Fsc Part 2 Mathematics (Complete Solutions)

Q1
Evaluate each limit by using theorems of limits:

(i)           Lim
x → 3
  (2x + 4)
(ii)          Lim
x → 1
  (3x2 - 2x + 4)
(iii)         Lim
x → 3
  x2 + x + 4
(iv)         Lim
x → 2
  xx2 - 4
(v)          Lim
x → 2
  (x3 + 1 - x2 + 5)
(vi)         Lim
x → -2
  
2x3 + 5x
3x - 2

Solution

(i)           Lim
x → 3
  (2x + 4)
Lim
x → 3
(2x)  +  Lim
x → 3
(4)
= 2 Lim
x → 3
(x)  +  Lim
x → 3
(4)
Now by applying limit we get:
= 2(3) + 4
= 6 + 4
= 10

(ii)          Lim
x → 1
  (3x2 - 2x + 4)
Lim
x → 1
(3x2)  -  Lim
x → 1
(2x)  +  Lim
x → 1
(4)
= 3 Lim
x → 1
(x)2  - 2 Lim
x → 1
(x)  +  Lim
x → 1
(4)
Now by applying limit we get:
= 3(1)2 - 2(1) + 4
= 3(1) - 2 + 4
= 3 + 2
= 5

(iii)         Lim
x → 3
  x2 + x + 4
Lim
x → 3
(x2 + x + 4)1/2
Now by applying limit we get:
= [(3)2 + 3 + 4]1/2
= [9 + 7]1/2
= [16]1/2
= (42)1/2
= (4)2 x 1/2
= 4

(iv)         Lim
x → 2
  xx2 - 4
= ( Lim
x → 2
x) ( Lim
x → 2
x2 - 4)
Now by applying limit we get:
= 2(2)2 - 4
= 24 - 4
= 20
= 2(0)
= 0

(v)          Lim
x → 2
  (x3 + 1 - x2 + 5)
Lim
x → 2
x3 + 1  -  Lim
x → 2
x2 + 5
Now by applying limit we get:
= (2)3 + 1 - (2)2 + 5
= 8 + 1 - 4 + 5
= 9 - 9
= 0

(vi)         Lim
x → -2
  
2x3 + 5x
3x - 2
Lim
x → -2
(2x3 + 5x)
Lim
x → -2
(3x - 2)
Now by applying limit we get:
2(-2)3 + 5(-2)
3(-2) - 2
2(-8) - 10
-6 - 2
-16 - 10
-8
-26
-8
13
4

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