2
√1 + 0 + √1 + 0
=
2
√1 + √1
=
2
1 + 1
 
 
=
2
2
 
 
 = 1