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UAE_Math_Grade_12_Vol_1_SE_718383_ch3
Sometimes an application of l’Ĥopital’s Rule must be followed by some simplifi-
cation, as we see in example 2.5.
EXAMPLE 2.5 Simplification of the Indeterminate Form ∞
∞
y ln x
Evaluate lim .
x→0 csc x
+
0.4
Solution First, notice that this limit has the form ∞ . From the graph in Figure 4.18,
∞
+
0.2 it appears that the function tends to 0 as x → 0 . Applying l’Ĥopital’s Rule, we have
d 1
x ln x dx (ln x) x ( ∞ )
0.4 0.8 1.2 lim = lim = lim .
+
+
x→0 csc x x→0 + d x→0 −csc x cot x ∞
-0.2 (csc x)
dx
This last limit still has the indeterminate form ∞ , but rather than apply l’Ĥopital’s
-0.4 ∞
Rule again, observe that we can rewrite the expression. We have
FIGURE 4.18 ln x 1 x ( sin x )
y = ln x lim = lim = lim − tan x = (−1)(0) = 0,
+
+
csc x x→0 csc x x→0 −csc x cot x x→0 + x
as expected, where we have used the fact that
lim sin x = 1.
x→0 x
(You can also establish this by using l’Ĥopital’s Rule.) Notice that if we had simply
1
continued with further applications of l’Ĥopital’s Rule to lim x , we
x→0 −csc x cot x
+
would never have resolved the limit. (Why not?)

Other Indeterminate Forms

0
There are five additional indeterminate forms to consider: ∞−∞, 0 ⋅ ∞, 0 , 1 ∞ and
0
∞ . Look closely at each of these to see why they are indeterminate. When evaluating
a limit of this type, the objective is to somehow reduce it to one of the indeterminate
forms 0 or ∞ , at which point we can apply l’Ĥopital’s Rule.
0 ∞

EXAMPLE 2.6 The Indeterminate Form ∞−∞
[ ]
Evaluate lim 1 − 1 .
x→0 ln (x + 1) x
y
Solution In this case, the limit has the form (∞−∞). From the graph in Figure 4.19,
1.0 it appears that the limit is somewhere around 0.5. If we add the fractions, we get a
form to which we can apply l’Ĥopital’s Rule. We have
0.8 [ ] x − ln (x + 1) ( )
lim 1 − 1 = lim 0
0.6 x→0 ln (x + 1) x x→0 ln (x + 1)x 0
0.4 d [x − ln (x + 1)]
dx
= lim By l’Ĥopital’s Rule.
0.2 x→0 d
dx [ln (x + 1)x]
x
-1 1 2 3 1
1 − ( )
FIGURE 4.19 = lim ( ) x + 1 0 . Copyright © McGraw-Hill Education
0
y = 1 − 1 x→0 1 x + ln (x + 1)(1)
ln (x + 1) x x + 1

244 | Lesson 4-2 | Indeterminate Forms and L’Hôpital’s Rule