Fsc Part 2 Mathematics (Complete Solutions)
Q1
Evaluate each limit by using theorems of limits:
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Solution
| (i) | Lim
x → 3
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(2x + 4) |
| = | Lim
x → 3
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(2x) | + | Lim
x → 3
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(4) |
| = 2 | Lim
x → 3
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(x) | + | Lim
x → 3
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(4) |
Now by applying limit we get:
= 2(3) + 4
= 6 + 4
= 10
| (ii) | Lim
x → 1
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(3x2 - 2x + 4) |
| = | Lim
x → 1
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(3x2) | - | Lim
x → 1
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(2x) | + | Lim
x → 1
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(4) |
| = 3 | Lim
x → 1
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(x)2 | - 2 | Lim
x → 1
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(x) | + | Lim
x → 1
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(4) |
Now by applying limit we get:
= 3(1)2 - 2(1) + 4
= 3(1) - 2 + 4
= 3 + 2
= 5
| (iii) | Lim
x → 3
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√x2 + x + 4 |
| = | Lim
x → 3
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(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
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x√x2 - 4 |
| = ( | Lim
x → 2
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x) ( | Lim
x → 2
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√x2 - 4) |
Now by applying limit we get:
= 2√(2)2 - 4
= 2√4 - 4
= 2√0
= 2(0)
= 0
| (v) | Lim
x → 2
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(√x3 + 1 - √x2 + 5) |
| = | Lim
x → 2
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√x3 + 1 | - | Lim
x → 2
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√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
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Now by applying limit we get:
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