Page 6 - Pra U STPM 2022 Penggal 2 - Mathematics

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Solution: (a) lim – f(x) = lim – (2 + e )
x
x → 1 x → 1
= 2 + e
1 lim lim
x → 1 + f(x) = x → 1 _ (1 + e – x)
= 1 + e – 1
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= e
Since lim – f(x) ≠ lim + f(x), lim f(x) does not exist.
x → 1 x → 1 x → 1
(b) lim – f(x) = lim – (x – 2) 2
x → 2 x → 2
= (2 – 2) 2
= 0
+ 1 –
lim
lim + f(x) = x → 2 1 2 2
x → 2 x
= 1 – 2 2
= 0
Since lim – f(x) = lim + f(x) = 0, lim f(x) exist.
x → 2 x → 2 x → 2

Limits at infinity

+
For n  Z , x → ∞ x 1 n = 0.

lim
Example 3

Evaluate
(a) lim 2 + x
x → ∞ 3x + 1
2
(b) lim 3n – n + 1
2
n → ∞ n + 1
Solution: We cannot substitute x = ∞ as ∞ is undefined.
∞
Instead, both the numerator and the denominator are divided by the highest power
of x. 2
1 2
lim

(a) lim 2 + x = x → ∞1 x + 1
x → ∞ 3x + 1 3 + x
= 1
3
1 2
n
2
2
lim

lim
(b) n → ∞1 3n – n + 1 = n → ∞1 3 – 1 + n 1 2
n + 1
2
1 +
= 3 n 2
4

01 STPM Math(T) T2.indd 4 28/01/2022 5:30 PM