Calculus- Early Transcendentals, 2021a

3.7. Continuity 109
Solution. We will use a continuity argument to justify that direct substitution can be applied. By the list
above, √ x,sinx,1,x and cosx are all continuous functions at π. Then √ x +sinx and 1+x+cosx are both
continuous at π. Finally,
√ x + sinx
1 + x + cosx
is a continuous function at π since 1 + π + cosπ ≠ 0. Hence, we can directly substitute to get the limit:
√ √ √ x + sinx π + sinπ π
lim
x→π 1 + x + cosx = 1 + π + cosπ = π = √ 1 . π
Continuity is also preserved under the composition of functions.
Theorem 3.51: Continuity of Function Composition
If g is continuous at a and f is continuous at g(a), then the composition function f ◦g is continuous
at a.
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Example 3.52: Continuity with Composition of Functions
Determine where the following functions are continuous:
1. h(x)=cos(x 2 )
2. H(x)=ln(1 + sin(x))
Solution.
1. The functions that make up the composition h(x)= f (g(x)) are g(x)=x 2 and f (x)=cos(x). The
function g is continuous on R since it is a polynomial, and f is also continuous everywhere. Therefore,
h(x)=(f ◦ g)(x) is continuous on R by Theorem 3.51.
2. We know from Example 3.49 that f (x) =lnx is continuous and g(x) =1 + sinx are continuous.
Thus by Theorem 3.51, H(x) = f (g(x)) is continuous wherever it is defined. Now ln(1 + sinx) is
defined when 1 + sinx > 0. Recall that −1 ≤ sinx ≤ 1, so 1 + sinx > 0 except when sinx = −1,
which happens when x = ±3π/2,±7π/2,.... Therefore, H has discontinuities when x = 3πn/2,
n = 1,2,3,... and is continuous on the intervals between these values.
Intermediate Value Theorem
Whether or not an equation has a solution is an important question in mathematics. Consider the following
two questions:
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