2.5 Continuity 91
THEOREM 9―Composite of Continuous Functions If / is continuous at c and g is continuous
at /(c), then the composite g ° / is continuous at c .
Intuitively, Theorem 9 is reasonable because ifx is close to c , then f { x ) is close to /(c), and since g is
continuous at /(c), it follows that g ( f ( x ) ) is close to g(/(c)).
The continuity of composites holds for any finite number of functions. The only requirement is that each
function be continuous where it is applied. For an outline of a proof of Theorem 9, see Exercise 6 in Appendix
5.
EXAMPLE 8 Show that the following functions are continuous everywhere on their respective domains.
(a) y =
Vx 2 -
- 2 /3
2x
(b) y
x — 2
(c) y =
(d) y =
2
1+
x smx
+2
Solution
(a) The square root function is continuous on [0, o o ) because it is a root of the continuous identity function
0.4 -
f ( x ) = x (Part 7, Theorem 8). The given function is then the composite of the polynomial f ( x ) = x2 —
2x — 5 with the square root function g ( t ) = V t ,and is continuous on its domain.
/(X3
(b) The numerator is the cube root of the identity function squared; the denominator is an everywhere-positive
polynomial. Therefore, the quotient is continuous.
/ O.E
(c) The quotient (x ― 2)/(x2 ― 2 ) is continuous for all x乒士 and the function is the composition of this
quotient with the continuous absolute value function (Example 7).
i / 0.1
-2TT -
77 0
77 llT
FIGURE 2.43 The graph suggests that y =
I ( x sinx)/(x2 + 2) | is continuous (Example
8d).
(d) Because the sine function is everywhere-continuous (Exercise 70), the numerator term x sin x is the
product of continuous functions, and the denominator term x2 + 2 is an everywhere-positive polynomial.
The given function is the composite of a quotient of continuous functions with the continuous absolute
value function (Figure 2.43). ■
Theorem 9 is actually a consequence of a more general result which we now state and prove.
THEOREM 10―Limits of Continuous Functions
f ( x ) = b , then
If ^ is continuous at the point b and limA.—c
limx—cg(/(x)) = g { b ) = g { \ i m x ^ c f ( x ) ) .
Proof
Let e 〉 0 be given. Since g is continuous at b , there exists a number §i > 0 such that
\g(y)
一 g ( b ) \ < e whenever 0 < |^ — Z?| < 8 \ .
Since limv^c f ( x ) = b , there exists a 5 > 0 such that
\f(x)
—b
\ < 8\ whenever 0 < \ x — c \ < 8.
If we let y = f ( x ) , we then have that
\y — b\ < d ]
whenever 0 < |x — c | < 6,
which implies from the first statement that \ g ( y ) ― g ( b ) \ = | g ( f ( x ) ) 一 g ( b ) \ < e whenever 0 < \ x
— c \ < 8. From the definition of limit, this proves that lim^cg(/(x)) = g ( b ) . m