Page 9 - Pra U STPM 2022 Penggal 2 - Mathematics
P. 9
Mathematics Semester 2 STPM Chapter 1 Limits and Continuity
Solution: (a)
y
8 1
6
4
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2
x
–2 –1 0 1 2 3 4 5 6 7
–2
f is a continuous function since the graph is unbroken in its domain.
(b)
y
4
x
– 3 –1 0 2
–1
– 4
f is not a continuous function.
Example 5
The function f is defined by
x
e + 3 , x , 1,
f(x) = 4 , x = 1,
e – x + 4 , x . 1.
lim
lim
(a) Find x → 1 – f(x) and x → 1 + f(x). Hence, determine whether f is continuous at
x = 1.
(b) Sketch the graph of f.
Solution: (a) x → 1 – f(x) = x → 1 – (e + 3)
lim
lim
x
= e + 3
lim + f(x) = lim + (e – x + 4)
x → 1 x → 1
= e + 3
lim f(x) = e + 3
x → 1
But f(1) = 4
lim
Since x → 1 f(x) ≠ f(1)
\ f is not continuous at x = 1.
7
01 STPM Math(T) T2.indd 7 28/01/2022 5:30 PM
