LESSON 28: Other Indeterminate Forms: Specific Objectives

255
LESSON 28: Other Indeterminate Forms 0 , , 0 ,

and 00
SPECIFIC OBJECTIVES:
At the end of this lesson, the students are expected to accomplish the
following:
transform given rational functions to equivalent ones;
use, apply the theorem LHopitals Rule to evaluate the limit of the
rational function which is of the form
0
and
0
; and
have mastered the differentiation formulas.
INSTRUCTIONAL STRATEGIES:
Since the students already know limits and derivatives of functions, at the
end of the previous lesson students will be given an advance organizer about
indeterminate forms for them to have an overview of the topics to be discussed.
Collaborative learning approach together with classroom discussion will also be
used for further enhancement of learning on the various indeterminate forms and
its theorems.
PERFORMANCE ASSESSMENT:
Each group will be given exercises for them to work on within a given time
limit. A representative from each group will be required to discuss on the board
one problem from the exercises given to measure the students understanding of
the concept. Correct solutions will be given additional points as incentive.
256
DISCUSSION OF THE TOPICS:
a. Indeterminate Form 0 and
Definition:
i.
If f(x) and g(x) are two differentiable functions such that

lim f (x) 0 and lim g(x) (which could be signed or
x a
x a
unsigned), the product of f(x) and g(x) is undefined having the
form 0 or 0 as x approaches its limit. To evaluate such
limit, their product is transformed to an equivalent one. Hence,
lim f(x) g(x) lim
x a
x a
f(x)
1
g( x )
then the limit is evaluated which may result to
0
or
0
. In
either of the case, LHopitals Rule applies.

ii.
If
lim f (x) , and the
x a
lim g(x)
x a
(which are both
f(x) - g(x) is said to be indeterminate

positive), then the xlim
a
of the form
. That is,
lim f(x) - g(x) lim f(x) lim g(x) -
x a
x a
x a
The limit could be evaluated by transforming the difference into

an equivalent quotient whose limit when evaluated may result to
0
or
0
. By then, LHopitals Rule applies.
b. Indeterminate Forms: 0 0 , 0 and 1

Definitions: Given two functions f(x) and g(x), and if:
lim f(x) 0 , and lim g(x) 0 , or
x a
x a
x a
x a
lim f(x) , and lim g(x) 0 , or

x a
lim f(x) 1, and lim g(x)
(or as x approaches
x a
or ), then the expression
lim
x a
f(x) g x
assumed the indeterminate forms 0 0 , 0 and 1 , respectively. These

indeterminate forms can be evaluated by letting a variable y for the
function, then apply the properties of logarithm.
257
Illustration:
Let:
y f ( x )
g( x )
ln y ln f ( x )
g( x )
ln y g( x ) ln f ( x )
ln f(x)
1
g(x)
y k , it follows that,
If the limit of xln
a
lim y e k
x a
(which is also a constant)
Since y f ( x ) g( x ) , therefore;
lim f(x)
x a
g( x )
ek
x csc 2x
Example 28.1. Evaluate : xlim
0
Solution:
lim x csc 2x 0 csc 0 0
x 0
Transforming the given function to an equivalent rational function;

lim x csc 2x lim
x 0
x0
x
0
0
sin2x sin 0 0
Apply LHR:
lim
x 0
d
x
dx
x
1
1
1
1
lim
lim
lim
0
x
0
x
0
d
sin2x
cos2x 2
2cos2x 2(cos 0) 2
sin2x
dx
x csc 2x 1
xlim
0
2
258
x ln x
0
Solution:
lim x ln x 0 ln 0 0
x 0
Transforming the given function to an equivalent function;

lim
x0
ln x lim
x0
ln x ln 0
1
1
x
0
Apply LHR:
d
1
ln x
(1)
ln x
lim
lim dx
lim x
lim x 0
x 0
1
x 0
x 0 1
x 0
d 1
2
x
x
dx x
lim x ln x 0
x 0

1 ln x
x 1
Solution:
1
1
1
1 1
1
lim
x 1 ln x
x 1
ln 1 1 1 0 0
Transforming to simple fraction;

1
( x 1) ln x (1 1) ln 1 0
1
lim
lim
x 1 ln x
x 1 (x - 1) ln x
x 1
(1 - 1) ln 1
0
Apply LHR:
1
d
1 (1)
( x 1) ln x
1
x
1
lim
lim dx
lim
x 1 ln x
x 1
d
x 1
(x - 1) ln x x 1 x 1 1 (1) ln x (1)
dx
x
259
x -1
11
0
x
lim
x
1
x
ln
x
x 1
1 1 (1) ln (1) 0
x
Again, apply LHR:
lim
x 1
lim
x 1
d
x - 1
dx
(x - 1)
1
lim
lim
x
1
x
1
d
1
x - 1 x ln x
x - 1 x ln x
1 x (1) (ln x )(1)
dx
x
1
1
1
2 ln x 2 ln(1) 2
1
1
1
lim
2
ln x x 1
x 1
1
1
Example 28.4. Evaluate : lim 2 2

x 0 x
x sec 2x
Solution:
1
lim
x 0
1
1
1 1
0 0(sec 0) 0 0 -
x sec 2x
Transforming to the equivalent function:

1
lim
x 0
x sec 2 x
1
2
xlim
cos 2 x
x
1 - cos2x
xlim
1 cos 2(0) 0
A
0
0
pply LHR:
d
1- cos2x
sin 2 x 2 sin 2(0) 0
dx
lim
xlim
d 2
0
x 0
2x
0
0
x
dx
1 - cos2x
lim
x 0
x2
260
Again, apply LHR:
sin 2x
lim
x 0
d
sin 2x
cos 2x (2)(1) lim 2 cos 2x 2 cos 0 2
dx
lim
lim
x 0
d
x 0
x 0
1
x
dx
1
1
lim 2 2
2
x 0 x
x sec 2x
2x
0
Solution:
lim
x 0
2x x 2(0) 0 0 0
Let:
y 2x
ln y ln 2 x
ln y x ln 2 x
ln 2 x
1
x
Apply limit on both sides as x 0 ;
ln 2 x ln 2(0) ln 0
1
x0 1
x
0
lim ln y lim
x 0
Apply LHR:
d
1
ln 2x
( 2)
ln 2x
dx
2
x
lim
lim
lim x 0
x 0
x 0
x 0
1
1
d 1
2
x
dx x
x
ln 2x
0
x 0
x 0
1
x
limln y 0
limln y lim
x 0
Take the inverse function of both sides

lim y e0
x 0
lim y 1
x 0
since y 2x
lim 2x 1
x
x 0
261
Example 28.6. Evaluate : lim ( x ) x 1
x 1
Solution:
lim
x 1
Let
1
x
( x ) 1
= 1 111 10 1
y ( x ) x 1
1
1
ln x ln x
x 1
x 1
Apply the limit on both sides as x 1 ,

ln y ln( x ) x 1
ln x
x 1
x 1 x 1
ln1 0
1 1 0
Apply LHR on the right member:
lim ln y lim
d
ln x
ln x
dx
lim
lim
x 1 x 1
x 1 d
x 1
dx
1
1
1 1
x
lim
lim 1
x 1
x 1 x
1
1
Thus,
lim
x 1
ln x
lim ln y 1 , take the inverse function of both sides
x 1 x 1
lim y e1
x 1
But
1
x
( x ) 1
lim ( x ) x 1 e or 2.718
x 1
(cot x ) x
0
Solution:
262
lim (cot x ) x (cot 0)0 0
x 0
Let
y (cot x ) x
ln y ln(cot x ) x x ln cot x
Apply the limit on both sides,
ln cot x
1
x
ln cot x
1
x 0
x
ln cot(0) ln( )
lim ln y lim
x 0
Apply LHR on the right member:

d
1
(ln cot x )
( csc 2 x )(1)
ln cot x
dx
cot
x
lim
lim
lim
1
1
x 0
x 0
x 0
d 1
2

x
dx x
x
sin x 1
1
2 x2
cos x sin 2 x
lim
lim sin x cos x lim
1
1
x 0
x 0
x 0 2 sin x cos x
2
2
x
x
lim
x 0
2x 2
2(0)
0
sin 2x sin(0) 0
Again, apply LHR:

d
(2x 2 )
2x 2
4x
2x
dx
lim
lim
lim
lim
x 0 sin 2 x
x 0 d
x 0 (cos 2 x )( 2)
x 0 cos 2 x
(sin 2x )
dx
2(0)
0
0
cos(0) 1
Hence,
lim
x 0
ln cot x
lim ln y 0
1
, take inverse function of both sides
x 0
x
lim y e 0 1
x o
Again, since y (cot x ) x

lim (cot x ) x 1
x 0
263
EXERCISES:
Evaluate the following limits and simplify:
1) lim
x 0
2x tan x
sin 4 x
2) lim
x 0
2x
sin 1 x
y2
ln cos y
3x 2
2x
e
x 0
12)
3x
sin 1 x csc x
7) lim
x 0
8) lim cos x ln x
x 0
4
x
lim 1 sin x
x 0
13) lim e x 3 x
x 0
ln cos 2x
6) lim ln tan 2x
x
11) lim 1 x 2
ln 2x 3
5) lim
10) lim
x 0 ln1 x
tan 1 2x
3) lim
y 0
4) xlim

2
9) lim
y 0 sin 2 y
y
2
x
14) lim cos 2x sin 2x x

x 0
1 e x
15) xlim
2
x2

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255

LESSON 28: Other Indeterminate Forms 0 , , 0 ,

have mastered the differentiation formulas.

If f(x) and g(x) are two differentiable functions such that

then the limit is evaluated which may result to

either of the case, LHopitals Rule applies.

lim f (x) , and the

(which are both

f(x) - g(x) is said to be indeterminate

lim f(x) - g(x) lim f(x) lim g(x) -

The limit could be evaluated by transforming the difference into

. By then, LHopitals Rule applies.

b. Indeterminate Forms: 0 0 , 0 and 1

lim f(x) , and lim g(x) 0 , or

lim f(x) 1, and lim g(x)

or ), then the expression

assumed the indeterminate forms 0 0 , 0 and 1 , respectively. These

(which is also a constant)

Transforming the given function to an equivalent rational function;

Transforming the given function to an equivalent function;

Example 28.3. Evaluate : xlim

Transforming to simple fraction;

Again, apply LHR:

Example 28.4. Evaluate : lim 2 2

Transforming to the equivalent function:

Again, apply LHR:

Apply limit on both sides as x 0 ;

Take the inverse function of both sides

Example 28.6. Evaluate : lim ( x ) x 1

Apply the limit on both sides as x 1 ,

Apply the limit on both sides,

Apply LHR on the right member:

Again, apply LHR:

Again, since y (cot x ) x

14) lim cos 2x sin 2x x

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