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lim
𝑥→𝑐
𝑎 = 𝑎
Examples:
a. lim
𝑥→2
4 = 4 b. lim
𝑥→6
9 = 9
lim
𝑥→𝑐
𝑥 = 𝑐
Examples:
a. lim
𝑥→5
𝑥 = 5 b. lim
𝑥→−8
𝑥 = −8
lim
𝑥→𝑐
𝑎 ∙ 𝑓 𝑥 = 𝑎 lim
𝑥→𝑐
𝑓(𝑥)
Examples:
a. lim
𝑥→2
4𝑥 = 4lim
𝑥→2
𝑥 b. lim
𝑥→−3
5𝑥 = 5 lim
𝑥→−3
𝑥
= 4 2 = 5(−3)
= 8 = −15
lim
𝑥→𝑐
𝑓 𝑥 + 𝑔(𝑥) = lim
𝑥→𝑐
𝑓(𝑥) + lim
𝑥→𝑐
𝑔(𝑥)
Examples:
a. lim
𝑥→2
(𝑥 + 7) = lim
𝑥→2
𝑥 + lim
𝑥→2
7
= 2 + 7
= 9
b. lim
𝑥→−6
(𝑥 + 8) = lim
𝑥→−6
𝑥 + lim
𝑥→−6
8
= −6 + 8
= 2
lim
𝑥→𝑐
𝑓 𝑥 − 𝑔(𝑥) = lim
𝑥→𝑐
𝑓(𝑥) − lim
𝑥→𝑐
𝑔(𝑥)
Examples:
a. lim
𝑥→3
(𝑥 − 5) = lim
𝑥→3
𝑥 − lim
𝑥→3
5
= 3 − 5
= −2
b. lim
𝑥→−2
(𝑥 − 6) = lim
𝑥→−2
𝑥 − lim
𝑥→−2
6
= −2 − 6
= −8
lim
𝑥→𝑐
𝑓 𝑥 ∙ 𝑔(𝑥) = lim
𝑥→𝑐
𝑓(𝑥) ∙ lim
𝑥→𝑐
𝑔(𝑥)
Examples:
a. lim
𝑥→2
3(2𝑥 + 1) = lim
𝑥→2
3 ∙ 2lim
𝑥→2
𝑥 + lim
𝑥→2
1
= 3 ∙ [ 2 2 + 1 ]
= 3 ∙ 5
= 15
b. lim
𝑥→3
𝑥(5𝑥 + 4) = lim
𝑥→3
𝑥 ∙ 5lim
𝑥→3
𝑥 + lim
𝑥→3
4
= 3 ∙ [ 5 3 + 4 ]
= 3 ∙ 19
= 57
lim
𝑥→𝑐
𝑓(𝑥)
𝑔(𝑥)
=
lim
𝑥→𝑐
𝑓(𝑥)
lim
𝑥→𝑐
𝑔(𝑥)
provided that lim
𝑥→𝑐
𝑔(𝑥) ≠ 0
Examples:
a. lim
𝑥→2
2𝑥+5
𝑥+3
=
lim
𝑥→2
(2𝑥+5)
lim
𝑥→2
(𝑥+3)
=
2lim
𝑥→2
𝑥 + lim
𝑥→2
5
lim
𝑥→2
𝑥 + lim
𝑥→2
3
=
2 2 + 5
2 + 3
=
9
5
b. lim
𝑥→4
5𝑥+5
𝑥+1
=
lim
𝑥→4
(5𝑥+5)
lim
𝑥→4
(𝑥+1)
=
5lim
𝑥→4
𝑥 + lim
𝑥→4
5
lim
𝑥→4
𝑥 + lim
𝑥→4
1
=
5 4 + 5
4 + 1
=
25
5
or 5
lim
𝑥→𝑐
𝑓 𝑥 𝑛 = lim
𝑥→𝑐
𝑓 𝑥
𝑛
Examples:
a. lim
𝑥→1
(3𝑥 + 4)3 = lim
𝑥→1
(3𝑥 + 4)
3
= 3 lim
𝑥→1
𝑥 + lim
𝑥→1
4
3
= 3(1) + 4 3
= (7)3
= 343
a. lim
𝑥→−3
(4𝑥 + 9)2
= lim
𝑥→−3
(4𝑥 + 9)
2
= 4 lim
𝑥→−3
𝑥 + lim
𝑥→−3
9
2
= 4(−3) + 9 2
= (−3)2
= 9
lim
𝑥→𝑐
𝑛
𝑓(𝑥) = 𝑛
lim
𝑥→𝑐
𝑓(𝑥)
Examples:
a. lim
𝑥→2
3
𝑥2 + 4 = 3
lim
𝑥→2
(𝑥2 + 4)
= 3
lim
𝑥→2
𝑥2 + lim
𝑥→2
4
=
3
(2)2+ 4
=
3
8
= 2
b. lim
𝑥→4
𝑥 + 12 = lim
𝑥→4
(𝑥 + 12)
= lim
𝑥→4
𝑥 + lim
𝑥→4
12
= 4 + 12
= 16
= 4
lim
𝑥→𝑐
𝑓(𝑥) = 𝑓 𝑐
Provided that 𝑓(𝑐) is a real number
Examples:
a. lim
𝑥→5
(2𝑥 − 8) = 2lim
𝑥→5
𝑥 − lim 8
𝑥→5
= 2 5 − 8
= 2
Since 2𝑥 − 8 is a polynomial and 5 is a real
number, you could also readily get the limit by
getting f(5); provided that f(5) is a real number
𝑓 𝑥 = 2𝑥 − 8
𝑓 5 = 2 5 − 8
= 2
b. lim
𝑥→3
(4𝑥2
+ 2𝑥 − 1) = 4lim
𝑥→3
𝑥2
+ 2lim
𝑥→3
𝑥 − lim
𝑥→3
1
= 4(3)2+ 2 3 − 1
= 41
Since 4𝑥2
+ 2𝑥 − 1 is a polynomial and 3 is a real number, you
could also readily get the limit by getting f(3); provided that
f(3) is a real number.
𝑓 𝑥 = 4𝑥2 + 2𝑥 − 1
𝑓 3 = 4(3)2
+2(3) − 1
= 41
Therefore, 𝒍𝒊𝒎
𝒙→𝟓
𝒇 𝒙 = 𝒇 𝟓 = 𝟐 Therefore, 𝒍𝒊𝒎
𝒙→𝟑
𝒇 𝒙 = 𝒇 𝟑 = 𝟒𝟏
Limit_Laws

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Limit_Laws

  • 1.
  • 2. lim 𝑥→𝑐 𝑎 = 𝑎 Examples: a. lim 𝑥→2 4 = 4 b. lim 𝑥→6 9 = 9
  • 3. lim 𝑥→𝑐 𝑥 = 𝑐 Examples: a. lim 𝑥→5 𝑥 = 5 b. lim 𝑥→−8 𝑥 = −8
  • 4. lim 𝑥→𝑐 𝑎 ∙ 𝑓 𝑥 = 𝑎 lim 𝑥→𝑐 𝑓(𝑥) Examples: a. lim 𝑥→2 4𝑥 = 4lim 𝑥→2 𝑥 b. lim 𝑥→−3 5𝑥 = 5 lim 𝑥→−3 𝑥 = 4 2 = 5(−3) = 8 = −15
  • 5. lim 𝑥→𝑐 𝑓 𝑥 + 𝑔(𝑥) = lim 𝑥→𝑐 𝑓(𝑥) + lim 𝑥→𝑐 𝑔(𝑥) Examples: a. lim 𝑥→2 (𝑥 + 7) = lim 𝑥→2 𝑥 + lim 𝑥→2 7 = 2 + 7 = 9 b. lim 𝑥→−6 (𝑥 + 8) = lim 𝑥→−6 𝑥 + lim 𝑥→−6 8 = −6 + 8 = 2
  • 6. lim 𝑥→𝑐 𝑓 𝑥 − 𝑔(𝑥) = lim 𝑥→𝑐 𝑓(𝑥) − lim 𝑥→𝑐 𝑔(𝑥) Examples: a. lim 𝑥→3 (𝑥 − 5) = lim 𝑥→3 𝑥 − lim 𝑥→3 5 = 3 − 5 = −2 b. lim 𝑥→−2 (𝑥 − 6) = lim 𝑥→−2 𝑥 − lim 𝑥→−2 6 = −2 − 6 = −8
  • 7. lim 𝑥→𝑐 𝑓 𝑥 ∙ 𝑔(𝑥) = lim 𝑥→𝑐 𝑓(𝑥) ∙ lim 𝑥→𝑐 𝑔(𝑥) Examples: a. lim 𝑥→2 3(2𝑥 + 1) = lim 𝑥→2 3 ∙ 2lim 𝑥→2 𝑥 + lim 𝑥→2 1 = 3 ∙ [ 2 2 + 1 ] = 3 ∙ 5 = 15 b. lim 𝑥→3 𝑥(5𝑥 + 4) = lim 𝑥→3 𝑥 ∙ 5lim 𝑥→3 𝑥 + lim 𝑥→3 4 = 3 ∙ [ 5 3 + 4 ] = 3 ∙ 19 = 57
  • 8. lim 𝑥→𝑐 𝑓(𝑥) 𝑔(𝑥) = lim 𝑥→𝑐 𝑓(𝑥) lim 𝑥→𝑐 𝑔(𝑥) provided that lim 𝑥→𝑐 𝑔(𝑥) ≠ 0 Examples: a. lim 𝑥→2 2𝑥+5 𝑥+3 = lim 𝑥→2 (2𝑥+5) lim 𝑥→2 (𝑥+3) = 2lim 𝑥→2 𝑥 + lim 𝑥→2 5 lim 𝑥→2 𝑥 + lim 𝑥→2 3 = 2 2 + 5 2 + 3 = 9 5 b. lim 𝑥→4 5𝑥+5 𝑥+1 = lim 𝑥→4 (5𝑥+5) lim 𝑥→4 (𝑥+1) = 5lim 𝑥→4 𝑥 + lim 𝑥→4 5 lim 𝑥→4 𝑥 + lim 𝑥→4 1 = 5 4 + 5 4 + 1 = 25 5 or 5
  • 9. lim 𝑥→𝑐 𝑓 𝑥 𝑛 = lim 𝑥→𝑐 𝑓 𝑥 𝑛 Examples: a. lim 𝑥→1 (3𝑥 + 4)3 = lim 𝑥→1 (3𝑥 + 4) 3 = 3 lim 𝑥→1 𝑥 + lim 𝑥→1 4 3 = 3(1) + 4 3 = (7)3 = 343 a. lim 𝑥→−3 (4𝑥 + 9)2 = lim 𝑥→−3 (4𝑥 + 9) 2 = 4 lim 𝑥→−3 𝑥 + lim 𝑥→−3 9 2 = 4(−3) + 9 2 = (−3)2 = 9
  • 10. lim 𝑥→𝑐 𝑛 𝑓(𝑥) = 𝑛 lim 𝑥→𝑐 𝑓(𝑥) Examples: a. lim 𝑥→2 3 𝑥2 + 4 = 3 lim 𝑥→2 (𝑥2 + 4) = 3 lim 𝑥→2 𝑥2 + lim 𝑥→2 4 = 3 (2)2+ 4 = 3 8 = 2 b. lim 𝑥→4 𝑥 + 12 = lim 𝑥→4 (𝑥 + 12) = lim 𝑥→4 𝑥 + lim 𝑥→4 12 = 4 + 12 = 16 = 4
  • 11. lim 𝑥→𝑐 𝑓(𝑥) = 𝑓 𝑐 Provided that 𝑓(𝑐) is a real number Examples: a. lim 𝑥→5 (2𝑥 − 8) = 2lim 𝑥→5 𝑥 − lim 8 𝑥→5 = 2 5 − 8 = 2 Since 2𝑥 − 8 is a polynomial and 5 is a real number, you could also readily get the limit by getting f(5); provided that f(5) is a real number 𝑓 𝑥 = 2𝑥 − 8 𝑓 5 = 2 5 − 8 = 2 b. lim 𝑥→3 (4𝑥2 + 2𝑥 − 1) = 4lim 𝑥→3 𝑥2 + 2lim 𝑥→3 𝑥 − lim 𝑥→3 1 = 4(3)2+ 2 3 − 1 = 41 Since 4𝑥2 + 2𝑥 − 1 is a polynomial and 3 is a real number, you could also readily get the limit by getting f(3); provided that f(3) is a real number. 𝑓 𝑥 = 4𝑥2 + 2𝑥 − 1 𝑓 3 = 4(3)2 +2(3) − 1 = 41 Therefore, 𝒍𝒊𝒎 𝒙→𝟓 𝒇 𝒙 = 𝒇 𝟓 = 𝟐 Therefore, 𝒍𝒊𝒎 𝒙→𝟑 𝒇 𝒙 = 𝒇 𝟑 = 𝟒𝟏