Limites

EJERCICIOS SOBRE CÁLCULO DE LÍMITES
1.

2.

5.

4.

a)

a)

a)

a)

i)

b)

b)

b)

b)

j)

c)

c)

c)

c)

k)

d)

d)

d)

d)

l)

e)

e)

m)

f)

f)

n)

g)

g)

o)

h)

h)

i)
j)
k)
1

1a

an > 0 ∀n ∈ N

an +1

lím a = L
n →∞
n


0 si L < 1
lím an = 
n →∞
∞ si L > 1

n +1

n
an =
n!

(n +1)

(n + 2)

n!

lim
=e
(n 1)
n → ∞ ( n + 1 )! n +

2

1b
3
2

n sen(n!)
an =
n +1

(−1)

n

3
2

3
2

(n + 1)

3
2

n
n sen(n!)
n
(−1)
≤
≤
n +1
n +1
n +1
3
2

2

3
2

3
2

n sen(n!)
≤
≤
n +1
3
2

n
n sen(n!)
(−1)
≤
≤
2
n +1
(n + 1)

n

3
2

(n + 1)

2

3
2

n
2
(n + 1)
3

n

3
2

3
2
3
2

n
nlim
=
= lim
2
n →∞ (n + 1)
n →∞ 

2
n
+ 2n 3 + 1 3 
3


2
2
2
n
n
n

= lim
n →∞

1
 n+ 2

+1


n
n n


=0

3
2

n sen(n!)
0 ≤ lim
≤0
n →∞
n +1
4

1
n

 π
an = n sen n 
 2
a2 n = (2n )

1
2n

(2n + 1)

1
n

lim n = 1
n →∞

1
 π
sen 2n  = (2n )2 n sen(nπ ) = 0
 2

a2 n +1 = (2n + 1)

1
2 n +1

1c

12 n +1

1
π
π


sen (2n + 1)  = (2n + 1)2 n +1 sen nπ + 
2
2


1
2 n +1

π  (2n + 1) → 1 si n es par

sen nπ +  =
1
2

− (2n + 1)2 n +1 → −1 si n es impar
5

1d
n!

2
an =
222 n

2

( ( n + 1 )! )

222

n

lim
=∞
( n + 1 ) ( n! )
n → ∞ 222
2

6

n 
an =  2
 n +1



2

sen ( n )

1e

  n 2  sen (n ) 
 n2  
 = sen( n) ln 

ln (an ) = ln  2 
 n +1
 n2 + 1  










  n2  
  n2  
lim ln 2   = ln lim 2   = ln 1 = 0
 n +1
 n →∞ n + 1  
n →∞




  n2  
 n2 
 n2 
( −1) lim ln 2   ≤ lim sen( n) ln 2  ≤ lim ln 2 
 n + 1   n →∞
 n + 1  n →∞  n + 1 
n →∞







 n2 
( −1).0≤ lim sen( n) ln 2  ≤ 0
 n +1
n→∞



lim an = lim e ln (an ) = lim e
n →∞

n →∞

n →∞

 n2 
sen ( n ) ln  2 
 n +1 



=e

 n2 
lim sen ( n ) ln  2 
 n +1 
n→∞



= e0 = 1

7

n 
an =  2
 n +1



2

n 
lim an =  2 
 n +1
n →∞


2

sen ( n )

sen ( n )

=e

 n2

lim  2 −1  sen ( n )

n→∞  n +1


=e

 n 2 − n 2 −1 
 sen ( n )
lim 
n→∞ 
n 2 +1 



=e

 −1 
lim  2  sen ( n )
n→∞  n +1 

= e0 = 1

8

1f

 n − 3n 
an = 
 n2 



2

ln( n )

 3 
lim an = lim 1 − 
n→∞
n →∞
 n 
=e

−3 lim

n→∞

1
ln( n n

)

=e

ln( n )

−n



3
 = lim 1 − 3 
 n  
 n →∞ 





1

−3 ln lim ( n n ) 
 n→∞




−

3 ln( n )
n

=e

ln( n )
n→∞ n

− 3 lim

=

= e −3 ln1 = e 0 = 1

9

sen(3n + 1) + 2n
an =
n 2 − 2n

2

1g

2n 2
sen(3n + 1)
lim an = lim 2
+ lim 2
n →∞
n →∞ n − 2n
n →∞ n − 2n
−1
sen(3n + 1)
1
lim 2
≤ lim 2
≤ lim 2
n →∞ n − 2n
n →∞
n →∞ n − 2n
n − 2n
sen(3n + 1)
0 ≤ lim 2
≤0
n →∞ n − 2n
2n 2
2
lim 2
= lim
=2n →∞ n − 2n
n →∞
2
1−
n
lim an = 0 + 2 = 2
n →∞

10

cos(3n + 1) + 2n
an =
n 2 − 2n

2

1h

cos(3n + 1)
2n 2
lim an = lim 2
+ lim 2
n →∞
n →∞ n − 2n
n →∞ n − 2n
−1
cos(3n + 1)
1
lim 2
≤ lim 2
≤ lim 2
n →∞ n − 2n
n →∞ n − 2n
n →∞ n − 2n
cos(3n + 1)
0 ≤ lim 2
≤0
n →∞ n − 2n
2n 2
2
lim 2
= lim
=2
n →∞ n − 2n
n →∞
2
1−
n
lim an = 0 + 2 = 2
n →∞

111i

n 2 +1

an =

cos(nπ )e 3n

2

 n3 + n + 2 


3


n



2n

Cálculo del límite del denominador

n +n+2

lim

n →∞
n3


3

2n





1
= lim1 + 3
n →∞
n


 n+2 

n

n+2 





1 

= lim 1 + 3
n →∞ 
n


 n + 2  




3

2n

n3 n+ 2
n+ 2 n3

=

( n + 2 )2 n
n3

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