UNIVERSIDAD NACIONAL AUTONOMA DE NICARAGUA

UNAN-LEON

FACULTAD DE CIENCIAS ECONÓMICAS Y EMPRESARIALES

Carrera

Componente

Modalidad

Parcial

Año

Profesor

Integrantes

Fecha

“A LA LIBERTAD POR LA UNIVERSIDAD”

Determine el limite si existe y cuando sea posible , indique los teorema de limite que se aplican.

2
1. lim (2 x −4 x +5)
x →3

2
= lim 2 x −lim 4 x + lim 5 Teorema 5
x →3 x →3 x →3

=¿ teorema 6,3

[ 2
= (2) lim x −( 4)(3)+ 5
x →3 ] teorema 3,7,4

= (2)¿ teorema 4

= 18−7

= 11

3 x+ 4
2. lim 8 x−1
x →2

lim 3 x+ 4
3 x+ 4
lim = x→2
teorema 8
x →2 8 x−1 lim 8 x−1
x→ 2

lim 3 x+ lim 4
x→2 x→ 2
= lim 8 x−lim 1 teorema 5
x→ 2 x →2

( lim 3 )( lim x )+4

x →2 x→2
= teorema 7,3
( lim 8 )(lim x )−1
x→ 2 x →2

3 (2)+ 4
= 8(2)−1 teorema 4,3

6+ 4 10
= 16−1 = 5

2
=3
 2 x +1
lim
3. x→−1 2
x −3 x +4

lim 2 x +1
2 x +1 x →−1
lim 2 = 2
x→−1 x −3 x +4 lim x −3 x +4
x →−1

lim 2 x + lim 1
x →−1 x →−1
= 2
lim x − lim 3 x+ lim + 4
x →−1 x →−1 x →−1

( lim 2 )( lim x )+1

x →−1 x →−1
=
( lim x )−( lim 3 )( lim x )+ 4
2

x →−1 x →−1 x →−1

= 2(−1)+1
¿¿

−2+1
= 1+ 3+4

1
=- 8

4. lim 3 2
x→ 4 √
x 2−3 x + 4
2 x −x−1

√
lim ( x −3 x + 4 )
2

√ √ √ √
2
3 x −3 x + 4 x 2−3 x + 4 3 x→ 4 2
3 4 −3(4 )+ 4 48 4 2
lim = 3
lim = = ¿= =3 =3
lim ( 2 x −x−1 ) 27
2 2
x→ 4 2 x −x−1 x → 4 2 x −x−1 2
2¿¿
x →4
 x−1
5. lim
x →1 √ x 2+ 3−2

lim x−1 lim ( x−1 )

x−1 x→ 1 x→ 1 1−1 0
lim = = = =0
√ x + 3−2 lim √ x + 3−2 lim √ ( x +3 )−2 √ 1 +3−2
2 2 2 2
x →1
x→ 1 x→ 1

6. lim √
x +4−2
x →0 x

lim √ x +4−2 lim √ ( x +4 )−2

lim
√ x +4−2 x → 0
= lim x = x→0 =√
0+4−2 0
=
x →0 x lim ( x ) 0 0
x →0 x →0

lim √ x +4−2 . √ x + 4+2

x →0 x √ x + 4+2

2
( √ x + 4 ) − ( 2 )2
= lim ❑
X→0 x . ( √ x + 4+2 )

lim 1
( x+ 4−4 ) 1 x →0 1 1
= lim =lim = = =
x →0 x ( √ x + 4+2 ) x →0 √ x +4 +2 lim √ x+ 4+ 2 √ 0+4 +2 4
x →0
7. lim
√ x−1
x →1 x−1

lim √ x−1 lim √ ( x )−1

lim
√ x−1 = x→ 1 = x→ 1 =
√1−1 = 0
x →1 x−1 lim x−1 lim ( x−1 ) 1−1 0
x→1 x→1

lim √ x−1 =lim ( √ x−1 ) ( x+ 1 )

x →1 x−1 x →1 ( x−1 ) ( √ x+ 1 )

2
( √ x ) −( 1 )2 lim x−1
= lim =
x →1 ( x−1 ) ( √ x+ 1 ) x →1 ( x−1 ) ( √ x+ 1 )

lim 1
1 x →1 1 1
= lim = = =
x →1 √ x +1 lim √ x+ 1 √ 1+1 2
x→ 1

( )
3 2
2 x +1
9. lim 2
x →1 x +2 x−8

2
lim ( 2 x +1 )
3

( )
3 2
2 x +1 x→ 1
lim 2 = 2
x →1 x +2 x−8 lim ( x 2 +2 x−8 )
x→ 1

3
lim 2 x + lim 1
x→ 1 x→ 1
=
lim x 2 +lim 2 x−lim 8
x→ 1 x→ 1 x →1
 ( lim 2 )( lim x ) +1
3
x→1 x→ 1
=
( lim x ) +( lim 2 )( lim x )−8
2

x→ 1 x→ 1 x→ 1

¿¿¿¿¿

( 3 )2
= 2
( 5)

9
=
25

2
3 s −8 s−16
10. lim 2
s→4 2 s −9 s+ 4

2
2 lim 3 s −8 s−16
3 s −8 s−16 s→ 4
lim =
s→4
2
2 s −9 s+ 4 lim 2 s 2−9 s+ 4
s→4

2
lim 3 s −lim 8 s−lim 16
s→ 4 s→4 s→4
= 2
lim 2 s −lim 9 s +lim 4
s→4 s→4 s →4

2
lim 3 lim s −lim 8 lim s−16
s→ 4 s→4 s→4 s→4
= 2
lim 2 lim s lim 9 lim s+ 4
s→4 s→ 4 s →4 s→4

3 ( 4 )2−8 ( 4 )−16 0
= 2
= Indeterminada
2 ( 4 ) −9 ( 4 )+ 4 0

2
3 s −8 s−16 ( 3 s +4 ) (1 s−4 )
lim = lim
2 s −9 s+ 4 s → 4 ( 2 s−1 )( 1 s−4 )
2
s→4
 3 s +4
= lim
s→4 2 s−1

lim 3 s +lim 4
s→4 s →4
=
lim 2 s−lim 1
s→ 4 s→4

lim 3 lim s+ 4
s→4 s→ 4
=
lim 2 lim s−1
s→ 4 s →4

3 ( 4) + 4
=
2 ( 4 ) −1

12+ 4 16
= =
8−1 7

3
t +8
11. lim
t →−2 t +2

3
3 lim t +8
t +8 t →−2
lim =
t →−2 t +2 lim t +2
t →−2

2
lim t + lim 8
t →−2 t →−2
¿
lim t + lim 2
t →−2 t →−2

( lim t ) +8
3
t →−2
=
( lim t )+ 2
t →−2
 (−2 )3+ 8 0
= = Indeterminada
−2+2 0

√
2
t −9
12. lim 2
t →−3 2 t +7 t+3

√ √ √
(t¿ ¿ 2¿−9)
√ (−3)2−9
2 2
t −9 t +9 lim ¿ ¿ ¿ 0
lim = lim 2 = lim ¿ ¿ = 2
= I
t →−3
2
2 t +7 t+3 t →−3 2 t +7 t+3 t →−3
t →−3
2(−3) +7 (−3)+ 3 0

√ √ √
lim t−3
lim
t →−3
t 2−9
2
2 t +7 t+3
= lim
(t−3)(t +3)
t →−3 (2 t+1)(t+ 3)
= lim
√
t−3
t →−3 2 t+ 1
= t →−3
lim 2 t+ 1
t →−3
¿
√−3−3
2(−3)+1
=

√ √
−6 = 30
−5 5

2
2 h −h−3
13. lim 3 2
h →−1 h +2 h +6+ 5

2
2 lim 2 h −h−3
2 h −h−3 h →−1
lim =
3
2
h →−1 h +2 h +6+ 5 lim h3 +2 h2 +6+ 5
h →−1
 2
lim 2h − lim h− lim 3
h →−1 h →−1 h →−1
= 3 2
lim h + lim 2 h + lim 6+ lim 5
h →−1 h →−1 h →−1 h →−1

( lim 2 )( lim h ) −( lim h )−3

2
→−1 →−1 →−1
=
( lim h ) +( lim 2 )( lim h ) +6+5
3 2

→−1 →−1 →−1

=2 ¿ ¿

2+1−3 0
= =
−1+ 2+6+5 12

=0

14. lim
√ x−√ a
x →a x−a

lim √ x−√ a lim √ (x ¿)− √(a)

lim √ x−√ a = x → a = x→a
¿=
√ a− √a =¿ 0 Indeterminada
x →a x−a lim x−a lim (x−a) a−a 0
x→a x→ a

( √ x−√ a )( √ x + √ a ) 2
( √ x ) −( √ a )
2
x−a
lim =lim =lim
x →a ( x−a ) ( √ x + √ a ) x →a ( x−a ) ( √ x + √ a ) x →a ( x−a ) ( √ x + √ a )

lim 1
1 x→a 1 1
=lim = = =
x →a ( √ x + √ a ) lim √ x + √ a √ a+ √ a 2 √a
x→a