Apuntes Resueltos Todo Calculo 1 1

8/17/2019 Apuntes Resueltos Todo Calculo 1 1

1/129


2/129


3/129


4/129


5/129

x3 + x2 ≥ x

x3 + x2 − x ≥ 0 ⇒ x(x2 + x − 1) ≥ 0

∆ = 1 + 4 = 5 x = 1±√

52

x

x −

1+√

52

x −

1−√ 5

2

≥ 0

] − ∞, 1−√

52 ] [

1−√ 52 , 0] [0,

1+√

52 ] [

1+√

52 , ∞[

x −

1+√

52

x

x −

1−√ 5

2

P (x)

S =[ 1−√

52

, 0]

[1+√

52

, ∞[

3x + 2

x + 1 ≤

x + 1

3x + 2 − 1x + 1

≤ 0 ⇒ 3x + 1x + 1

≤ 0 (3x + 1)(x + 1) ≤ 0 x = −1

S =] − ∞, −1[∪[− 13 , ∞[

x2 − x − 2 ≤ x

x2 − x − 2 ≥ 0(x − 2)(x + 1) ≥ 0 ⇒ x ∈] − ∞, −1] ∪ [2, ∞[


6/129

x < 0 x > 0

x2 − x − 2 ≤ x2−x − 2 ≤ 0

x ≥ −2

S = [2, ∞[

(x2 + 1)√

x − 1x2 − 7x + 10 0 ∀x ∈ R

1

x2 − 7x + 10 1


7/129

x − 1x − 2 > 1

x − 1x

−2 − 1 > 0

x − 1 − x + 2x − 2 > 0

1

x − 2 > 0x > 2

x − 1x − 2 < −1

x − 1x

−2

+ 1 < 0

x − 1 + x − 2x − 2 < 0

2x − 3x − 2 < 0

S 2 =]32 , 2[

S f =]3

2, 2[∪]2, ∞[

|x + 1| − 2|x − 3| ≥ x

x


8/129

x ≥ 3 x + 1 − 2(x − 3) ≥ x

x + 1 − 2x + 6 ≥ x−x + 7 ≥ x

2x ≤ 7x ≤ 7

2

S 3 = [3, 72 ]

, S f = [53 ,

72 ]

3x − 4|4 − 3x| − x2 ≤ 0

4 − 3x < 0 ⇒ x > 43

3x − 43x − 4 − x2 ≤ 0 ⇒

3x − 4x2 − 3x + 4 ≥ 0

∆ = 9 − 16 < 0 x2 − 3x + 4 > 0, ∀x ∈ R

3x − 4 ≥ 0 ⇒x ≥ 43

S 1 =

4

3, ∞

4

−3x

≥ 0

⇒ x

≤

4

33x − 4

4 − 3x − x2 ≤ 0 ⇒ 3x − 4

x2 + 3x − 4 ≥ 0

3x − 4(x + 4)(x − 1) ≥ 0

S p2 =] − 4, 1[∪

4

3, ∞

S 2 =] − 4, 1[∪{4/3}

S f =] − 4, 1[∪

4

3, ∞

||3x + 9| − |x|| ≥ 5


9/129

|3x + 9| − |x| ≥ 5 |3x + 9| − |x| ≤ −5

|3x + 9| = 3|x + 3| ∀x ∈ R

|x + 3| ≥ |x|

3|x + 3| ≥ |x + 3| ≥ |x| ⇒ 3|x + 3| − |x| ≥ 0

S (b) = ∅

|3x + 9| − |x| ≥ 5 ⇒ |3x + 9| ≥ 5 + |x|

x


10/129

−1 ≤ x3

+ 4

3xx

3

+ 4

3x

+ 1

≥ 0

x2 + 4 + 3x

3x ≥ 0

∆ = 9 − 16 < 0, x > 0

x

3 +

4

3x ≤ 1

x

3 +

4

3x − 1 ≤ 0

x2 + 4−

3x

3x ≤ 0

∆ = 9 − 16 < 0, x < 0

S 1 = φ x > 0;

x

3 − 4

3x

≤ 1

−1 ≤ x3 − 43x ≤ 1

−1 ≤ x3 − 4

3xx

3 − 4

3x + 1 ≥ 0

x2 − 4 + 3x3x

≥ 0(x + 4)(x − 1)

3x ≥ 0

]∞, −4] [−4, 0[ ]0, 1] [1, ∞[x + 4 − + + +

x − − + +x − 1 − − − +P (x) − + − +

S ∗ = [−4, 0[∪[1, ∞[


11/129

x

3 − 4

3x ≤ 1

x

3 − 4

3x − 1 ≤ 0

x2 − 4 − 3x3x

≤ 0(x − 4)(x + 1)

3x ≤ 0

] − ∞, −1] [−1, 0[ ]0, 4] [4, ∞[x + 1 − + + +

x − − + +x − 4 − − − +P (x) − + − +

S ∗∗ =]− ∞

,

−1]

∪]0, 4]

S 2 =S ∗ ∩ S ∗∗ = [−4, −1] ∪ [1, 4] x > 0 S 2c = [1, 4] S f = [1, 4]

x ∈ R |x − 2| + 4 − √ x − 6 < 2

|x − 2| + 4 ≥ 0

x ≥ 6 x ≥ 6 x ≥ 6

|x − 2| + 4 < 2 + √ x − 6

|x − 2| + 4 < 4 + 4√ x − 6 + x − 6 x ≥ 6 |x − 2| = x − 2

x − 2 < 4√ x − 6 + x − 64 < 4

√ x

−6

1 < √ x − 6

1 < x − 6 ⇒ x > 7

S =]7, ∞[


12/129

f (x) = |x + 1| + x2 − x − 1(x + 3)

√ x2 − 2x + 1 x ∈ R g(x) =

f (x)

g(x)

|x + 1| + x2 − x − 1(x + 3)

√ x2 − 2x + 1 ≥ 0

|x + 1| + x2 − x − 1(x + 3)

(x − 1)2 ≥ 0

√ a2 = |a|

|x + 1| + x2 − x − 1(x + 3)|x − 1| ≥ 0

|x + 1| =

x + 1 x ≥ −1−(x + 1) x


13/129

−1 ≤ x 0

⇔ |c|2 + 4|c + π|(c − e) > 0 |c| =

√ c2 ⇒ |c|2 = c2.

c2 + 4|c + π|(c − e) > 0


14/129

|c + π| =

c + π c ≥ −π−c − π c 0

⇔ c2 + 4(c2 − ec + πc − πe) > 0⇔ c2 + 4(c2 + (π − e)c − πe) > 0

⇔ 5c2 + 4(π − e)c − 4πe > 0 5c2 + 4(π − e)c − 4πe

c1 = − 4

10

(π

−e)

− 4πe

5

+ 4

10

(π

−e)

2

c2 =

−

4

10

(π

−e) +

4πe

5

+ 4

10

(π

−e)

2

5c2 + 4(π − e)c − 4πe = (c − c1)(c − c2) c1 < c2

(c − c1)(c − c2) > 0

S 1 = (] − ∞, c1[ ∪ ]c2, +∞[) ∩ [π, +∞[

c 0

⇔ c2 − 4(c + π)(c − e) > 0⇔ c2 − 4(c2 + πc − ec − πe) > 0⇔ c2 − 4(c2 + (π − e)c − πe) > 0⇔ −3c2 − 4(π − e)c + 4πe > 0⇔ 3c2 + 4(π − e)c − 4πe


15/129

f : [0, +∞[→ R f (x) = x + 1√ x2 + 1

f

x ∈ [0, +∞[ y > 0 ∀x ∈ Dom(f ) Rec(f )

x

y = x + 1√

x2 + 1

y2 = (x + 1)2

x2 + 1

x2y2 + y2 = x2 + 2x + 1

x2(y2 − 1) − 2x + (y2 − 1) = 0

x = (1 − y2) ± 4 − 4(y2 − 1)2

2(y2 − 1)

x = (1 − y2) ± 2

1 − (y2 − 1)2

2(y2 − 1)

y = {−1, 1} 1 − (y2 − 1)2 ≥ 0

|y2 − 1| ≤ 1−1 ≤ y2 − 1 ≤ 1

y2 ≥ 0y2 − 2 ≤ 0

S 2 = [−√

2,√

2]

y > 0

Rec(f ) =]0,√

2]

f (x) = |x + 1| − 2|x − 3| − x.

|x + 1| − 2|x − 3| − x ≥ 0|x + 1| − 2|x − 3| ≥ x

Dom(f ) = [5

3, 7

2]


16/129

f : A → B f (x) = x − 13x + 2

A B f (x)

Dom(f ) = R − {− 23} A = R − {−23} a, b ∈ Dom(f ) f (a) = f (b) ⇒ a = b

f (a) = a − 13a + 2

= b − 13b + 2

= f (b)

(a − 1)(3b + 2) = (b − 1)(3a + 2)3ab + 2a − 3b − 2 = 3ab + 2b − 3a − 2

5a = 5b

a = b

∀x ∈ A B = Rec(f ) x

f (x) = y

y = x − 13x + 2

⇔ y(3x + 2) = x − 1 ⇔ 3xy + 2y = x − 1

2y + 1 = x(1 − 3y) ⇔ x = 2y + 11 − 3y

B = Rec(f ) = R − { 13}

f −1(x)

f −1(x) = 2x + 1

1 − 3x

f (x) = x2 − 1 g(x) =

2x + 3 x ≤ 0√

x 0 < x ≤ 8x3 x ≥ 8

g(f (x))

g(f (x)) =

2f (x) + 3 f (x) ≤ 0

f (x) 0 < f (x) ≤ 8(f (x))3 f (x) ≥ 8

g(f (x)) =

2(x2 − 1) + 3 x2 − 1 ≤ 0 (1)√

x2 − 1 0 < x2 − 1 ≤ 8 (2)(x2 − 1)3 x2 − 1 ≥ 8 (3)


17/129

x2 − 1 ≤ 0 ⇔ x ∈ [−1, 1] x2 − 1 ≥ 8 ⇔ x2 − 9 ≥ 0 ⇔ x ∈] − ∞, −3] ∪ [3, ∞[ 0 < x2 − 1 ≤ 8

0 < x2 − 1 ⇔ x ∈] − ∞, −1[∪]1, ∞[

x2 − 1 ≤ 8 ⇔ x2 − 9 ≤ 0 ⇔ x ∈ [−3, 3] S (2) = [−3, −1[∪]1, 3]

g(f (x)) =

2(x2 − 1) + 3 x ∈ [−1, 1]√ x2 − 1 x ∈ [−3, −1[∪]1, 3]

(x2 − 1)3 x ∈] − ∞, −3] ∪ [3, ∞[

f (x) =

2 − √ 2x + 1

Dom(f )

2 − √ 2x + 1 ≥ 0

x ≥ −1

2

2 ≥ √ 2x + 1

4 ≥ 2x + 13 ≥ 2xx ≤ 3

2

Dom(f ) =

−1

2, 3

2

f (x)


18/129

a, b ∈ Dom(f ) a < b a < b

2a < 2b

2a + 1 < 2b + 1

√ 2a + 1 < √ 2b + 1−√ 2a + 1 > −

√ 2b + 1

2 − √ 2a + 1 > 2 − √ 2b + 1 2 − √ 2a + 1 >

2 −

√ 2b + 1

f (a) > f (b)

f [0, 1[→ R f (x) = 1√ 1 − x2 Rec(f )

x

y = 1√

1 − x2y2 =

1

1 − x2y2 − x2y2 = 1

x2 = y2 − 1

y2

x =

y2 − 1

y2

x ∈

[0, 1[

0 ≤

y2−1y2


19/129

2tan x

1 + tan2 x =

2 · sinxcos x1 + sin

2 xcos2 x

=

2sin x

cosxcos2 x+sin2 x

cos2 x

= 2sin x

cos x · cos2 x

= 2 sin x cos x

= sin(2x)

f (x) =

sin(2x)

Dom(f )

sin(2x) ≥ 0

sin(2x) ≥ 0 ⇔ 2x ∈ [0, π] ∪ [2π, 3π] ∪ [4π, 5π] ∪ ....⇔ x ∈

0,

π

2

∪

π, 3π

2

∪

2π, 5π

2

∪ ....

⇔ x ∈k∈Z

kπ,

(2k + 1)π

2

f (x) = 1+sin x1−cosx

f

cos x = 1

x = 2kπk ∈ Z

Dom(f ) = R− {2kπ} f

f (x) = 0 ⇔ 1 + sin x = 0⇔ sin x = −1⇔ x =

3π

2 ,

7π

2 ,

11π

2 ,...

⇔ x = 3π

2 + 2kπ

k ∈ Z


20/129

| sin x| ≤ 1 | cos x| ≤ 1

1 + sin x ≥ 0 ∧ 1 − cos x ≥ 0

1 + sin x1 − cos x ≥ 0

f

f (x) = a cos(wx) + b sin(wx)

f

f (x) =√

a2 + b2 sin(wx + φ)

√ a2 + b2

f (x)√ a2 + b2

= a√

a2 + b2 cos(wx) +

b√ a2 + b2

sin(wx)

φ

cos φ = b√

a2 + b2

sin φ = a√

a2 + b2

f (x)√ a2 + b2

= sin φ cos(wx) + cos φ sin(wx)

f (x) =

a2 + b2 sin(wx + φ).

f (x) = sin x − cos x

f (x)√ 2

= sin x√

2− cos x√

2

cos φ = 1√

2

sin φ = 1√ 2

φ = π

4

f (x)√ 2

= cos π

4 sin x − cos x sin π

4

f (x) =√

2sin

x − π4


21/129

f (x) = 3sin x cos2 x − sin3 x + 12 cos(3x)

f (x) = 2 sin x cos2 x + sin x cos2 x − sin3 x + 12

cos(3x)

= 2 sin x cos x cos x + sin x(cos2 x − sin2 x) + 12

cos(3x)

= sin(2x)cos x + sin x cos(2x) + 1

2 cos(3x)

= sin(3x) + 1

2 cos(3x)

54

f (x) 54

= sin(3x)

54

+ cos(3x)

2

54

sin φ = 1

2

54

cos φ = 1

54

tan φ = 12 ⇒ φ = arctan12

f (x) =

5

4 sin

3x + arctan

1

2

limn→∞

n3 − 3n2 + 5n − 1(n − 2)2

limn→∞

n3 − 3n2 + 5n − 1(n − 2)2 = limn→∞

n3 − 3n2 + 5n − 1n2 − 4n + 4 = limn→∞

n3

1 − 3n + 5n2 − 1n3

n3

1n − 4

n2 + 4

n3

= ∞ lim

n→∞P (n)

Q(n)

P (n) Q(n)

P (n) = a0 + a1n + a2n2 + a3n

3 + ... + aknk

Q(n) = b0 + b1n + b2n2 + b3n

3 + ... + bknk

limn→∞

P (n)

Q(n) = lim

n→∞a0 + a1n + a2n

2 + a3n3 + ... + akn

k

b0 + b1n + b2n2 + b3n3 + ... + bknk


22/129

k

limn→∞

nk

a0nk

+ a1nk−1

+ a2nk−2

+ a3nk−3

+ ... + ak

nk b0nk

+ b1nk−1

+ b2nk−2

+ b3nk−3

+ ... + bk = lim

n→∞

a0nk

+ a1nk−1

+ a2nk−2

+ a3nk−3

→0

+... + ak

b0

nk +

b1

nk−1 +

b2

nk−2 +

b3

nk−3 →0

+... + bk

= akbk

an bn cn an = bn − n+1n cn = 2bn3 limn→∞an = 2 cn

limn→∞

an = limn→∞

bn − limn→∞

n + 1

n

2 = limn→∞

bn − 1 ⇒ limn→∞

bn = 3

cn = 2bn3

limn→∞

cn =2 · lim

n→∞bn

3 =

2 · 33

= 2

cn

limn→∞

n

k=1

2k

n2

7n

limn→∞

n

k=1

2k

n2

7n

= limn→∞

1

n2 · 2 · n(n + 1)

2

7n= lim

n→∞

n + 1

n

7n= lim

n→∞

1 +

1

n

7n= e7

ĺımn→∞

1n

2

+

2n

2

+

3n

2

+ ... +

nn

2

n

1

n

2+

2

n

2+

3

n

2+ ... +

nn

2=

1 + 22 + 32 + ... + n2

n2

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

6

n2

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

6n2


23/129

ĺımn→∞

1n

2+

2n

2+

3n

2+ ... +

nn

2n

= ĺımn→∞

n(n+1)(2n+1)6n2

n

= ĺımn

→∞

(n + 1)(2n + 1)

6n2

= 1

3

ĺımn→∞

1 + 5 + 52 + 53 + ... + 5n−1

5n + n

1 + 5 + 52 + 53 + ... + 5n−1 =n

k=1

5k−1

= 1 − 5n

1 − 5

ĺımn→∞

1 + 5 + 52 + 53 + ... + 5n−1

5n + n = ĺım

n→∞

1−5n1−5

5n + n

= −14

ĺımn→∞

1 − 5n5n + n

= −14

ĺımn→∞

15n − 11 + n5n

= 1

4

ĺımn→∞

yn yn = P n n

r

b

P n = n · b


24/129

b n r h

sina

2

=

b

2r

n

a

na = 2π

a = 2π

n

b = 2r sinπ

n

P n = 2nr sinπ

n

ĺımn→∞

P n = ĺımn→∞

2nr sinπ

n

t = πn n → ∞ ⇒ t → 0

ĺımn→∞

2nr sinπ

n

= 2r ĺım

t→0π sin t

t= 2πr

(an : n ≥ 1) an = (2n)!(2θ + 1)2n

(n!)2 θ ∈ R

limn→∞ a

n+1an


25/129

α ∈ R limn→∞

nk=1

(α2 − 8)k

limn→∞

n

k=1 ark−1 = a

1 − r

|r| < 1

limn→∞

nk=1

(α2 − 8)k = limn→∞

k=1

(α2 − 8)(α2 − 8)k−1 = α2 − 8

1 − (α2 − 8) ⇔ |α2 − 8| < 1

|α2 − 8| < 1 ⇔ −1 < α2 − 8 < 1

−1 < α2 − 8 ⇔ 0 < α2 − 7 ⇔ α ∈] − ∞, −√ 7[∪]√ 7, ∞[ α2

−8 < 1

⇔ α2

−9 < 0

⇔ α

∈]

−3, 3[

α ∈] − 3, −√

7[∪]√

7, 3[

α

L = α2 − 8

1 − (α2 − 8) = α2 − 89 − α2

(P n)n∈N P 0 > 0 P n+1 = bP na + P n

a b

(P n)n∈N b − a

P n+1 = bP na + P n

limn→∞

limn→∞

P n+1 =b limn→∞

P n

a + limn→∞

P n⇒ L = bL

a + L ⇒ L(a + L) = bL

L(a + L − b) = 0 ⇒

L1 = 0L2 = b − a

a > b (P n)n∈N

a > b P n+1 − P n

P n+1 − P n = bP na + P n

− P n = bP n − aP n − P 2n

a + P n=

P n(b − a) − P 2na + P n

a > b b − a


26/129

a < b 0 < P 0 < b − a 0 < P n < b − a, ∀n ∈ N (P n)n∈N

0 < a < b 0 < P 0 < b − a

0 < P n < b − a

0 < P n+1 < b − a

0 < P n < b − a b

0 < bP n < b(b − a)

a + P n

0 < bP na + P n

< b(b − a)

a + P n

0 < bP na + P n

< b(b − a)

a + P n< b − a ⇒ 0 < P n+1 < b − a

P n+1P n

P n+1P n

= bP n

(a + P n)P n=

b

a + P n

P n < b − a ⇒ a + P n < b, P n+1

P n> 1 ⇒ P n+1 > P n

limn→∞

P n

b − a.

xn =n

k=0

1

k!

yn = xn + 1

n · n!

xn+1 − xn

xn+1 − xn =n+1k=0

1

k! −

nk=0

1

k!

xn+1 − xn =n

k=0

1

k! +

1

(n + 1)! −

nk=0

1

k!

xn+1 − xn = 1(n + 1)!

> 0


27/129

xn

yn+1 − yn = xn+1 + 1(n + 1) · (n + 1)! −

xn +

1

n · n!

yn+1 − yn = xn+1 − xn + 1(n + 1)

·(n + 1)!

− 1n

·n!

yn+1 − yn = 1(n + 1)!

+ 1

(n + 1) · (n + 1)! − 1

n · n!yn+1 − yn = 1

(n + 1) · n! + 1

(n + 1)2 · n! − 1

n · n!yn+1 − yn = n(n + 1) + n − (n + 1)

2

n(n + 1)2n!

yn+1 − yn = n2 + n + n − n2 − 2n − 1

n(n + 1)2n! = − 1

n(n + 1)2n! 0x2 − 2x + 1 > 0

−1 + 2x − x2


28/129

an ∈]0, 1[ n = 1 a1 ∈]0, 1[ n = k

ak ∈]0, 1[

n = k + 1

ak+1 ∈]0, 1[

0 < ak < 1

0 < a2k < ak

0 > −a2k > −ak2ak > 2ak − a2k > ak

ak < ak(2 − ak) < 2ak ak(2

−ak) < 1 ak

∈]0, 1[

0 < ak(2 − ak) < 1ak+1 ∈]0, 1[

{an}n∈N

an+1 − an an+1 − an = an(2 − an) − an = an(2 − an − 1) = an(1 − an)

an ∈

]0, 1[ an ∈ R+ 1

−an > 0

an+1 − an = an(1 − an) > 0 ⇒ an+1 − an > 0 ⇒ an+1 > an {an}n∈N

L = limn→∞

an L

an+1 = an(2 − an)

limn→∞

an+1 = limn→∞

an(2 − limn→∞

an)

L = L(2 − L)L = 2L − L2L2 − L = 0

L(L − 1) = 0 ⇒

L1 = 0

L2 = 1


29/129

L = 1

limn→∞

1 − n2

n2√

n2 + 1+

2 − n2n2

√ n2 + 2

+ ... + n − n2n2

√ n2 + n

an = 1−n2n2√

n2+1 + 2−n

2

n2√

n2+2 + ... + n−n

2

n2√

n2+n

1 − n2n2

√ n2 + n

+ 2 − n2

n2√

n2 + n+ ... +

n − n2n2

√ n2 + n

≤ an ≤ 1 − n2

n2√

n2 + 1+

2 − n2n2

√ n2 + 1

+ ... + n − n2n2

√ n2 + 1

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

√ n2 + n

≤ an ≤ (1 + 2 + ... + n) − (n2 + n2 + ... + n2)

n2√

n2 + 1n(n+1)

2 − n3n2

√ n2 + n

≤ an ≤n(n+1)

2 − n3n2

√ n2 + 1

n2 + n − 2n32n2

√ n2 + n

≤ an ≤ n2 + n − 2n3

2n2√

n2 + 1

limn→∞

n2 + n − 2n32n2

√ n2 + n

≤ limn→∞

an ≤ limn→∞

n2 + n − 2n3n2

√ n2 + 1

limn→∞

n3

1n

+ 1n2

− 2n3

2

1 + 1n

≤ limn→∞

an ≤ n3

1n

+ 1n2

− 2n3

2

1 + 1n2

−1 ≤ lim

n→∞an ≤ −1 ⇒ lim

n→∞an = −1

limn→∞

n + sin

nπ2

2n + 1

sinnπ2

≤ 1−1 ≤ sin

nπ2

≤ 1

n − 1 ≤ n + sinnπ

2

≤ n + 1

n − 12n − 1 ≤

n + sin

nπ2

2n + 1

≤ n + 12n + 1

limn→∞n

−1

2n + 1 ≤ limn→∞n + sin nπ2 2n + 1 ≤ limn→∞ n + 12n + 1

1

2 ≤ lim

n→∞n + sin

nπ2

2n + 1

≤ 12 ⇒ lim

n→∞n + sin

nπ2

2n + 1

= 1

2

limn→∞

an an = [α]+[2α]+[3α]+...+[nα]

n2

α > 0 [.]

ψ − 1 < [ψ] ≤ ψ


30/129

(α − 1) + (2α − 1) + ... + (nα − 1) < [α] + [2α] + ... + [nα] ≤ α + 2α + ... + nα(α − 1) + (2α − 1) + ... + (nα − 1)

n2 < [α]+[2α]+...+[nα]

n2 ≤ α + 2α + ... + nα

n2

limn→∞

α nk=1

k− nn2

< limn→∞ an ≤ limn→∞

α nk=1

kn2

limn→∞

αn(n+1)2 − n

n2 < limn→∞ an ≤ lim

n→∞αn(n+1)

2

n2

limn→∞

αn + α − 22n

< limn→∞ an ≤ limn→∞

αn(n + 1)

2n2α

2 < limn→∞ an ≤ α

2

limn→∞

[α] + [2α] + ... + [nα]

n2

= α

2

limn→∞

1

n

e

1n +

e

1n

2+

e1n

3+ ... +

e

1n

n

1

n

e

1n +

e

1n

2+

e1n

3+ ... +

e

1n

n =

1

n

nk=1

e

1n

k=

e1n

n

nk=1

e

1n

k−1

= e

1n

n ·

1 −

e

1n

n

1 − e1

n

= e1n (1 − e) ·

1n

1 − e 1n

limn→∞

1

n

e

1n +

e

1n

2+

e1n

3+ ... +

e

1n

n = lim

n→∞e

1n (1 − e) ·

1n

1 − e 1n

u = 1n

n → ∞ u → 0

limu→0

eu(1 − e) u1 − eu = (1 − e) limu→0 e

u · −ueu − 1

= e − 1

(wn)n∈N (γ n)n∈N

γ n = w1 + 2w2 + ... + nwn

1 + 2 + ... + n

γ n

wn M > 0

|wn| ≤ M


31/129

|γ n| =

nk=1

kwk

n

k=1 k

|wk| ≤ M

|γ n| ≤

nk=1

kM

nk=1

k

≤ =

M n

k=1

k

nk=1

k

|γ n| ≤ M

(γ n)n∈N

limx→

1

x2 − 2x + 1

x3

− x (00)

limx→1

x2 − 2x + 1x3 − x = limx→1

(x − 1)2x(x − 1)(x + 1)

limx→1

x − 1x(x + 1)

= 0

limx→0

√ 1 + x2 − 1

x

limx→0

√ 1 + x2 − 1x

·√ 1 + x2 + 1√

1 + x2 + 1= lim

x→01 + x2 − 1

x√

1 + x2 + 1

= limx→0

x2

x√

1 + x2 + 1 = lim

x→0x√

1 + x2 + 1= 0

limx→1

3√

7 + x3 − √ 3 + x2x − 1


32/129

limx→1

3√

7 + x3 − √ 3 + x2

x − 1 = lim

x→1

3√

7 + x3 − √ 3 + x2 + 2 − 2x − 1

= limx→1

3√

7 + x3 − 2x − 1 −

√ 3 + x2 − 2

x − 1

= limx→1

3√

7 + x3 − 2x − 1 − limx→1

√ 3 + x2 − 2

x − 1

= limx→1

3√

7 + x3 − 2x − 1 ·

3

(7 + x3)2 + 2 3√

7 + x3 + 43

(7 + x3)2 + 2 3

√ 7 + x3 + 4

− limx→1

√ 3 + x2 − 2

x − 1 ·√

3 + x2 + 2√ 3 + x2 + 2

= limx→1

x3 − 1(x − 1)

3

(7 + x3)2 + 2 3√

7 + x3 + 4 − lim

x→1x2 − 1

(x − 1) √ 3 + x2 + 2= lim

x→1x2 + x + 1

3

(7 + x3)2 + 2 3√

7 + x3 + 4− lim

x→1x + 1√

3 + x2 + 2=

3

3 · 4 − 2

4 = −1

4

limx→0

(1 + mx)n − (1 + nx)mx2

limx→0

(1 + mx)n − (1 + nx)mx2

= limx→0

nk=0

n

k

(mx)k −

mk=0

m

k

(nx)k

x2

= limx→0

n(n−1)2 m2x2 + ... + mnxn −

m(m−1)2

n2x2 + ... + nmxm

x2

= lim

x→0 n(n − 1)m2

2 + .. + mnxn−2 −

m(m − 1)n2

2 + .. + nmxm−2

=

nm(n − m)2

limx→1

xn+1 − (n + 1)x + n(x − 1)2


33/129

u = x − 1 x → 1 ⇒ u → 0

limx→1

xn+1 − (n + 1)x + n(x − 1)2

u=x−1 = lim

u→0(u + 1)n+1 − (n + 1)(u + 1) + n

u2

= limu→0

n+1

k=0n + 1

k u(n+1)−k − u(n + 1) − (n + 1) + nu2

= limu→0

n+1n+1

+n+1n

u +

n+1n−1

u2 + ... + un+1 − u(n + 1) − (n + 1) + n

u2

= limu→0

1 + u(n + 1) + 12n(n + 1)u2 + ... + un+1 − u(n + 1) − (n + 1) + n

u2

= limu→0

1

u2

1

2n(n + 1)u2 + ... + un+1

= limu→0

n(n + 1)

2 +... + un−1

→0

= n(n + 1)2

limx→0

sin(7x) − sin(5x)sin(3x) − x

limx→0

sin(7x) − sin(5x)sin(3x) − x = limx→0

7 sin(7x)7x − 5 sin(5x)5x3 sin(3x)

3x − 1=

7 − 53 − 1 = 1

limx→0

x sin x

1 − cos x

limx→0

x sin x

1 − cos x = limx→0x sin x · (1 + cos x)

sin2 x

= limx→0

sinxx

· (1 + cos x)sinxx

2= 2

limx→0

tan x − sin xx3

limx→0

tan x − sin xx3

= limx→0

sinxcosx − sin x

x3

= limx→0

sin x − sin x · cos xx3 cos x

= limx→0

sin x · (1 − cos x)x3 cos x

· 1 + cos x1 + cos x

= limx→0

sin3 x

x3 cos x(1 + cos x) =

1

2


34/129

limx→0

√ 1 + sin x − √ 1 − tan x

sin(2x)

lim

x→0

√ 1 + sin x − √ 1 − tan x

sin(2x)

= lim

x→0

√ 1 + sin x − √ 1 − tan x

sin(2x) ·

√ 1 + sin x +

√ 1 − tan x

√ 1 + sin x + √ 1 − tan x= lim

x→01 + sin x − 1 + tan x

sin(2x)√

1 + sin x +√

1 − tan x= lim

x→0sin x + tan x

2sin x · cos x √ 1 + sin x + √ 1 − tan x= lim

x→0

sin x cos x+sinxcosx

2sin x cos x√

1 + sin x +√

1 − tan x= lim

x→0sin x(cos x + 1)

2sin x cos2 x√

1 + sin x +√

1 − tan x= lim

x→0cos x + 1

2cos2 x √

1 + sin x +√

1

−tan x= 2

2(2) = 1

2

limx→π3

2cos x − 1x − π3

u = x − π3 x → π3 ⇒ u → 0

limx→π3

2cos x − 1x − π3

u=x−π3 = lim

u→02cos

u + π3

− 1u

= limu→0

2

cos u · cos π3 − sin u · sin π3− 1

u

= limu→0

212 cos u − √ 32 sin u− 1u

= limu→0

cos u − 1

u −

√ 3 · sin u

u

= −

√ 3

limx→2

sin(πx)

(x − 2) cos(πx)

t = x − 2 x → 2 u → 0

limx

→2

sin(πx)

(x−

2)cos(πx) = lim

t

→0

sin(π(t + 2))

t cos(π(t + 2))

= limt→0

sin(πt + 2π)

t cos(πt + 2π)

= limt→0

π sin(πt)

πt · 1

cos(πt + 2π)= π

limx→π4

cos x − sin xex−

π

4 − 1


35/129

u = x − π4 x → π4 ⇒ u → 0

limx→π4

cos x − sin xex−

π

4 − 1u=x−π4

= limu→0

cos

u + π4− sin u + π4

eu − 1

= limu→0

cos u cos π4 − sin u sin π4 − sin u cos π4 + cos u sin

π4 eu − 1

= limu→0

1u

√ 2

2 (cos u − sin u − sin u − cos u)

eu−1u

=

√ 2

2 limu→0

− 2 sinuu

eu−1u

= −√

2

limx→0

ex − e−xln(1 − x)

limx

→0

ex − e−xln(1

−x)

= limx

→0

ex − e−x + 1 − 1ln(1

−x)

= limx→0

ex−1x

− e−x−1x

ln(1 − x) 1x= −2

limx→0

ex2 − 1

ln(1 + x2)

x2 = t x → 0 t → 0

limt→0

et − 1ln(1 + t)

= limt→0

et−1t

1t ln(1 + t)

= limt→0

et−1t

ln(1 + t)1t

= 1

r

F

AB

ABCD

limAB→0

AF B

ABCD


36/129

x = r cos θ2 z = r sin θ2

AR = 4xz = 4r2 cos θ2 · sin θ2 AS = r

2θ2

AB → 0 ⇒ θ → 0

limAB→0

AF B

ABCD = lim

θ→0r2θ

2 · 1

4r2 cos θ2 ·

sin θ2

= 1

4 limθ→0

θ2

sin θ2· sec θ

2 =

1

4

r F AB C AB D E C F A F B

limAB→0

ABDE

AB

AB = 2r sinθ

2

DE = 2r tan

θ

2

h

h = r

1 − cos

θ

2

Atrapecio =

1

2 (DE + AB)

Atrapecio = r2

1 − cos

θ

2

tan

θ

2

+ sin

θ

2

As = r2θ

2

AB → 0 ⇒ θ → 0

limAB→0

ABDE

AB = lim

θ→01 − cos θ2

θ2

tan

θ

2

+ sin

θ

2

= 0

θ < β x2 − 2ax + b2 = 0 a > b > 0 limb→a

aβ − b2aθ − b2

θ β

x2 − 2ax + b2 = 0 ⇔ x = 2a ±√

4a2 − 4b22

⇔ x = a ±

a2 − b2


37/129

θ = a − √ a2 − b2β = a +

√ a2 − b2

limb→a

aβ − b2aθ − b2 = limb→a

a

a +√

a2 − b2− b2a

a − √ a2 − b2− b2= lim

b→aa2 − b2 + √ a2 − b2a2 − b2 − √ a2 − b2

= limb→a

√ a2 − b2 √ a2 − b2 + 1√ a2 − b2 √ a2 − b2 − 1

= limb→a

√ a2 − b2 + 1√ a2 − b2 − 1 = −1

f (x) = x2e

2x

x − 1

limx→1+

f (x) limx→1−

f (x) limx→0+

f (x) limx→0−

f (x)

limx→1+

x2e2x

x − 1 = +∞

lim

x→1−

x2e2x

x − 1 =

−∞ x = 1

limx→0+

x2e2x

x − 1

1x=u = lim

u→+∞

eu

u

21u − 1 = ∞

limx→0−

x2e2x

x − 1

1x=u = lim

u→−∞

eu

u

21u − 1 = 0

+∞

f (x)

f (x) y = mx+n m = limx→+∞

f (x)

x n = lim

x→+∞(f (x) − m

m = limx→+∞

f (x)

x = lim

x→+∞x2e

2x

x(x − 1)

= limx→+∞

xe2x

x − 1 = 1


38/129

n = limx→+∞ (

f (x) − mx) = limx→+∞

x2e

2x

x − 1 − x

= limx→+∞x2e

2x

−x(x

−1)

x − 1 = lim

x→+∞

x2

e2x − 1

x − 1

+

x

x − 1

v = 1x

x → +∞ ⇒ v → 0

limx→+∞

x2

e2x − 1

x − 1

+

x

x − 1

v= 1x

= limv→0

1

v2

e2v − 1

1−vv

+

1v

1−vv

= limv→0

e2v−1

v

1 − v + 1

1 − v

= limv→0

21 − v

e2v − 12v

+ 11 − v

= 3

n = 3 y = x + 3

f (x) = x√ x2 − 1

Dom(f ) = x ∈] − ∞, −1[∪]1, +∞[

limx→−1

f (x) = limx→−1

x√ x2

−1

= −∞ x = −1

limx→1

f (x) = limx→1

x√ x2 − 1

= +∞ x = 1 ±∞

limx→±∞

f (x) = limx→±∞

x√ x2 − 1

= limx→±∞

x

|x| 1 − 1x2

x → −∞ |x| = −x

limx→−∞

x

|x|

1 − 1x2= lim

x→−∞x

−x

1 − 1x2

= limx→−∞

−1 1 − 1

x2

= −1


39/129

x → −∞ y = −1 x → +∞ |x| = x

limx→+∞

x

|x|

1 − 1x2

= limx→+∞

x

x

1 − 1x2

= limx→+∞ 1

1 − 1x2

= 1

x → +∞ y = 1

f (x) = x2 + 2x − 1

x2

f (x) = x2 + 2x − 1

x2 Dom(f ) = R − 0 x = 0,

limx→0

f (x)

limx→0

x2 + 2x − 1x2

= −10

= −∞

x = 0 lim

x→±∞f (x)

limx→±∞

x2 + 2x − 1x2

= 1

y = 1 ∞ −∞

f (x) = 2(x − 1)2√

4x2 + 2x + 1

Dom(f ) = R

x limx→±∞

f (x) = ∞

m = limx→±∞

f (x)

x = lim

x→±∞2x2 − 4x + 2

x√

4x2 + 2x + 1= lim

x→±∞x2

2 − 4x

+ 2x2

x|x|

4 + 2

x + 1

x2

m1 = limx→−∞

x2 2 − 4x + 2x2 x(−x)

4 + 2

x + 1

x2

= limx→−∞

− 2 − 4x + 2x2 4 + 2

x + 1

x2

= −1

m2 = limx→∞

x2

2 − 4x

+ 2x2

x(x)

4 + 2x +

1x2

= limx→∞

2 − 4

x + 2

x2

4 + 2x +

1x2

= 1

n n = lim

x→±∞(f (x) − mx)


40/129

n1 = limx→−∞

2x2 − 4x + 2√

4x2 + 2x + 1+ x

= lim

x→−∞

2x2 − 4x + 2 + x√ 4x2 + 2x + 1√

4x2 + 2x + 1

= limx→−∞

x2 2 − 4x +

2x2 +

1x

√ 4x2 + 2x + 1x2 4

x2 + 2

x3 + 1

x4

= limx→−∞

2 − 4x

+ 2x2

+

4 + 2x

+ 1x2

4x2 +

2x3 +

1x4

= ∞

n2 = limx→∞

2x2 − 4x + 2√

4x2 + 2x + 1− x

= lim

x→∞

(2x2 − 4x + 2) − x√ 4x2 + 2x + 1√

4x2 + 2x + 1

· (2x

2 − 4x + 2) + x√ 4x2 + 2x + 1(2x2 − 4x + 2) + x√ 4x2 + 2x + 1

n2 = limx→∞

4(x − 1)4 − x(4x2 + 2x + 1)√ 4x2 + 2x + 1

(2x2 − 4x + 2) + x√ 4x2 + 2x + 1

= limx→∞

4(x − 1)4 − x(4x2 + 2x + 1)(2x2 − 4x + 2)√ 4x2 + 2x + 1 + x(4x2 + 2x + 1)

= ∞ f

y = ± bxa

x2

a2 − y2

b2 = 1

y = mx + n f

limx→±∞

(f (x)

−(mx + n)) = 0

x2

a2 − y

2

b2 = 1

y = b

a

x2 − a2

x → +∞

limx→+∞

b

a

x2 − a2 − bx

a

= lim

x→+∞b

a

x2 − a2 − x

·√

x2 − a2 + x√ x2 − a2 + x

= limx→+∞

b

a −a2

√ x2 − a2 + x= 0

x → −∞

limx→−∞

b

a

x2 − a2 −

−bx

a

= lim

x→∞b

a

x2 − a2 − x

= 0


41/129

a b f (x) =

ax + 2b, x ≤ 0x2 + 3a − b, 0 < x ≤ 23x − 5, x > 2

x = 0f (0) = 2b

limx→0−

(ax + 2b) = 2b, limx→0+

(x2 + 3a − b) = 3a − b

2b = 3a − b ⇒ 3b − 3a = 0 x = 2

f (2) = 4 + 3a − b

limx→2

−(x2 + 3a

−b) = 4 + 3a

−b, lim

x→2+

(3x

−5) = 1

3a − b = −3

3a − b = −33b − 3a = 0

a = −32

b = − 32

a b f (x) = x2

√ 1+x2

−1 x < 0

ax + b 0 ≤ x ≤ 2x−√ x+2√ 4x+1−3 x > 2

f x = 0 x = 2

limx→0−

f (x) = limx→0−

x2√ 1 + x2 − 1

(00)

limx→0−

x2

√ 1 + x2 − 1 = limx→0− x

2

√ 1 + x2 − 1 · √ 1 + x

2

+ 1√ 1 + x2 + 1

= limx→0−

x2√

1 + x2 + 1

1 + x2 − 1= 2

limx→0+

(ax + b) = b


42/129

b = 2 x = 2

limx→2−

(ax + b) = 2a + b

limx→2+x

−

√ x + 2

√ 4x + 1 − 3 = limx→2+x

−

√ x + 2

√ 4x + 1 − 3 · x +

√ x + 2

x + √ x + 2 ·

√ 4x + 1 + 3

√ 4x + 1 + 3= lim

x→2+

x2 − x − 2 √ 4x + 1 + 3(4x + 1 − 9) x + √ x + 2

= limx→2+

(x − 2)(x + 1) √ 4x + 1 + 34(x − 2) x + √ x + 2

= limx→2+

(x + 1)√

4x + 1 + 3

4

x +√

x + 2 = 18

16 =

9

8

2a + b = 9

8

2a = 98 − 2

a = − 716

f (x) =

√

x2 + 9 − 3x2

, x < 0

sin(x)

6x , x > 0

limx→0

f (x)

limx→0−

√ x2 + 9 − 3

x2 ·

√ x2 + 9 + 3√ x2 + 9 + 3

= limx→0−

1√ x2 + 9 + 3

= 1

6

limx→0+

sin x

6x =

1

6 limx→0+

sin x

x =

1

6

limx→0

f (x) = 1

6

c ∈ R f (x) x = 0 f (x) =

x3−πxx+sin(3x)

x = 0c x = 0

x = 0

limx→0

f (x) = f (0) = c

limx→0

f (x) = limx→0

x3 − πxx + sin(3x)

= limx→0

x(x2 − π)x

1 + sin(3x)x

= lim

x→0x2 − π

1 + 3·sin(3x)

3x

= −π4

= c


43/129

f : Dom(f ) ⊂ R → R f (x) =

√

x+3−√ 3x+1√ 1−x x π

f

x < 1

x + 3 ≥ 0 ⇔ x ≥ −33x + 1 ≥ 0 ⇔ x ≥ −1

31 − x > 0 ⇔ x


44/129

aπ + b = 3

a + b = 0

aπ + b = 3

a = 3

π−1b = − 3π−1

f R f (x) =

1−sin2(x2 )(π−x)2 x = π

b x = π

b

f

x = π

f (π) = limx→π f (x)

f (π) = b

limx→π

1 − sin2 x2 (π − x)2 = limx→π

cos2

x2

(π − x)2

u = π − x x → π ⇒ u → 0

limu→0

cos212(π − u)

u2

= limu→0

cos π2 cos

u2 + sin

π2 sin

u2

u

2

= limu→0 12 sin u2u2

2

= 1

4

f (π) = limx→π

f (x)

b = 1

4

f : R → R f (x) =

sin((1−a)x)x

x < 0

b(x − a)2 0 ≤ x ≤ 1sin(a(x

−1))

lnx x > 1

a

b

f x = {0, 1}

x = 0

limx→0−

f (x) = limx→0+

f (x)


45/129

limx→0−

f (x) = limx→0−

sin((1 − a)x)x

= limx→0−

(1 − a) sin((1 − a)x)(1

−a)x

= (1 − a)

limx→0+

f (x) = limx→0+

b(x − a)2

= a2b

1 − a = a2b x = 1

limx→1−

f (x) = limx→1+

f (x)

limx→1−

f (x) = limx→1−

b(x − a)2

= b(1 − a)2

limx→1+

f (x) = limx→1+

sin(a(x − 1))ln x

u = x − 1 x → 1+ ⇒ u → 0+

limu

→0+

sin(au)

ln(1 + u) = lim

u

→0+

a sin(au)au

1u

ln(1 + u)

= limu→0+

a sin(au)au

ln(1 + u)1u

= a

b(1 − a)2 = a

b(1 − a)2 = a1 − a = a2b ⇒

b(1 − a)2 = a1−aa2

= b⇒ (1−a)3

a2 = a ⇒ 1 −3a+3a2−a3 =

a3 ⇒ 2a3 − 3a2 + 3a − 1 = 0 p(a) = 2a3 − 3a2 + 3a − 1 p(12) = 0

(2a3 − 3a2 + 3a − 1) ÷ a − 12 = 2a2 − 2a + 2−(2a3 − a2)

(−2a2 + 3a − 1)−(−2a2 + a)

(2a − 1)−(2a − 1)

0


46/129

p(a) =

a − 1

2

(2a2 − 2a + 2)

p(a) = 0

a − 1

2

=0 (2a2 − 2a + 2) = 0

a = 12

b = 1 − 12

14

=1214

= 2

f (x) =√

x + 1 ⇒ f (x) = 12√

x+1

f (x) = limh→0

f (x + h) − f (x)h

f (x) = limh→0

(x + h) + 1 − √ x + 1

h ·

(x + h) + 1 +

√ x + 1

(x + h) + 1 +√

x + 1

= limh→0

x + h + 1 − x − 1h

(x + h) + 1 +√

x + 1

= limh→0

1√ x + h + 1 +

√ x + 1

= 1

2√

x + 1

g(x) = f (x + c) ⇒ g(x) = f (x + c)

g(x) = limh→0

g(x + h) − g(x)h

= limh→0

f ((x + h) + c) − f (x + c)h

= f (x + c)

f x = a limh→0

f (a + αh) − f (a + βh)h

limh→0f (a + αh)

−f (a + βh)

h = limh→0f (a + αh)

−f (a + βh) + f (a)

−f (a)

h

= limh→0

f (a + αh) − f (a) − (f (a + βh) − f (a))h

= limh→0

f (a + αh) − f (a)h

− limh→0

f (a + βh) − f (a)h

= α

limh→0

f (a + αh) − f (a)αh

− β

limh→0

f (a + βh) − f (a)βh

= αf (a) − βf (a) = (α − β )f (a)


47/129

f (x) =

x sin

1x

, x = 0

0, x = 0 x = 0 x = 0

limx→0

x sin1

x u = 1x x → 0 u → ∞

limu→∞

sin u

u = 0

f (0) = 0 f x = 0

f (0) = limh→0

f (h) − f (0)h

= limh→0

h

h sin

1

h

= lim

h→0sin

1

h

=

f f (x) =

sin x x < π

mx + b x ≥ π m b f x = π

f x = π f

limx

→π−

f (x) = limx

→π−

sin x

= 0

limx→π+

f (x) = limx→π+

(mx + b)

= mπ + b

mπ + b = 0

x = π limh

→0

f (π + h) − f (π)h

= f (π) f (π) = mπ +b

f −(π) = limh→0−

sin(π + h) − sin πh

= limh→0−

sin π cos h + cos π sin h

h

= cos π limh→0−

sin h

h= −1


48/129

f +(π) = limh→0+

m(π + h) + b − mπ − bh

= limh→0+

mπ + mh + b − mπ − bh

= limh→0+

mhh

= m

m = −1

−π + b = 0b = −π

f f (x) = x5 + x2 + 6 x


49/129

3a + b = 0

a + b = 6

3a + b = 0⇔

b = 6 − a3a + 6 − a = 0

a = −3b = 9

f

f (x) =

2x2

x+a x < a

2x − ex−a x ≥ a

f

x → a± a f a = 1

limx→a−

f (x) = limx→a−

2x2

x + a

= 2a2

2a= a

limx→a+

f (x) = limx→a+

2x − ex−a

= 2a − 1

a = 2a − 1a = 1

a = 1

a = 1 f (1) (1, f (1))

f (1)

f (x) =

2x2

x+1 x < 1

2x − ex−1 x ≥ 1


50/129

f −(1) = limh→0−

f (1 + h) − f (1)h

= limh→0

−

2(1+h)2

1+h+1 − 1

h= lim

h→0−2(h2 + 2h + 1) − h − 2

h(h + 2)

= limh→0−

2h2 + 4h + 2 − h − 2h(h + 2)

= limh→0−

2h2 + 3h

h(h + 2)

= limh→0−

2h + 3

h + 2

= 3

2

f +(1) = limh→0+

f (1 + h) − f (1)h

= limh→0+

2(1 + h) − eh − 1h

= limh→0+

2 + 2h − eh − 1h

= limh→0+

2 − e

h − 1h

= 1

x = 1

f f (x) =

1−cos x

x x < 0

ax + b 0 ≤ x 3

a

b

f

x = 0 x = 1

x = 0

limx→0−

f (x) = limx→0−

1 − cos xx

= 0

limx→0+

f (x) = limx→0+

(ax + b)

= b

b = 0


51/129

x = 1

limx→1−

f (x) = limx→1−

(ax + b)

= 2a + b

limx→1+

f (x) = limx→1+

(x2 + 2)

= 3

2a + b = 3

a = 3

2

f

f (x) =

1−cosx

x x < 0

32x 0 ≤ x 3

f

x = 3 x = 0 x = 4

x = 0

f −(0) = ĺımh→0−

f (h) − f (0)h

= ĺımh→0−

1−cos(h)h

− 0h

= ĺımh→0−

1 − cos(h)h2

· 1 + cos(h)1 + cos(h)

= ĺımh→0−

sin2(h)

h2 · 1

1 + cos(h)

= 1

2

f +(0) = ĺımh→0+

f (h) − f (0)h

= ĺımh→0+

32

h − 0h

= 3

2


52/129

f (0) x = 3

f −(3) = ĺımh→0−

f (3 + h) − f (3)h

= ĺımh→0

−

(3 + h)2 + 2 − 11h

= ĺımh→0−

9 + 6h + h2 − 11h

= ĺımh→0−

h(6 + h)

h= 6

f +(3) = ĺımh→0+

(3+h)3−3(3+h)22(3+h)2−18 − 32

h

= ĺımh→0+

(3+h)2((3+h)−3)2((3+h)

−3)((3+h)+3)

− 32

h

= ĺımh→0+

(3 + h)2 − 3((3 + h) + 3)2h((3 + h) + 3))

= ĺımh→0+

9 + 6h + h2 − 9 − 3h − 92h(6 + h)

= ĺımh→0+

h2 + 3h − 92h(6 + h)

= −∞

f (3) x = 4 x ∈]3, +∞[

f (x) = (3x2 − 6x)(2x2 − 18) − (x3 − 3x2)4x

(2x2 − 18)2

f (4) = (64 − 24)(32 − 18) − (64 − 48)16

(32 − 18)2

= 40 · 14 − 16 · 16

142

= 76

49

f +∞

m = ĺımx→∞

f (x)

x

= ĺımx→∞

x3−3x22x2−18

x

= ĺımx→∞

x3 − 3x22x3 − 18x

= 1

2


53/129

n = ĺımn→∞

(f (x) − mx)

= ĺımx→∞

x3 − 3x22x2

−18

− 12

x

=

1

2 ĺımx→∞

x3 − 3x2 − x3 + 9xx2 − 9

= −3

2

y = x

2 − 3

2

f (x) = 2√ x + x2

f (x) = 2 · 12√

x + 2x

= 1√

x + 2x

f (x) = ln x · sin x + tan xx2

f (x) = sin x

x + ln x · cos x + x

2 sec2 x − 2x tan xx4

= sin x

x + ln x · cos x + x sec

2 x − 2tan xx3

f (x) = (√

x+ 3√

x) cos xex tanx

f (x) =ddx ((

√ x + 3

√ x)cos x) · ex tan x − (√ x + 3√ x)cos x · ddx (ex tan x)

e2x tan2 x

=

12√

x + 2

3 3√

x2 cos x − (√ x + 3√ x)sin x ex tan x − (√ x + 3√ x)cos x · ex tan x + ex sec2 x

e2x tan2 x

=

12√

x + 2

3 3√

x2

cos x − (√ x + 3√ x)sin x

ex tan x − cos x · ex (√ x + 3√ x) tan x + sec2 x

e2x tan2 x

f (x) =

2xx2+1


54/129

f (x) = 1

2

2xx2+1

· ddx

2x

x2 + 1

=

1

2 2xx2+1 · 2(x2 + 1)

−2x

·2x

(x2 + 1)2

= 1

2xx2+1

· 1 − x2

(x2 + 1)2

= 1 − x2√

2x · (x2 + 1) 32

f (x) = ln

etanx√ 2x3+5

f (x) = ln etan x

−ln 2x3 + 5

= tan x − 12

ln(2x3 + 5)

f (x) = sec2 x − 6x2

2(2x3 + 5)

= sec2 x − 3x2

(2x3 + 5)

f (x) = 3

3√

x + 1−sin xcos(ln x)

f (x) = 2

3 3

3√

x + 1−sinxcos(lnx)2 · ddx

3√

x + 1 − sin xcos(ln x)

= 2

3 3

3√

x + 1−sinxcos(lnx)2 ·

32√

x +

− cos x · cos(ln x) + (1 − sin x) ·

sin(lnx)x

cos2(ln x)

=

2

3 3

3√

x + 1−sinxcos(lnx)2 ·

2

3√

x +

(1 − sin x) sin(ln x) − x cos x · cos(ln x)x · cos2(ln x)

f (x) = arctan ln(1 + tan(sin(3x2)))

f (x) = 1

1 + ln(1 + tan(sin(3x2))) · 1

2

ln(1 + tan(sin(3x2)))· 1

1 + tan(sin(3x2)) · sec2(sin(3x2)) · cos(3x2) · 6x

= 3x · sec2(sin(3x2)) · cos(3x2)

(1 + ln(1 + tan(sin(3x2))))

ln(1 + tan(sin(3x2))) · (1 + tan(sin(3x2)))


55/129

w(φ) = cos

1√ φ

exp

θφ2

+ φθ ln (θ − φ) dwdφ

dw

dφ =

d

dφ

cos

1√

φ exp

θφ2

+ d

dφ

φ

θ ln (θ − φ)

= − sin 1√

φ exp θφ2 · d

dφ 1√

φ exp θφ2+ 1

θ ln(θ − φ) − φ

θ(θ − φ)

= − sin

1√ φ

exp

θφ2 ·− 1

2

φ3exp

θφ2

+

2θφ√ φ

exp

θφ2

+ 1

θ ln(θ − φ) − φ

θ(θ − φ)

= sin

1√

φ exp

θφ2

· exp θφ2 · 12

φ3− 2θφ√

φ

+

(θ − φ)ln(θ − φ) − φθ(θ − φ)

w = w(t) w = ln √

u − eu u(t) = sin(5t) d2w

dt2 t

w = ln √

u − eu

w = 1

2 ln(

√ u − eu)

dw

dt =

1

2

1√ u − eu

1

2√

u

du

dt − eu du

dt

dw

dt =

1

2 · du

dt · 1√

u − eu

1

2√

u − eu

d2w

dt2 =

1

2 d2udt2 1√ u − eu 12√ u − eu−dudt 2

1

(√ u − eu)2 12√ u − eu+ dudt · 1√ u − eu −14√ u3 dudt − eu dudt=

1

2

d2u

dt21√

u − eu

1

2√

u − eu

−

du

dt

21

(√

u − eu)2

1

2√

u − eu

−

du

dt

2· 1√

u − eu

1

4√

u3+ eu

u(t) = sin(5t) du

dt = 5 cos(5t)

d2u

dt2 = −25 sin(5t)

y + xy3 = 2 +√

xy

y = y(x)

y + y3 + 3xy2y = 1

2√

xy · (1 + xy)

y + 3xy2y − xy

2√

xy =

1

2√

xy − y3

y =1

2√

xy − y3

1 + 3xy2 +√

x2√

y


56/129

sin(x2 + y) − 2xy3 = arctan(y − x) + 2

cos(x2 + y) · (2x + y) − 2(y3 + 3xy2y) = 11 + (y − x)2 · (y

− 1)

2x cos(x2 + y) + y cos(x2 + y) − 2y3 − 6xy2y = y1 + (y − x)2 −

11 + (y − x)2

y

cos(x2 + y) − 6xy2 − 11 + (y − x)2

= 2y3 − 1

1 + (y − x)2 − 2x cos(x2 + y)

y =2y3 − 11+(y−x)2 − 2x cos(x2 + y)cos(x2 + y) − 6xy2 − 11+(y−x)2

w(x) = uv u = u(x) v = v(x)

dw

dx

ln(·)

w = uvln(·) ⇒ ln w = v · ln ud

dx(·) ⇒ 1

w

dw

dx =

dv

dx · ln u + v · 1

u · du

dx

⇒ dwdx

= w

dv

dx · ln u + v

u · du

dx

⇒ dw

dx = uv

dv

dx · ln u + v

u · du

dx

f (x) =

3x2 +

√ x + ex cos x

arctan x

u v

u = 3x2 +√

x + ex cos x ⇒ dudx

= 6x + 1

2√

x + ex(cos x − sin x)

v = arctan x ⇒ dvdx

= 1

1 + x2

f (x) =

3x2 +√

x + ex cos xarctan x· 11+x2 · ln 3x2 + √ x + ex cos x+ arctanx3x2+√ x+ex cos 6x + 12√ x + ex(cos x − si

dy

dx

y = xx2

ln()

ln y = x2 ln x

1

yy = 2x ln x + x

y = xx2

(2x ln x + x)


57/129

y = ln(1−sinhx)(1 + tan x)

ln y = (1 − sinh x) ln(ln(1 + tan x))1

y y = − cosh x · ln(ln(1 + tan x)) + (1

−sinh x)sec2 x

(1 + tan x) ln(1 + tan x)

y = ln(1−sinhx)(1 + tan x) ·− cosh x · ln(ln(1 + tan x)) + (1 − sinh x)sec

2 x

(1 + tan x) ln(1 + tan x)

y =

1 + cosh x

1 − exp(x2)

ln y = ln

1 + cosh x

1 − exp(x2)

ln y = 1

2 ln

1 + cosh x

1−

exp(x2)ln y =

1

2

ln (1 + cosh x) − ln 1 − exp(x2)

1

yy =

1

2

sinh x

1 + cosh x +

2x exp(x2)

1 − exp(x2)

y = 1

2

1 + cosh x

1 − exp(x2) ·

sinh x

1 + cosh x +

2x exp(x2)

1 − exp(x2)

T

L

T = 2π

L

g

g u L

dL

du = kL

T (u) = kT

2 .

T = 2π

L

g ⇒ T = π

Lg

· 1g · dL

du

dL

du = kL

T = π

Lg

· 1g · kL ⇒ T = kπ

L

g

T

2 = π

L

g

T = kT

2


58/129

M y = M x2e2x d2y

dx2 − 4 dy

dx + 4y = 6e2x

y y

y = M x2e2x

dydx =

M (2xe2x + 2x2e2x)

d2y

dx2 = M

2(e2x + 2xe2x) + 2(2xe2x + 2x2e2x)

= 2Me2x

1 + 4x + 2x2

d2y

dx2 − 4 dy

dx + 4y = 2Me2x(1 + 4x + 2x2) − 4Me2x(2x + 2x2) + 4Mx2e2x

= Me2x(2 + 8x + 4x2 − 8x − 8x2 + 4x2)= 2Me2x

d

2

ydx2 − 4 dydx + 4y = 6e2x

2M e2x = 6e2x

M = 3

d2u

dx2 + x2u = 0 (∗)

u = y√

x (∗) x2 d2y

dx2 + x

dy

dx + y

x4 − 1

4

= 0 (∗∗)

u = y√

x

du

dx = x1

2

dy

dx + 1

2 yx−1

2

d2u

dx2 =

1

2x−

12

dy

dx + x

12

d2y

dx2 +

1

2

dy

dx · x− 12 − 1

2yx−

32

(∗)

d2u

dx2 + x2u =

1

2x−

12

dy

dx + x

12

d2y

dx2 +

1

2

dy

dx · x− 12 − 1

2yx−

32

+ x2 · (yx 12 ) = 0

= 1

2x−

12

dy

dx + x

12

d2y

dx2 +

1

2

dy

dx · x− 12 − 1

4yx−

32 + yx

52 = 0

= x12

d2y

dx2 + x−

12

dy

dx + y

x

52 − 1

4x−

32

= 0

x32

x2d2y

dx2 + x

dy

dx + y

x4 − 1

4

= 0

w = x2

2 (∗∗)

w2 d2y

dw2 + w

dy

dw + y

w2 − 1

16

= 0


59/129

y = f (w(x))

dy

dx =

dy

dw · dw

dx

dwdx = x

dy

dx = x

dy

dwd2y

dx2 =

dy

dw + x

d2y

dw2 · dw

dx

= dy

dw + x2

d2y

dw2

(∗∗)

x2d2y

dx2 + x

dy

dx + y

x4 − 1

4 = x2

dy

dw + x2

d2y

dw2+ x2

dy

dw + y

x4 − 1

4= 0

= x4 d2y

dw2 + 2x2

dy

dw + y

x4 − 1

4

= 0

x2 = 2w

4w2 d2y

dw2 + 4w

dy

dw + y

4w2 − 1

4

= 0

w2 d2y

dw2 + w

dy

dw + y

w2 − 1

16

= 0

x(t) = x0 cos k

mt

k m

d2x

dt2 +

k

mx = 0

dx

dt = −x0

k

m sin

k

mt

d2x

dt2 = −x0 · k

m cos km t

d2x

dt2 +

k

mx = −x0 · k

m cos

k

mt

+ x0

k

m cos

k

mt

d2x

dt2 +

k

mx = 0


60/129

f, g : R → R f (0) = 0 g(0) = 1 f (x) = g(x) g(x) = f (x). h(x) = (f (x))2 − (g(x))2

h(x)

h(x) = 2f (x)f (x)−

2g(x)g(x)

= 2 (f (x)f (x) − g(x)g(x))

h(x) = 2 (f (x)g(x) − g(x)f (x))= 0

h(x) = 0 h(x) x = 0

h(0) = f (0) − g(0)= −1

h(x) = −1

f (x) = x2 + x

x0 ∈ Dom(f )

Documents

Apuntes Resueltos Todo Calculo 1 1