calculo unid 3 evidencia

7/23/2019 calculo unid 3 evidencia

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Límite y continuidad

f ( x )= x

3−8

x−2 =0

x----2

X 1.5 1.75 1.9 1.99 1.999

2 2.001

2.01 2.1 2.25 25

Y 9.25 10.56

11.41

11.94

11.999

i ,12

12.006

12.060

12.61

13.56

15.25

Nota: de a ta!a ante"io" deducimo# $ue e ímite de %&x' cuando x-----2 e# 12

(i%e"encia de cu!o de a %o"ma &x-Y'

x¿¿¿

'

f ( x )= x

3−8

x−2 )

( x−2)( x2+2 x+4)( x−2) ) x

2+2 x+ y2=(2)2+2 (2 )+4=4+4+4=12



*n a %unci+n

2x

X Y3 63.5 73. 7.63.9 7.3.95 7.93.99 7.94 2x3

X y

5 134.5 124.3 11.64.1 11.24.05 11.14.01 11.02

*em/o 3

f ( x )= x3

X) 2 de#de a i$uie"da

X y1 11.5 3.3751.9 6.591.95 7.4141.99 7.1.995 7.941..999 7.9X) 2 de#de a de"eca

x y3 272.5 15.622.1 9.262.05 .612.01 .122.005 .06012.001 .0120

x

X



acuo de ímite# de %uncione# &eem/o#'.

1-

2-lim x→ 2

5=5

3-

lim

x→ 3

x=3

4- lim

x→ 2

4 x=4lim x→ 2

x=4 (2 )=8

5- lim

x→ 2

x . x=lim x →2

x . lim x →2

x=2.2=4

6-

x2+¿2lim

x→ 5

x=3 (52 )+2 (5 )=85

(3 x2+2 x )=¿3 lim

x →5

¿

lim x→ 5

¿

7- lim x→ 2

5 x3−3

2 x+1 =

5lim x →2

x3−lim x→ 2

5

2 lim x →2

x+ lim x →2

1 =

5(2)3−5

2 (2 )+1=

37

5

- lim

x→ 2

√ 2 x2+1=√ 2lim

x →2

x2+ lim

x →2

1=√ 2(2)2+1=3

acuo de ímite# de %uncione#

acua e ímite de a %unci+n 1

f ( x )= lim x→ 1/2

x3+5 x

4 x−6 =

(1

2 )

3

+5 (1

2 )

4 ( 12 )−6

=

1

8+5

2

2−6 =

−21

32 =−0.65


y=lim x →2

x2+2 x−1= (2 )2+2 (2 )−1=7


f ( x )= lim x→ 1/4

x2+2 x−3

x+1=(1

4)2

+2( 1

4 )−3

( 14 )+1=

1

16+(

1

2 )−3

5

4

=−39

20




f ( x )= lim x→−2

2 x2−3 x+1 x+2

=2(−2)2−3 (−2 )+1

(−2 )+2=

15

0 =+∞

LL *L L* (* L 8NN

lim x →4

y= x2− x−12

x−4 = (4 )

2

−(4 )−12

(4 )−4=0

0

!#e"aci+n #i %actu";#emo# en e nume"ado"

y= x

2− x−12

x−4 =

( x+3)( x−4) x−4

= x+3=4+3=7

<i=uiente ca#o:

lim x→ 2

y= x

2+ x−6

x2−4=

(2 )2 (2 )−6

(2)2−4=

0

0

<e %acto"i#a( x−2)( x+3)( x−2)( x+2)

=( x+3)( x+2)

=5

4

<i=uiente ca#o:

lim x→ 2

y=√ x−√ 2

x−2 =√ x−√ 2

x−2 .

√ x+√ 2

x+2 = √ ( x)

2−√ (2)2

( x−2)(√ x+√ 2)=

( x−2)

( x−2)(√ x √ 2)=

1

(√ x √ 2)=

1

√ 2√ 2

>inomio conu=ado "adica

acua e ímite de %&x':

lim x→ 0

f ( x)= x√ ( x−1 )−1

=0



√ ( x+1)2−1

¿¿

lim x→ 0

f ( x)= x

√ ( x−1 )−1.√ ( x−1 )+1

√ ( x−1 )+1=

( x √ ( x−1 )+1)

(√ ( x−1 )−1)(√ ( x−1 )+1)=( x √ ( x−1 )+1)

¿

<i=uiente ca#o:

lim x→ 2

y= 4− x

2

3−√ x2+5=

4− x2

3−√ x2+5. 3−√ x2+5

3−√ x2+5

x2+5

3+√ ¿¿ x

x2+5

3+√ ¿¿

x2+5

3+√ ¿¿

4− x2 ¿

4− x2 ¿

9−(¿¿ 2+5)=¿4− x

2 ¿¿¿

?NL@N

3ab4

√ a2b2c

3ab4

√ a2

b3

c

.

4

√ a2bc

4

√ a2

b2

c

=3ab ( 4√ a2

b c3 )

4

√ a2

b3

c a2

bc3=3 ab( 4√ a2

b c3)

4

√ a4

b4

c4

=3ab ( 4√ a2

b c3 )

abc

¿3 (

4

√ a2b c

3)c

?acionaiaci+n mediante conu=aci+n:



6

√ 5+√ 3=

6

√ 5+√ 3.√ 5−√ 3

√ 5−√ 3=

6 (√ 5−√ 3 )

√ 52−√ 32=6 (√ 5−√ 3 )

5−3

¿6 (√ 5−√ 3)

2 =3 (√ 5−√ 3)

no de o# imite# m;# im/o"tante# de c;cuo di%e"encia,/a"a cua$uie" %unci+n #e dete"mina /o" a %o"mua.

lim x→ 0

f ( x+h )−f ( x )h

(ado %&x' )

ax2

,hallar el limh →0

f ( x+h )−f ( x )

<i f ( x )=a x3

f ( x+h )=a ( x+h )3=a ( x3+3 x2+h+3 x h

2+h3 )

lim x→ 0

f ( x+h )−f ( x )h

=a x

3+3ax2

h+3axh2+ah

3

h =3 x

3+3axh2+a h

2=3 x3+3ax (0 )2+a (0 )2=3a x

2

(ado f ( x )= x2−3 x+5

si f ( x )= x2−3 x+5

si f ( x+h )=( x+h )2−3 ( x+h )+5

x2+2hx+h

2−3 x−3h+5−( x2−3 x+5)h

=2hx+h

2−3h

h =

2hx+(o)2−3(0)h

=2 x−3

(ado %&x') mx2

hallar ellimite lim x→ 0

f ( x+h )−(m x2)

h , hallar el lim

x →0

f ( x+h )− f ( x)h

si f ( x )=m x2



f ( x+h )=m( x+h)2=m ( x2+2 xh+h

2 )−(m x2)

h =

x2

m+2 xhm+h2

m−m x2

h =2 xhm+h

2m

h =2 xm+hm=2

(ado %&x') mx2


1

√ 4− x, hallar el lim

x →0

f ( x+h )−f ( x)h

lim x→ 0

1

√ 4−( x+h)

4− x−4− x−h

lim x→ 0

f ( x+h )−f ( x )h

=

1

√ 4−( x+h)−

1

√ 4− x

h =

√ 4− x−4( x+h)

√ 4−( x+h ) .√ 4− x

h = √ ¿

√ 4− ( x+h )(4− x)¿h=√ −2 x−h

16+ x2 =√

(ado %&x') mx2


√ x2−1 ,hallarel lim x→ 0

f ( x+h )−f ( x)h

(ado %&x') √ x2−1

8&x') √ ( x+h)2−1

x

√ (¿¿2+2hx+h2)−1−√ x2−1

h =√ x2+2hx+h

2−1− x2+1

h =

√ h(2 x+h)

h =√ 2 x

f ( x+h )−f ( x)h

=√ ( x+h)2−1−√ x2−1

h =¿¿

lim x →0

¿

NN((

a' lim x→ 0

1

x x)0 e Limite no exi#te



<i "e=i#t"amo# ao"e# c"eciente# y de c"eciente#tenemo#A

XB0x y

2 0.51 10.5 20.1 10

0.01 1000.001 10000.000

110000

XC0x y

-2 -0.5-0.5 -2

-0.1 -10-0.01 -100-0.001 -1000

-0.000

1

-10000

#íntota e"tica im/a".

sea f ( x )= 3 x

x+1

X -100 -10 -7 -5 -3 -1 0 1 3 5 7 10 100

y 3.03 3.33 3.5

3.75

4.5 i 0 1.5

2.25 2.50 2.62

2.75 2.97



(ete"mina" =";Dcamente a a#íntota e"tica y o"ionta.

f ( x )=2 x−5

3 x+2

X -100

-10 -7

-5 -3 -1

0 1 3 5 7 10 100

y 0.67

0.92

1 1.15

1.57

7 -2.5

-0.6

0.090

0.294

0.39

0.46

0.645

E?@NL

lim x →−∞

f ( x )=0.6664 lim x →+∞

f ( x )=0.6667



lim x →+∞

2 x−5

3 x+2=o .6667lim

x →−∞

2 x−5

3 x+2=o .6664

calculo unid 3 evidencia

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