Fsc Part 2 Mathematics (Complete Solutions)

Q2
Evaluate each limit by using algebric techniques.

(i)           Lim
x → -1
  
x3 - x
x + 1
(ii)          Lim
x → 1
  (
3x3 + 4x
x2 + x
)
(iii)         Lim
x → 2
  
x3 - 8
x2 - x - 6
(iv)         Lim
x → 1
  
x3 - 3x2 + 3x - 1
x3 - x
(v)          Lim
x → -1
  (
x3 + x2
x2 - 1
)
(vi)         Lim
x → 4
  
2x2 - 32
x3 - 4x2
(vii)        Lim
x → 2
  
x - 2
x - 2
(viii)       Lim
h → 0
  
x + h - x
h
(ix)         Lim
x → α
  
xn - αn
xm - αm

Solution

(i)           Lim
x → -1
  
x3 - x
x + 1
Lim
x → -1
x(x2 - 1)
x + 1
Lim
x → -1
x(x - 1)(x + 1)
x + 1
Lim
x → -1
x(x - 1)
Lim
x → -1
x2 - x
Now by applying limit we get:
= (-1)2 - (-1)
= 1 + 1
= 2

(ii)          Lim
x → 1
  (
3x3 + 4x
x2 + x
)
Lim
x → 1
x(3x2 + 4)
x(x + 1)
Lim
x → 1
(3x2 + 4)
(x + 1)
Lim
x → 1
(3x2 + 4)
Lim
x → 1
(x + 1)
Now by applying limit we get:
3(1)2 + 4
1 + 1
3(1) + 4
2
3 + 4
2
7
2

(iii)         Lim
x → 2
  
x3 - 8
x2 - x - 6
Lim
x → 2
(x)3 - (2)3
x2 + 3x - 2x - 6
By applying formula a3 - b3 = (a - b)(a2 + ab + b2)
Lim
x → 2
(x - 2)[x2 + 2x + (2)2]
x(x + 3) - 2(x + 3)
Lim
x → 2
(x - 2)(x2 + 2x + 4)
(x - 2)(x + 3)
Lim
x → 2
x2 + 2x + 4
x + 3
Lim
x → 2
(x2 + 2x + 4)
Lim
x → 2
(x + 3)
Now by applying limit we get:
(2)2 + 2(2) + 4
2 + 3
4 + 4 + 4
5
12
5

(iv)         Lim
x → 1
  
x3 - 3x2 + 3x - 1
x3 - x
Lim
x → 1
x3 - 1 - 3x2 + 3x
x(x2 - 1)
By applying formula a3 - b3 = (a - b)(a2 + ab + b2) and a2 - b2 = (a - b)(a + b)
Lim
x → 1
(x - 1)[x2 + (x)(1) + (1)2] - 3x(x - 1)
x(x - 1)(x + 1)
Lim
x → 1
(x - 1)(x2 + x + 1) - 3x(x - 1)
x(x - 1)(x + 1)
Lim
x → 1
(x - 1)(x2 + x + 1 - 3x)
x(x - 1)(x + 1)
Lim
x → 1
x2 + x + 1 - 3x
x(x + 1)
Lim
x → 1
x2 - 2x + 1
x2 + x
Lim
x → 1
x2 - 2x + 1
Lim
x → 1
x2 + x
Now by applying limit we get:
(1)2 - 2(1) + 1
(1)2 + 1
1 - 2 + 1
1 + 1
0
2
= 0

(v)          Lim
x → -1
  (
x3 + x2
x2 - 1
)
By applying formula a2 - b2 = (a - b)(a + b)
Lim
x → -1
[
x2(x + 1)
(x - 1)(x + 1)
]
Lim
x → -1
x2
Lim
x → -1
x - 1
Now by applying limit we get:
(-1)2
-1 - 1
= -
1
2

(vi)         Lim
x → 4
  
2x2 - 32
x3 - 4x2
Lim
x → 4
2x2 - 32
x3 - 4x2
Lim
x → 4
2(x2 - 16)
x2(x - 4)
Lim
x → 4
2[x2 - (4)2]
x2(x - 4)
By applying formula a2 - b2 = (a - b)(a + b)
Lim
x → 4
2(x - 4)(x + 4)
x2(x - 4)
Lim
x → 4
2(x + 4)
x2
Lim
x → 4
2x + 8
x2
Lim
x → 4
2x + 8
Lim
x → 4
x2
Now by applying limit we get:
2(4) + 8
(4)2
8 + 8
16
16
16
= 0

(vii)        Lim
x → 2
  
x - 2
x - 2
Multiplying and dividing by x + 2
Lim
x → 2
x - 2
x - 2
 x 
x + 2
x + 2
Lim
x → 2
(x - 2)(x + 2)
(x - 2)(x + 2)
By applying formula (a - b)(a + b) = a2 - b2
Lim
x → 2
(x)2 - (2)2
(x - 2)(x + 2)
Lim
x → 2
x - 2
(x - 2)(x + 2)
Lim
x → 2
1
x + 2
Lim
x → 2
1
Lim
x → 2
x + 2
Now by applying limit we get:
1
2 + 2
1
22

(viii)       Lim
h → 0
  
x + h - x
h
Multiplying and dividing by x + h + x
Lim
h → 0
x + h - x
h
 x 
x + h + x
x + h + x
Lim
h → 0
(x + h - x)(x + h + x)
(h)(x + h + x)
By applying formula (a - b)(a + b) = a2 - b2
Lim
h → 0
(x + h)2 - (x)2
(h)(x + h + x)
Lim
h → 0
x + h - x
(h)(x + h + x)
Lim
h → 0
h
(h)(x + h + x)
Lim
h → 0
h
(h)(x + h + x)
Lim
h → 0
1
x + h + x
Lim
h → 0
1
Lim
h → 0
x + h + x
Now by applying limit we get:
1
x + 0 + x
1
x + x
1
2x

(ix)         Lim
x → α
  
xn - αn
xm - αm
Dividing by x - α
Lim
x → α
xn - αn
x - α
xm - αm
x - α
Lim
x → α
xn - αn
x - α
 .  Lim
x → α
1
xm - αm
x - α
Now by applying limit we get:
n - 1  . 
1
m - 1
n
m
 . 
αn - 1
αm - 1
n
m
 .  αn - 1  .  α-(m - 1)
n
m
 .  αn - 1 - m + 1
n
m
 .  αn - m

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