matematika kalkulus ( limit )
TRANSCRIPT
1
Group : 1 ( Page 1-8 Kalkulus )
Name :
1. Azhari Rahman2. Muhammad Pachroni Suryana3. Yudiansyah
Class : 1EA
Exercise 1.1
Compute the value of f(x) when x has the indicated values given in (a) and (b). For (c), make an observation based on your results in (a) and (b).
1. f(x)= x+2x−5
a. x= 3.001Solutions :
f(3.001)= x+2x−5
f(x) = 3.001+23.001−5 =
−5.0011.999 = - 2.501
b. x= 2.99Solutions :
f(2.99)= x+2x−5
f(x)= 2.99+22.99−5 =
−4.992.01 = - 2.482
c. Observation?It appears than when x is close to 3 in value, then f(x) is close to -2.5 in value.
2. f(x)= x−54 x
a. x= 1.002
f(1.002)= x−54 x
2
f(x)= 1.002−54 (1.002) =
−3.9984.008 = - 0.997
b. x= .993
f(.993)= x−54 x
f(x)= .993−54 (.993) =
−4.0073.972 = - 1.008
c. Observation?It appears than when x is close to 1 in value, then f(x) is close to -1 in value.
3. f(x)= 3xx
2
a. x= .001
f(.001)= 3xx
2
f(x)= 3(.001).001
2
=0.000003.001 = 0.003
b. x= -.001
f(-.001)= 3xx
2
f(x)= 3(−.001)−.001
2
=−0.000003.001 = - 0.003
c. Observation?It appears than when x is close to 0 in value, f(x) is not close to any fixed number in value.
Exercise 1.2
Find the following limits or indicate nonexistence.
1. limx→3
x2−4x+1
Solutions : limx→3
x2−4x+1
= limx→3
x2−4
limx→3
x+1
3
= 54
2. limx→2
x2−9x−2
Solutions : limx→2
x2−9x−2
= −50 =
3. limx→1
√x3+7
Solutions : limx→1
√x3+7 = √8
= 2√2
4. limx→π¿
Solutions : limx→π¿ = 5 π2+9
5. limx→0
5−3 xx+11
Solutions : limx→0
5−3 xx+11 =
limx→0
5−3 x
limx→0
x+11
= 511
6. limx→0
9+3x2
x3+11
Solutions : limx→0
9+3x2
x3+11= limx→0
9+3x2
limx→0
x3+11
= 911
7. limx→1
x2−2 x+1x2−1
4
Solutions: limx→1
x2−2 x+1x2−1
= limx→1
(x−1)(x−1)( x−1 )(x+1)
= limx→1
x−1x+1
= 02
= 0
8. limx→ 4
6−3 xx2−16
Solutions : limx→ 4
6−3 xx2−16
= limx→4
6−3 x(x−4 )(x+4)
= −60
=
9. limx→−2
√4 x3+11
Solutions : limx→−2
√4 x3+11= √−32+11
= √21
10. limx→−6
8−3 xx−6
Solutions : limx→−6
8−3 xx−6 =
limx→−6
8−3 x
limx→−6
x−6
= −2612
= −136
5
Exercise 2.1
Evalute the following limits.
1. limx→3
x−3x2+x−12
Solutions : limx→3
x−3x2+x−12
= limx→3
x−3( x−3 )(x+4)
= limx→3
1(x+4)
= 17
2. limh→ 0
(x+h)2−x2
h
Solutions : limh→0
(x+h)2−x2
h = lim
h→ 0
x2+2hx+h2−x2
h
= limh→0
2hx+h2
h
= limh→0
h (2x+h)h
= limh→ 02x+h
= 2x
3. limx→4
x3−64x2−16
Solutions : limx→4
x3−64x2−16
= limx→4
x3−64x2−16
= limx→4
(x−4 )(x¿¿2+4 x+16)( x−4 )(x+4)
¿
= limx→ 4
x2+4 x+16x+4
= 488
6
= 6
4. If f(x) = 5x+8, find limh→0
f (x+h )−f (x)h
Solutions : limh→0
f (x+h )−f (x)h
= limh→0
(5 ( x+h )+8)−(5 x+8)h
= limh→0
(5 x+5h+8)−(5 x+8)h
= limh→0
5hh
=
5. limx→−3
5 x+7x2−3
Solutions : limx→−3
5 x+7x2−3
= limx→35 x+7
limx→3
x2−3
= −86
= - 43
6. limx→25
√ x−5x−25
Solutions : limx→25
√ x−5x−25 = lim
x→25
(√x−5)( x−25)
(√ x+5)(√ x+5)
= limx→25
x−25x √x+5 x−25√ x−125
= 00
=
7. If g(x) =x2 , find limx→2
g ( x )−g(2)x−2
Solutions : limx→2
g ( x )−g(2)x−2
= limx→2
x2−4x−2
7
= limx→2
(x+2)(x−2)x−2
= limx→2x+2
= 4
8. limx→0
2 x2−4 xx
Solutions : limx→0
2 x2−4 xx
= limx→0
(2x−4) xx
= limx→0
(2x−4 )
= -4
9. limr→0
√ x+r−√xr
Solutions : limr→0
√ x+r−√xr
= limr→0¿¿¿ ¿¿
= limr→ 0
x+r−xr (√x+r+√x )
= 00
=
10.limx→ 4
x3+6x−4
Solutions : limx→ 4
x3+6x−4
= limx→ 4
x3+6x−4
x+4x+4
= x4+4 x3+6 x+24(x−4 )(x+4)
= 700
=
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