EJERCICIOS PROPUESTOS - LIMITES AL INFINITO1)limx→∞ [ x3x2+2− x2

x+2 ]= limx→∞( x+2 ) (x3 )−(x2+2 )(x2)

(x2+2 )(x+2)

¿limx→∞

x4+2x3−x4−2 x2

(x2+2 )(x+2)

limx→∞

2x3−2 x2

(x2+2 )( x+2)=limx→∞

2 x3

x3−2 x

2

x3

( x2

x2+2

x2 )( xx + 2x )=limx→∞

2−0

1=2

2)limx→∞ [ 3 x22 x+1

−(2 x−1 ) (3x2+x+2 )

4 x2 ]= limx→∞ 3 x2

2 x+1−

(6 x3+x2+3 x−2 )4 x2

limx→∞

4 x2 (3 x2 )−(2x+1 ) (6 x3−x2+3x−2 )

(2 x+1 ) (4 x2 )limx→∞

12x4−12x4+2x3−6 x2+4 x−6 x3−x2−3 x+2

(2x+1 )(4 x2)

¿limx→∞

−4 x3−7 x2+x+2

(2 x+1 )(4 x2)

limx→∞

−4 x3

x3−7 x

2

x3+ xx3

+ 2x3

( 2 xx +1x )( 4 x

2

x2 )=−48

=−12

3)limx→∞ [ x3

2 x2−1− x2

2 x+1 ]=limx→∞

(2 x+1 ) (x3 )−(2 x2−1 )(x2)

(2 x2−1 )(2 x+1)

limx→∞

2x4+ x3−(2 x4−x2)

(2x2−1 ) (2x+1)=

limx→∞

x3+x2

(2 x2−1 )(2 x+1)

¿limx→∞

x3

x3+ x

2

x3

( 2x2

x2−1

x2 )( 2 xx +1x )

=limx→∞

1+0

(2 )(2)=14

4)limx→∞ [ 3 x2−22x+1

÷x2−4 xx−3 ]= limx→∞

(3 x2−2 )(x−3)

(2 x+1 )( x2−4 x )

¿limx→∞

3x3−9 x2−2x+6

(2x+1 )(x2−4 x)=limx→∞

3x3

x3−9 x

2

x3−2 xx3

+ 6x3

( 2xx +1x )( x

2

x2−4 x

x2 )=32

5)limx→∞

[√ x2+ x−√x2+9 ]=limx→∞

√ x2+x−√x2+9(√ x2+x+√x2+9√ x2+x+√x2+9 )

limx→∞

x2+x−x2−9

√ x2+x+√ x2+9=

limx→∞

x−9

√x2+x+√x2+9

¿limx→∞

xx−9x

√ x2x2+ xx2+√ x2x2+ x 9x2=12

6)limx→−∞ [√2x2+1x+3 ]= lim

x→−∞|x|√2+1

x+3=limx→−∞

−x √2x

+ 1x

xx+ 3x

=−√2

7)limx→∞

(√( x+a ) ( x+b )−x )

¿ limx→∞

√ ( x+a ) ( x+b )−x (√ ( x+a ) ( x+b )+x√ ( x+a ) ( x+b )+x )

limx→∞

( x+a ) ( x+b )−x2

√ ( x+a ) ( x+b )+x=limx→∞

x2+bx+ax+ab−x2

√( x+a ) ( x+b )+x

limx→∞

b xx

+ a xx

+ abx

√( bx + xx )( ax + xx )+ xx2

=a+b2

8)Hallar las constantes k y b, tal que cumpla lim

x→∞ [kx+b− (x3+1 )x2+1 ]= 0

limx→∞ [ kx (x2+1 )+b (x2+1 )−x3−1

x2+1 ]¿ limx→∞ [ k x3+kx+b x2+b−x3−1x2+1 ]

limx→∞ [ k x

3

x3+ kxx3

+ b x2

x3− x

3

x3− 1x3

x2

x3+ 1x3

]=limx→∞ [ k+ kx2+ bx−1− 1x31x+ 1x3

]=limx→∞ k−1k−1=0k=1b=0

9)Determinar M+N, tal que lim

x→∞ [Mx+N− x3

x2+1 ]=0limx→∞ [Mx (x2+1 )+N (x2+1 )−x3

x2+1 ]= limx→∞ [M x3+Mx+N x2+N−x3

x2+1 ]

limx→∞ [ M x

3

x3+Mxx3

+ N x2

x3+ Nx3

− x3

x3

x2

x3+ 1x3

]= limx→∞ [M+Mx2

+ Nx

+ Nx3

−1

1x+ 1x3

]=limx→∞ M−1

M−1=0M=1N=0

10)

limx→∞

(√x2−2 x−1−√x2−7 x+3 )

limx→∞

√ x2−2x−1−√ x2−7 x+3(√x2−2 x−1+√ x2−7 x+3√x2−2 x−1+√ x2−7 x+3 )

limx→∞

x2−2x−1−x2+7 x−3

√x2−2 x−1+√ x2−7 x+3=

limx→∞

5x−4

√x2−2 x−1+√ x2−7 x+3limx→∞

5 xx

−4x

√ x2−2 x−1x2+√ x2−7 x+3x2

=limx→∞

5−0

√ x2x2−2 xx2 −1

x2+√ x2x2−7 xx2 +

3

x2

=52

11)limx→−∞

(√x2−2 x−1−√x2−7 x+3 )

limx→−∞

√ x2−2 x−1−√ x2−7 x+3(√x2−2 x−1+√ x2−7 x+3√x2−2 x−1+√ x2−7 x+3 )

limx→−∞

x2−2x−1−x2+7x−3

√ x2−2x−1+√x2−7 x+3=

limx→∞

5x−4

√x2−2 x−1+√ x2−7 x+3

limx→∞

5 xx

−4x

√ x2(1−2x− 1

x2 )x

+ √ x2(1−7x + 3x2 )x

¿limx→∞

5

(−x)√(1− 2x− 1x2 )x

+(−x)√(1−7x + 3x2 )

x

=−52

12)limx→∞

(√x (x+a)−x )= limx→∞

√ x(x+a)−x (√x (x+a)+x√x (x+a)+x )limx→∞

x ( x+a )−x2

√x (x+a)+x=limx→∞

x2+ax−x2

√ x2+ax+x=

limx→∞

ax

√x2+ax+x

limx→∞

a xx

√ x2x2+ axx2 + xx=a2

13)limx→∞

(√x+√2x−√x−√2 x )=limx→∞

√ x+√2 x−√ x−√2 x(√ x+√2 x+√ x−√2x√ x+√2 x+√ x−√2x )

limx→∞

x+√2 x−x+√2x

√x+√2x+√ x−√2x=

limx→∞

2√ 2 xx√ xx +√ 2xx2 +√ xx−√ 2 xx2

=2√22

=√2

14)limx→∞

[ x (√ x2+1−x)]=limx→∞

x (√ x2+1−x)(√ x2+1+x√ x2+1+x )limx→∞

x (x2+1−x2)

√x2+1+x=

limx→∞

xx

√ x2x2 + 1x2+ xx=12

15)limx→−∞

[ x (√x2+1−x) ]= limx→−∞

x (√ x2+1−x)(√ x2+1+x√ x2+1+x )limx→−∞

x (x2+1−x2)

√ x2+1+x=

limx→∞

xx

(−x)√1+ 1x2x

+ xx

=1

(−1 ) (1 )+1=10

16)limx→∞

¿

a)limx→∞

3√ x3+2x2+3− 3√1−¿)limx→∞

x3+2x2+3−x3

(3√ x3+2x2+3)2+x 3√ x3+2 x2+3+x2

limx→∞

2x2

x2+ 3x2

( 3√ x3x3 +2 x2x3 + 3x3

2

)+ xx2 3√ x3+2x2+3+ x2

x2

= 21+1

=1

b)limx→∞

x2+4 x+1−x2

√ x2+4 x+1+ x=

4 xx

+ 1x

√ x2x2 + 4 xx2 +1

x2+xx

=41+1

=2

→1−2=−1

17)limx→∞

[ 3√8 x3+x2− 3√x3+x2x ]

limx→∞

8 x3+x2−x3−x2

x¿¿

limx→∞

7 x3

x¿¿

¿limx→∞

7 x2

¿¿

¿limx→∞

7 x2

x2

¿¿