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 x√x2 - 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 x√x2 - 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