Criterio de Stolz PDF

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Criterio de Stolz...

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ln(n) sen(n) (i) l´ım n→∞ n

e n + 2n (ii) l´ım n→∞ 5n

3 n (iii) l´ım n→∞ ln( n ) n−1

√ √ √ 1 + 2+··· + n √ (iv) l´ım n→∞ n n

•

ln(n) sen(n) = (i) l´ım n→∞ n

(ii) l´ım

n→∞

3 n n→∞ ln( n ) n−1

l´ım

#

sen(n)

=0×

=0

³ e ´n ³ 2 ´n e n + 2n = l´ ım + = [( n→∞ 5 5n 5

(iii) l´ım

n→∞

"

= l´ım

n→∞

3 n

ln(1 +

1 ) n−1

)∞ ] = 0 + 0 = 0. 3 n n→∞ 1 n−1

=[

] = l´ım

=

3(n − 1) =3 n

(iv) l´ım

n→∞

√ bn = n n

√ √ √ 1+ 2 +··· + n √ = ··· n n an = bn

√ n+1 an+1 − an √ l´ım = l´ım √ = [× n→∞ bn+1 − bn n→∞ (n + 1) n + 1 − n n

÷

√ √ √ 1 + 2 + ··· + n ]=

√ ¡ √ √ ¢ n + 1 (n + 1) n + 1 + n n √ √ l´ım ¡ √ ¢¡ √ ¢= n→∞ (n + 1) n + 1 − n n (n + 1) n + 1 + n n √ p n2 + 2n + 1 + n n2 + n 2 (n + 1)2 + n n(n + 1) l´ım = . = l´ım n→∞ n→∞ 3n2 + 3n + 1 3 ((n + 1)3 − n3 )

(i) l´ım

n→∞

n − sen(n) n 2

(iv) l´ım

n→∞

ln(n + n) n

µ ¶ 1 n→∞ n µ µ ¶¶ 1 (v) l´ım n 1 − cos n→∞ n (ii) l´ım n sen

(iii) l´ım n tan n→∞

µ

1 +1 n

¶

•

µ ¶ 1 n − sen(n) (i) l´ım = l´ım 1 − sen(n) = [sen(n) n→∞ n→∞ n n

] =1+0×

µ ¶ 1 µ ¶ sen 1 n = l´ım =[ (ii) l´ım n sen 1 n→∞ n→∞ n n µ ¶ 1 π (iii) l´ım n tan + 1 = ∞ × = ∞. n→∞ n 4 (iv) l´ım

n→∞

ln(n2 + n) =[ n

] = 1.

] = 0.

µ µ ¶¶ 1 (v) l´ım n 1 − cos =[ n→∞ n

cos(n) ln(n2 − n + 1) (i) l´ım n→∞ n

] = l´ım n n→∞

´ ³ 1 (ii) l´ım 2n e n − 1 n→∞

(iii) l´ım

1 = 0. 2n2

n→∞

µ

µ ¶ ¶n 1 1 sen +1 n n

•

cos(n) ln(n2 − n + 1) (i) l´ım = n→∞ n

=1

"

cos(n)

#

=0×

=0

³

1

´

(ii) l´ım 2n e n − 1 = l´ım 2 n→∞

(iii) l´ım

n→∞

µ

n→∞

³

1

en − 1 1 n

´

=[

] = 2.

¶n µ ¶ l´ım sen( 1n ) 1 1 = en→∞ +1 = e0 = 1 sen n n e

e n − 2n n ¡ ¢ sen(n) − tan 1n (iii) l´ım n→∞ n

(i) l´ım

n→∞

(ii) l´ım

n→∞

¡ ¡ ¢¢n 1 + sen 1n

13 + 23 + · · · + n 3 n→∞ n4

(iv) l´ım •

en e n − 2n = l´ım n→∞ n→∞ n n

(i) l´ım

µ µ ¶n ¶ 2 1− =[ e

] =∞×1=∞

µ µ ¶¶n 1 l´ım n sen( n ) 1 1 + sen = [1∞ ] = en→∞ n→∞ n

(ii) l´ım l´ım

sen

= en→∞

( 1n )

1 n

sen(n) − tan (iii) l´ım n→∞ n

(iv) l´ım

n→∞

] = e.

=[ ¡1 ¢ n

= l´ım

n→∞

¡ ¢! tan n1 1 = [0 × sen(n) − n n

Ã

n→∞

0 = 0. ∞

13 + 23 + · · · + n 3 = ··· n4 an = 13 + 23 + · · · + n3

bn l´ım

]−

bn = n4

1 an+1 − an (n + 1)3 (n + 1)3 = . = l´ım = l´ım 3 4 4 n→∞ 4n + 6n2 + 4n + 1 n→∞ (n + 1) − n 4 bn+1 − bn

cos(n) − sen(n) √ n→∞ n

(i) l´ım

(ii) l´ım

n→∞

³

1

en − •

1 ´n 2

³ ³ 1´ 1´ (iii) l´ım n2 ln 1 + sen n→∞ n n

cos(n) − sen(n) √ = l´ım n→∞ n→∞ n

(i) l´ım

µ

cos(n) sen(n) √ − √ n n

¶

= [0 ×

] =0−0=0

¶+∞ µ ¶+∞ µ ¶n µ 1 1 1 1 n (ii) l´ım e − = 0. = = 1− n→∞ 2 2 2 (iii) l´ım n2 ln(1 + n→∞

1 ln(1 + 1 ) sen( ) = l´ım 1 n→∞ n n n

1 1 ) ) n sen( n 1 n

¶n 4n3 + 2n − 1 n→∞ 4n3 + n2 − 2n + 1 4n (iii) l´ım n n→∞ 5 cos(1) + cos(2) + · · · + cos(n) (v) l´ım n→∞ n2 (i) l´ım

µ

=[

] = 1.

arctan(n) ln(n2 + 2) n→∞ n ³√ ´ √ (iv) l´ım n2 − n − n2 + 4n

(ii) l´ım

n→∞

•

(i) l´ım

n→∞

µ

3

4n + 2n − 1 4n 3 + n 2 − 2n + 1

¶n

l´ım

= e n→∞

n

µ

4n3 + 2n − 1 −1 4n3 + n2 − 2n + 1

¶

=

−n3 + 4n2 − 2n 1 3 2 e n→∞ 4n + n − 2n + 1 = e− 4 l´ım

e arctan(n) ln(n2 + 2) (ii) l´ım = n→∞ n 4n = l´ım n→∞ 5n n→∞

(iii) l´ım (iv) l´ım

n→∞

³p

·

arctan(n)

¸

=

× 0 = 0.

µ ¶n 4 =0 5

n2 − n −

p

´ n2 + 4n = [×

÷

]=

³√ ´³√ ´ √ √ n2 − n − n2 + 4n n2 − n + n2 + 4n −5n ³√ ´ ´ = = l´ım ³√ l´ım √ √ n→∞ n→∞ 2 2 2 n − n + n + 4n n − n + n2 + 4n −5 2

(v) l´ım

n→∞

cos(1) + cos(2) + · · · + cos(n) n2

bn = n2 l´ım

n→∞

an = cos(1) +cos(2) +· · ·+ cos(n)

bn

cos(n + 1) cos(n + 1) an+1 − an = l´ım = l´ım =[ 2 2 n→∞ n→∞ (n + 1) − n 2n + 1 bn+1 − bn

× 0] = 0.

¡ ¢ ¶n ln n1 + n n4 + 2n3 − 1 (i) l´ım (ii) l´ım √ n→∞ n→∞ n4 + 2n2 − 2n + 3 n ¡ √ ¢ sen(1) + sen(2) + · · · + sen(n) (iii) l´ım ln(1 + n + n) − ln(n) (iv) l´ım n→∞ n→∞ n2 µ

•

(i) l´ım

n→∞

µ

4

3

n + 2n − 1 n4 + 2n2 − 2n + 3

¶n

l´ım

= e n→∞

n

µ

n4 + 2n3 − 1 −1 n4 + 2n2 − 2n + 3

¶

=

2n4 − 2n3 + 2n2 − 4n e n→∞ n4 + 2n2 − 2n + 3 = e2 l´ım

e (ii) l´ım

ln

n→∞

¡1 ¢

n +n √ = l´ım n→∞ n

µ

− ln(n) √ √ + n n

¶

µ √ ¶ ¶ 1 1 n + n) + √ +1 = = l´ım ln n→∞ n→∞ n n n ³ ´ ¶ ln 1n + √1n + 1 µ 1 1 + √ = 0. ] = l´ım 1 n→∞ n + √1n n n

¡ √ ¢ (iii) l´ım ln(1 + n + n) − ln(n) = l´ım ln n→∞

[

(iv) l´ım

n→∞

bn = n2 l´ım

n→∞

] = 0 + ∞ = ∞.

=[ µ

1+

sen(1) + sen(2) + · · · + sen(n) n2 bn

an = sen(1)+sen(2) +· · · +sen(n)

sen(n + 1) sen(n + 1) an+1 − an = l´ım =[ = l´ım n→∞ (n + 1)2 − n2 n→∞ 2n + 1 bn+1 − bn

× 0] = 0.

µ ¶ sen(n) ln(n) 1 (i) l´ım n sen 1 + (ii) l´ım n→∞ n→∞ n n µ ¶ ln(1 + 1) + ln(1 + 12 ) + · · · + ln(1 + n1 ) 1 (iv) l´ım (iii) l´ım n tan n→∞ n→∞ n n2 •

¶ µ 1 = ∞ × sen(1) = ∞. (i) l´ım n sen 1 + n→∞ n · sen(n) ln(n) (ii) l´ım = sen(n) n→∞ n ¡ ¢ µ ¶ tan n1 1 = l´ım =[ (iii) l´ım n tan 1 n→∞ n→∞ n n (iv) l´ım

n→∞

ln(1 +

1 ) n

¸

= 0.

] = 1.

ln(1 + 1) + ln(1 + 21 ) + · · · + ln(1 + n1 ) . n2

bn = n

2

an = ln(1 + 1) + ln(1 + 12 ) + · · · +

bn 1

l´ım

ln(1 + n+1 ) an+1 − an = l´ım =[ n→∞ bn+1 − bn 2n + 1

l´ım

1 = 0. (n + 1)(2n + 1)

n→∞

n→∞

(i) l´ım

n→∞

(iv) l´ım

n→∞

µ

n3 − 2n n3 + 3n2 − 2n − 1

¶n

]=

³ ´ n cos 1n (ii) l´ım √ n→∞ n2 − 2n + 7

n2 sen (iii) l´ım

n→∞

ln(1 + a) + ln(1 + a2 ) + · · · + ln(1 + an ) ;a > 0 ( n

(i) l´ım

n→∞

3

n − 2n n3 + 3n2 − 2n − 1

¶n

l´ım

= e n→∞

n

µ

n3 − 2n −1 3 n + 3n2 − 2n − 1

1 n

en

a).

•

µ

³ ´

¶

=

−3n3 + n e n→∞ n3 + 3n2 − 2n − 1 = e−3 l´ım

e n cos

³ ´ 1 n

³1 ´ n = 1. = l´ım √ (ii) l´ım √ cos n→∞ n n2 − 2n + 7 n→∞ n2 − 2n + 7 ³ ´ n2 sen n1 (iii) l´ım ] = 0. =[ n→∞ en (iv) l´ım

n→∞

ln(1 + a) + ln(1 + a2 ) + · · · + ln(1 + an ) . n an =

ln(1 + a) + ln(1 + a2 ) + · · · + ln(1 + an )

bn = n

bn

 a ∈ (0, 1),  0 ln(1 + an ) an+1 − an = l´ım ln(1 + an ) = = l´ım ln(2) a = 1, l´ım n→∞ (n + 1) − n n→∞ bn+1 − bn n→∞  +∞ a > 1.

(i) l´ım

n→∞

µ

n4 − 3n3 − 2n2 + 1 n4 − 1

1+ (iv) l´ım

n→∞

¶n

(ii) l´ım

n→∞

µ

1 1 + ···+ n a a ;a > 0 ( ln(n)

n+1 n3 + 2

¶1 n

(iii) l´ım

n→∞

sen(n2 + 3n + 1) n2 + 1

a).

•

(i) l´ım

n→∞

µ

4

3

2

n − 3n − 2n + 1 n4 − 1

¶n

l´ım

= en→∞

n

µ

n4 − 3n3 − 2n2 + 1 −1 n4 − 1

¶

=

−3n4 − 2n3 + 2n n4 − 1 = e−3 e n→∞ l´ım

e

(ii) l´ım

n→∞

µ

n+1 n3 + 2

¶ 1n

0

= [0 ].

l´ım

n→∞

µ

n+1 n3 + 2

¶ 1n

= A,

µ ¶ ¶ µ ln(n3 + 2) n+1 1 ln(n + 1) − ln(A) = l´ım ln 3 = l´ım =[ n→∞ n n n→∞ n +2 n

] = 0.

A = e0 = 1 sen(n2 + 3n + 1) =[ n→∞ n2 + 1

(iii) l´ım

1+ (iv) l´ım

n→∞

× 0] = 0.

1 1 + ···+ n a . a ln(n) 1 1 an = 1+ +· · ·+ n a a

bn

1 n+1 1 an+1 − an a = = l´ım n+1 l´ım = l´ım n→∞ bn+1 − bn n→∞ ln(n + 1) − ln(n) n→∞ a ln(1 + n1 ) ( 1 +∞, a ≤ 1 n n l´ım = n→∞ ln(1 + 1 ) an+1 0, a>1 n

(i) l´ım

n→∞

µ

n+1 n−1

¶bn

; b ∈ IR

(ii) l´ım

n→∞

´ 1 n3 − n + 2 ³ n+7 e −1 2 n − 2n + 1

(iii) l´ım

•

(i) l´ım

n→∞

µ

n+1 n−1

¶bn

= 1,

b = 0.

b 6= 0 l´ım

n→∞

µ

n+1 n−1

¶bn

µ

n+1 l´ım bn −1 n→∞ n −1 =e

¶

l´ım

=e

´ n3 − n + 2 ³ n+7 1 − 1 =[ e n→∞ n2 − 2n + 1 ´ ³ 1 n+7 − 1 3 e n −n+2 = 1. l´ım 1 n→∞ (n2 − 2n + 1)(n + 7) n+7

(ii) l´ım

nn =[ n→∞ en

(iii) l´ım

] = ∞.

n→∞

µ

2bn n−1

¶

= e2b . ]=

nn

n→∞ en

bn = ln(n)

1 ln(n2 + 1)n (i) l´ım 2 (n − sen(n)) (ii) l´ım n→∞ n→∞ n n2 •

¡ ¡ ¢¢ sen tan 1n (iii) l´ım ¡ ¢ n→∞ 1 − cos 1 n

¶ µ 1 1 sen(n) =0× − (n − sen(n)) = l´ ım n→∞ n2 n→∞ n n2

= 0.

(i) l´ım

ln(n2 + 1)n n ln(n2 + 1) ln(n2 + 1) = l´ ım = l´ ım =[ n→∞ n→∞ n→∞ n2 n2 n ¡ ¢ ¡ ¡ ¢¢ ¡ ¡ ¢¢ 1 sen tan 1n tan n1 sen tan n1 2n2 ¡ ¢ ¡ 1¢ ¡ 1 ¢ = l´ım = (iii) l´ım 1 1 n→∞ n→∞ 1 − cos 1 − cos 1n tan n n n 2n

(ii) l´ım

[

]=0

] = l´ım 2n = ∞. n→∞

xn+1 = xn2, x0 =

(xn )n∈N 1/2 (xn )

1 2

0

(xn )

• x0 = 1/2 ≥ 0

xn+1 = xn2 ≥ 0

(xn )

(xn ) xn+1

x0 = 1/2 ≤ 1/2

xn+2 ≤ xn+1

x n ≤ 1/2

0 1 2

xn+1 = xn2 ≤ 1/4 ≤ 1/2, (xn ) x0 = 1/2 ≥ 1/4 = x1

xn+1 ≤ xn

2 xn+2 = xn+1 ≤ xn2 = xn+1 ,

xn

l´ım xn = l.

n→∞

l2 = l =⇒ l = 0

l = 1.

l=0

xn ≤ 1/2, {xn }n∈N xn+1 =

1¡ α ¢ xn + , x0 = α. 2 xn √ α

{xn }n∈N {xn }n∈N

• √ xn ≥ α ¡ ¢ √ √ √ α xn ≥ α ⇐⇒ 12 xn−1 + xn−1 ≥ α ⇐⇒ x2n−1 + α ≥ 2xn−1 α ⇐⇒ √ √ x2n−1 − 2xn−1 α + α ≥ 0 ⇐⇒ (xn−1 − α)2 ≥ 0,

xn+1 ≤ xn x2n

xn+1 ≤ xn ¡ ¢ ⇐⇒ 21 xn + xα ≤ xn ⇐⇒ n

+ α ≤ 2x2n ⇐⇒ α ≤ xn2 ⇐⇒

√ α ≤ xn ,

(xn ) ℓ = l´ım xn

(xn )

n→∞

1¡ α¢ ℓ+ ⇐⇒ 2 ℓ √ 2ℓ2 = ℓ2 + α ⇐⇒ ℓ2 = α ⇐⇒ ℓ = α, ¡ l´ım xn+1 = l´ım 21 xn +

α xn

¢

⇐⇒ ℓ =

(xn ) l´ım yn = a

(yn )

n→∞

y2 yn y1 + + ··· + 2 n . l´ım 1 n→∞ ln(n)

• ln(n) y yn y1 2 + + ··· + n = l´ım 2 l´ım 1 n→∞ n→∞ ln(n)

yn+1 yn+1 µn + 1 ¶ = l´ım µ n→∞ n+1 ln (n + 1) ln 1 + n ¶ µ 1 ln 1 + n

1 n 1 n

(xn ) xn+1 = d + xn , n ≥ 1.

yn+1 ¶ = l´ım = a, n→∞ n + 1 n l´ım yn = a

n→∞

d ∈ IR

xn = x1 + (n − 1)d, ∀n ∈ N x1 + xn x1 + · · · + xn = n . 2 ³ 1 n´ 2 l´ım + 2 + ··· + 2 . 2 n→∞ n n n •

(xn ) xn+1 = d + xn , n ≥ 1.

xn = x1 + (n − 1)d, ∀n ∈ N

n+1

x2 = d + x1 = x1 + (2 − 1)d n xn = x1 + (n − 1)d xn+1 = xn + d = x1 + (n − 1)d + d = x1 + nd. n∈N x1 + xn x1 + · · · + xn = n 2

x1 + x2 2 x1 + xn x1 + · · · + xn = n n n+1 2 x1 + xn+1 − d x1 + xn + xn+1 + xn+1 = n x1 + · · · + xn + xn+1 = n 2 2 x1 + xn+1 dn 2x1 x1 + xn+1 dn + + + x1 + nd = n − =n 2 2 2 2 2 x1 + xn+1 dn + x1 x1 =n + + 2 2 2 x1 + xn+1 xn+1 x1 x1 + xn+1 =n + + = (n + 1) . 2 2 2 2 n = 2 x1 + x2 = 2

n=2

1 + 2 + ··· + n = n

1+n n(n + 1) = . 2 2

³ 1 n´ 2 + 2 + · · · + 2 = l´ım l´ım 2 x→∞ n→∞ n n n

n(n + 1) 1 2 = . 2 n2...