Page 5 - Pra U STPM 2022 Penggal 2 - Mathematics

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Example 1

Evaluate 1
2
2
(a) lim (x – x – 2) (b) lim x + 2x – 3
2
x → 1 x → 2 x + x + 1
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3
2
(c) lim 36 – x 2 (d) lim x – 2x + 3x
x → 6 6 – x x → 0 x – 2x
2
Solution: (a) lim (x – x – 2) = lim x – lim x – lim 2
2
2
x → 1 x → 1 x → 1 x → 1
= 1 – 1 – 2
2
= –2
Alternative Method
Substitute x = 1,
2
2
lim (x – x – 2) = 1 – 1 – 2
x → 1
= –2
2
2
(b) lim x + 2x – 3 = 2 + 2(2) – 3
2
x → 2 x + x + 1 2 + 2 + 1
2
= 5
7
(c) x → 61 36 – x 2 2
lim
6 – x
The function f(x) = 36 – x 2 is not defined when x = 6
6 – x
If x ≠ 6, then lim 36 – x 2 = lim (6 – x)(6 + x)
x → 6 6 – x x → 6 6 – x
= lim (6 + x)
x → 6
= 6 + 6
= 12
Note : We can take values of x as near as possible to 6, but not equal to 6.
2
2
3
(d) x → 0 1 x – 2x + 3x 2 = x → 0 x(x – 2x + 3)
lim
lim

x – 2x
x (x – 2)
2
2
lim

= x → 0 1 x – 2x + 3 2 , x ≠ 0
x – 2

= 0 – 0 + 3
0 – 2
= – 3
2
Example 2
Determine whether the limit exists in each of the following cases.
2 + e , x , 1

(a) lim f(x) where f(x) = x
x → 1 1 + e – x , x . 1

2
(x – 2) , x < 2
(b) lim f(x) where f(x) =  2
x → 2 1 – , x . 2
x
3

01 STPM Math(T) T2.indd 3 28/01/2022 5:30 PM