Calculus 2nd Edition Rogawski

72 CHAPTER 2 LIMITS
Other indeterminate forms are 1 ∞ , ∞ 0 ,
and 0 0 . These are treated in Section 7.7.
We say that f (x) has an indeterminate form (or is indeterminate) atx = c if the
formula for f (c) yields an undefined expression of the type
0
0 , ∞
, ∞ · 0, ∞−∞
∞
Our strategy, when this occurs, is to transform f (x) algebraically, if possible, into a
new expression that is defined and continuous at x = c, and then evaluate the limit by
substitution (“plugging in”). As you study the following examples, notice that the critical
step is to cancel a common factor from the numerator and denominator at the appropriate
moment, thereby removing the indeterminacy.
x 2 − 4x + 3
EXAMPLE 1 Calculate lim
x→3 x 2 + x − 12 .
Solution The function has the indeterminate form 0/0 at x = 3 because
Numerator at x = 3: x 2 − 4x + 3 = 3 2 − 4(3) + 3 = 0
Denominator at x = 3: x 2 + x − 12 = 3 2 + 3 − 12 = 0
Step 1. Transform algebraically and cancel.
x 2 − 4x + 3
x 2 + x − 12
(x − 3)(x − 1)
= =
(x − 3)(x + 4)
} {{ }
Cancel common factor
x − 1
x + 4
} {{ }
Continuous at x = 3
Step 2. Substitute (evaluate using continuity).
Because the expression on the right in Eq. (1) is continuous at x = 3,
EXAMPLE 2 The Form ∞ ∞
x 2 − 4x + 3
lim
x→3 x 2 + x − 12 =
Calculate lim
x→ π 2
x − 1
x + 4 = 2 7
lim
x→3
} {{ }
Evaluate by substitution
tan x
sec x .
(if x ̸= 3) 1
Solution As we see in Figure 2, both tan x and sec x have infinite discontinuities at x = π 2 ,
so this limit has the indeterminate form ∞/∞ at x = π 2 .
y
y = sec x
1
−1
y = tan x
π
2
x
FIGURE 2
Step 1. Transform algebraically and cancel.
( 1
)
(sin x)
tan x
sec x = cos x
1
cos x
= sin x (if cos x ̸= 0)