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 2√2

(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 2√x

(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 mαm - 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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