James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 2.5 Continuity 117
but
lim f sxd − lim v x b − n 2 1 ± f snd
x ln2 x ln 2
n
3 Definition A function f is continuous on an interval if it is continuous at
every number in the interval. (If f is defined only on one side of an endpoint of the
interval, we understand continuous at the endpoint to mean continuous from the
right or continuous from the left.)
y
1
ƒ=1-œ„„„„„ 1-≈
ExamplE 4 Show that the function f sxd − 1 2 s1 2 x 2 is continuous on the
interval f21, 1g.
SOLUtion If 21 , a , 1, then using the Limit Laws, we have
lim f sxd − lim (1 2 s1 2 x 2 )
x l a x l a
− 1 2 lim
x l a
s1 2 x 2 (by Laws 2 and 7)
− 1 2 s lim
x l a s1 2 x 2 d (by 11)
− 1 2 s1 2 a 2 (by 2, 7, and 9)
− f sad
Thus, by Definition l, f is continuous at a if 21 , a , 1. Similar calculations show that
lim f sxd − 1 − f s21d and lim
x l211 x l1 2
f sxd − 1 − f s1d
-1
0
1
x
so f is continuous from the right at 21 and continuous from the left at 1. Therefore,
according to Definition 3, f is continuous on f21, 1g.
The graph of f is sketched in Figure 4. It is the lower half of the circle
FIGURE 4
x 2 1 sy 2 1d 2 − 1
n
Instead of always using Definitions 1, 2, and 3 to verify the continuity of a function as
we did in Example 4, it is often convenient to use the next theorem, which shows how to
build up complicated continuous functions from simple ones.
4 Theorem If f and t are continuous at a and c is a constant, then the following
functions are also continuous at a:
1. f 1 t 2. f 2 t 3. cf
f
4. ft 5. if tsad ± 0
t
Proof Each of the five parts of this theorem follows from the corresponding Limit
Law in Section 2.3. For instance, we give the proof of part 1. Since f and t are continuous
at a, we have
lim f sxd − f sad and lim tsxd − tsad
x l a x l a
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