Page 77 - Calculus Demystified

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EXAMPLE 2.7 CHAPTER 2 Foundations of Calculus
Analyze the limit
2
x + 4
lim .
x→−2 x + 2
SOLUTION
The denominator tends to 0 while the numerator does not. According to
Theorem 2.3, the limit cannot exist.
2
You Try It: Use the Pinching Theorem to calculate lim x→0 x sin x.
x 2
You Try It: What can you say about lim x→−1 ?
2
x − 1

2.3 Continuity

Let f be a function whose domain contains the interval (a, b). Assume that c is a
point of (a, b). We say that the function f is continuous at c if

lim f(x) = f(c).
x→c
Conceptually, f is continuous at c if the expected value of f at c equals the actual
value of f at c.
EXAMPLE 2.8
Isthe function

2
2x − x if x< 2
f(x) =
3x if x ≥ 2
continuousat x = 2?
SOLUTION
We easily check that lim x→2 f(x) = 6.Also the actual value of f at 2, given
by the second part of the formula, is equal to 6. By the definition of continuity,
we may conclude that f is continuous at x = 2. See Fig. 2.5.

EXAMPLE 2.9
Where isthe function

1
if x< 4

g(x) = x − 3
2x + 3 if x ≥ 4

continuous?