Calculo Diferencial

DESARROLLO DE EJERCICIOS










1. Resolver los siguientes límites

a) lim 3(² - 4( + 5 = 3 (4 )²- 4 (4) +5 = 48-16+5
(→4 (³- 1 4³+ 1 64 – 1


lim 3(²-4( + 5 =37
(→4 (³- 1 63




b) lim ( 5(³ - 10( +16 =( 5 (0)³ -10 (0) +16
(→0


lim ( 5(³ - 10( +16 = ( 16 = 4
(→0


c) lim 4sen (2 y) -3cos (y) = 4 sen (2π) – 3 (cosπ)
y→π

lim 4sen 2y – 3 cosy = 3
y→π

2. Calcular el límite, si existen.

a) lim ( x + h )³ - x³ = (³ +3(²h +3xh²+ h³- x³
h→0 h h


lim 3(²h+3xh²+h³ ( lim( h (3x²+3xh +h²)
h→0 h h→0 h


lim (3(²+ 3xh +h²) ( 3(²+3( (0) + (0)² ( 3x²+3x
h→0


lim (x+h)³ ( 3(²+3( (3( (x+1)
h→0 hb) lim (²-( a-1)x-a ( lim (²- ax +x-a ( lim (² (+ax-a
x→a (³ - a³ (→a (³ - a³ (→a (³ - a³


lim x(x+1) – (x+1)a ( lim (x+1)(x-a)
x→a (x-a) x²+ax+a²) (→a (x-a)(x²+ax+a²)


lim x+1 ( a+1 ( a+1
x→a (²+ax+a² a²+a²+a² 3a²


c) lim x4 +(³+(²+x-4
x→1 x-1


(4 +(³+(²+(-4 (-1
-X4+X³x³+2(²+3(+4
/ 2(³+(²+(-4
-2(³+2(²
/ 3(²+(-4
- 3x²+3(
/ 4(-4
- 4(+4
/ /

Lim (4+(³+(²+(-4 ( lim (³+2(²+3(+4
(→1 (-1 (→1

((1)³+2(1)²+3(1) + 4(10


d) Lim (a+x - (a-x ( lim ((a+x - (a-x )((a+x + (a-x)
x→0 x x→0 x ((a+ x + ( a-x )

lim 2( ( lim 2 ( 2
x→0 x ((a +x + (a-x) x→0 (a+x + (a-x(a+0 +( a-0

( 2 ( 2 ( 2 ( 2(a ( (a
(a +0 +( a-0 (a + (a 2(a 2a a

Lim (a +x - ( a-x ( (a
(→0 x a



3. Resolver los siguientes límites infinitos.


a) lim 7(²-4(+10 ( lim 7(²/(² -4(+10/(²
(→∞ 4x²-5(-2 (→∞ 4(²/(²-5(/(²- ²/(²


lim 7- 4/x+10/(² ( lim 7-4 lim ( 1/() +10 lim (1/(²)
(→∞ 4-5/( - ²/(²(→∞ (→∞ (→∞
lim 4-5 lim ( 1/() -2 lim (1/(²)
(→∞ (→∞ (→∞


Como lim (1/x) ( 0 y lim ( 1/(²) ( 0 lim cte( cte
(→∞ (→∞ (→∞




Lim f(x) ( 7-0+0 ( 7
x→∞ 4-0-0 4


Entonces lim 7(²-4(+10 ( 7
(→∞ 4x²-5(-2 4






b) lim[((((²+9 -()] ( lim ((4+9(² - (
(→∞ (→∞


Lim (((+9 - lim x ( lim(x) lim ((²+9 - lim(x)
(→∞ (→∞ (→∞ (→∞ (→∞




Lim (x) [(lim (² + lim 9 - lim x] ( 0
x→∞ x→∞ x→∞ x→∞




c) lim 3h +4 ( (lim h→∞ 3h+4)/( lim h→∞(2h²+5 )
h→∞ (2h²-5


lim 3h+4 ( lim h→∞(3h) ( 3 limh→∞ h ( 3
h→∞ (2h²-5 lim h→∞(2h² (2 lim h→∞h (2

lim 3h+4 ( 3
h→∞ (2h²-5 (2


4. Hallar el límite de las funciones trigonométricas propuestas.


a) Lim sen(8x) + sen (4x)
(→0 sen (6x)
Utilizando la regla del Hopital ( L hopital)


Si f(a) ( g (a) (0 y existe el limite del cociente
f”(t)/g’(t) cuando t→a, setiene que:


lim f (t) ( lim f”(t) g’(t)(0
t→a g (t) t→a g’(t)


lim sen (8x) + sen(4x) ( lim [ sen(8x) + sen(4x)]'
x→0 sen (6x) x→0 [sen (6x)]'


lim 8cos (8x) +4cos(4x) ( 8cos(8*0)+4cos(4*0)
(→0 6cos (6x) 6cos (6*0)


Como cos(0) ( 1 →8(1)+4(1) ( 12 (
6(1) 6


Entonceslim sen (8x) + sen(4x) ( 2
(→0 sen (6 x)




b) lim tg y – sen y ( lim sec²y –cosy ( lim 2sec²y*tg +seny
y→0 y³ y→0 3y² y→0 6y


lim 2sec²y*tg +seny ( lim 2[sec4y+2sec²ytg²y] +cosy
y→0 6y y→0 6


2 [sec4 (0)+2 sec² (0)* tg(0)]+ cos(0) ( 2(1+2(1)(0)+1
6 6


(2+1( 3 ( 1/2...