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

Section 2.5 Continuity 119
Knowledge of which functions are continuous enables us to evaluate some limits very
quickly, as the following example shows. Compare it with Example 2.3.2(b).
ExamplE 5 Find lim
x l22
SOLUtion The function
x 3 1 2x 2 2 1
.
5 2 3x
f sxd − x 3 1 2x 2 2 1
5 2 3x
is rational, so by Theorem 5 it is continuous on its domain, which is h x | x ± 5 3 j.
Therefore
lim
x l22
x 3 1 2x 2 2 1
5 2 3x
− lim f sxd − f s22d
x l22
y
− s22d3 1 2s22d 2 2 1
5 2 3s22d
− 2 1
11 n
1
¨
0
FIGURE 5
P(cos ¨, sin ¨)
(1, 0)
x
It turns out that most of the familiar functions are continuous at every number in their
domains. For instance, Limit Law 10 (page 96) is exactly the statement that root functions
are continuous.
From the appearance of the graphs of the sine and cosine functions (Figure 1.2.18),
we would certainly guess that they are continuous. We know from the definitions of sin
and cos that the coordinates of the point P in Figure 5 are scos , sin d. As l 0, we
see that P approaches the point s1, 0d and so cos l 1 and sin l 0. Thus
6 lim
l 0 cos − 1
lim sin − 0
l 0
Another way to establish the limits in
(6) is to use the Squeeze Theorem with
the inequality sin , (for . 0),
which is proved in Section 3.3.
Since cos 0 − 1 and sin 0 − 0, the equations in (6) assert that the cosine and sine functions
are continuous at 0. The addition formulas for cosine and sine can then be used to
deduce that these functions are continuous everywhere (see Exercises 64 and 65).
It follows from part 5 of Theorem 4 that
tan x − sin x
cos x
is continuous except where cos x − 0. This happens when x is an odd integer multiple
of y2, so y − tan x has infinite discontinuities when x − 6y2, 63y2, 65y2, and
so on (see Figure 6).
y
1
_
3π
2
_π
π
_
2
0
π
2
π
3π
2
x
FIGURE 6
y − tan x
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