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APPLICATION OF THE L’HÔPITAL’S RULE
Umarov Habibullo Rakhmatullayevich
,
Gulistan State University, Lecturer, Department of Mathematics, Gulistan, Uzbekistan.
Shodmonkulova Shakhnoza Mirvaliyevna
,
Gulistan State University, Lecturer, Department of Mathematics, Gulistan, Uzbekistan.
https://doi.org/10.5281/zenodo.10969560
Abstract. This article studies the main types of uncertainties that arise when finding limits.
Some methods of disclosure of uncertainties are given.
Key words: Taylor series, limit point, L’Hôpital’s rules, logarithm, exponential.
ПРИМЕНЕНИЕ ПРАВИЛА ЛОПИТАЛЯ
Аннотация. В данной статье изучаются основные виды неопределенностей,
возникающих при нахождении пределов. Приведены некоторые методы раскрытия
неопределённостей.
Ключевые слова: Ряд Тейлора, предельная точка, правила Лопиталя, логарифм,
экспонент.
INTRODUCTION
Uncertainty disclosure – methods for calculating the limits of a function given by formulas,
which, as a result of a formal substitution of an argument in them, lose their meaning, that is, they
turn into expressions like: (
−
), (
/
), (
0
/
0
), (
0
), (
0
0
), (
1
), (
0
). Here,
0
is an
infinitesimal value, and ∞ is an infinitely large value.
Revealing uncertainties allows:
1. Simplification of the type of function (transformation using abbreviated multiplication
formulas, trigonometric formulas, etc.).
2. Use of wonderful limits.
3. Application of L’Hôpital’s rule.
4. Using the replacement of an infinitesimal expression with its equivalent.
OBJECT AND METHODS OF THE RESEARCH
The most powerful method is L’Hôpital’s rule, however, it does not allow calculating the
limit in all cases. A method for this kind of uncertainty was published in the 1696 textbook
“Analyse des Infiniment Petits” by Guillaume Lopital. The method was communicated to Lopital
in a letter by its discoverer Johann Bernoulli. In addition, it is directly applicable only to the second
and third of the listed types of uncertainties, that is, the relation, and in order to reveal other types,
they must first be reduced to one of these. Also, to calculate the limits, the expansion of the
expressions included in the uncertainty under study is often used in a Taylor series in the vicinity
of the limit point.
RESULTS AND THEIR DISCUSSION
L’Hôpital’s first rule
(uncertainty of the form
0
/
0
at
−
→
a
x
).
Theorem 1.
Let:
1)
( )
x
f
and
( )
x
g
are defined and differentiable in some interval
(
)
a
b
a
,
1
−
,
0
1
b
;
2)
( )
( )
0
lim
lim
=
=
−
→
−
→
x
g
x
f
a
x
a
x
;
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3)
( ) ( )
0
'
,
'
x
g
x
f
for all
x
belonging to
(
)
a
b
a
,
2
−
for some
0
2
b
;
4) there is a finite or infinite limit at
−
→
a
x
of the relation
( )
( )
x
g
x
f
'
'
, i.e.
( )
( )
x
g
x
f
a
x
'
'
lim
−
→
.
Then there is a limit of the relation
( )
( )
x
g
x
f
and the equality
( )
( )
( )
( )
x
g
x
f
x
g
x
f
a
x
a
x
'
'
lim
lim
−
→
−
→
=
.
The second L’Hôpital’s rule
(uncertainty of the form
/
for
a
x
→
).
Theorem 2.
Let the functions
( )
x
f
and
( )
x
g
be defined and differentiable in the interval
( )
b
a
,
;
( )
0
'
x
g
for all
x
belonging to
( )
b
a
,
;
( )
→
x
f
and
( )
→
x
g
at
a
x
→
; there
is a finite or infinite limit of
( )
( )
x
g
x
f
a
x
'
'
lim
→
, then
( )
( )
( )
( )
x
g
x
f
x
g
x
f
a
x
a
x
'
'
lim
lim
→
→
=
.
To reveal the uncertainties (
0
0
), (
1
), (
0
) we use the following method: we find the
limit of the (natural) logarithm of the expression containing the given uncertainty. As a result, the
type of uncertainty changes. After finding the limit, we take the exponent from it.
( ) ( )
( )
(
)
−
=
=
0
0
ln
0
0
0
e
e
;
( ) ( ) ( )
0
1
ln
1
=
=
e
e
;
( )
( )
(
)
( )
( )
=
=
0
ln
0
0
e
e
.
Example 1.
Is it possible to apply L’Hôpital’s rule to the limit
x
x
x
x
sin
1
sin
lim
2
0
→
The functions
( )
x
x
x
f
1
sin
2
=
and
( )
x
x
g
sin
=
,
0
\
R
x
are defined and
continuous in a neighborhood of the point
0
=
x
(excluding the point
0
=
x
); their derivatives
( )
x
x
x
x
f
1
cos
1
sin
2
'
−
=
and
( )
x
x
g
cos
'
=
simultaneously exist for
0
x
; their expression
( )
(
)
( )
(
)
x
x
x
x
x
x
x
g
x
f
1
sin
4
2
sin
2
1
cos
cos
'
'
2
2
2
2
2
2
+
−
+
=
+
at
0
x
and
( )
( )
x
x
x
x
x
g
x
f
x
x
cos
1
cos
1
sin
2
lim
'
'
lim
0
0
−
=
→
→
( )
1
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Since
(
)
0
cos
1
sin
2
lim
1
0
=
−
→
x
x
x
x
and
(
)
1
0
cos
1
cos
lim
−
→
x
x
x
does not exist, the
limit
( )
1
also does not exist. Therefore, the application of L’Hôpital’s rule in this example is
impossible. Note that
0
1
sin
sin
lim
0
=
→
x
x
x
x
x
.
Example 2.
Find the limit
cthx
x
x
e
w
+
=
→
2
1
lim
0
.
Uncertainty is brought to the form
0
0
e
, we get
( )
1
2
1
ln
2
1
−
+
=
+
thx
e
cthx
x
x
e
e
and applying L’Hôpital’s rule we have
2
1
2
1
2
lim
2
1
ln
lim
2
0
0
=
+
=
+
−
→
→
x
ch
e
e
thx
e
x
x
x
x
x
.
Thus,
e
e
w
=
=
2
1
.
Example 3.
Find the limit
(
)
x
x
x
x
x
x
−
−
+
+
→
1
1
ln
lim
1
1
.
Function
( )
(
)
x
x
x
x
f
x
−
+
=
+
1
ln
1
and
( )
x
x
g
−
=
1
,
0
x
,
1
x
satisfy the
following conditions:
1)
( )
( )
0
lim
lim
1
1
=
=
→
→
x
g
x
f
x
x
;
2) their derivatives
( )
(
)
1
ln
1
1
1
ln
'
1
−
+
+
+
+
=
+
x
x
x
x
x
x
x
x
f
,
( )
1
'
−
=
x
g
, exist for
0
x
;
3)
( )
( )
2
'
'
lim
1
−
=
→
x
g
x
f
x
exists;
4)
( )
(
)
( )
(
)
0
'
'
2
2
+
x
g
x
f
at
0
x
.
Therefore, the first L’Hôpital’s rule applies, according to which we have
(
)
( )
( )
2
'
'
lim
1
1
ln
lim
1
1
1
−
=
=
−
−
+
→
+
→
x
g
x
f
x
x
x
x
x
x
x
Example 4.
Find the limit
1
ln
lim
1
+
−
−
→
x
x
x
x
x
x
.
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The function
( )
x
x
x
f
x
−
=
and
( )
1
ln
+
−
=
x
x
x
g
,
0
x
,
1
x
satisfy the following
conditions:
1)
( )
( )
0
lim
lim
1
1
=
=
→
→
x
g
x
f
x
x
.
2) derivatives
( )
(
)
1
1
ln
'
−
+
=
x
x
x
f
x
,
( )
1
1
'
−
=
x
x
g
exists in a sufficiently small
neighborhood of the point
1
=
x
.
3) for in the indicated neighborhood.
4) according to the previous example, there is a finite limit.
( )
( )
(
)
2
1
1
ln
lim
'
'
lim
1
1
1
−
=
−
−
+
=
+
→
→
x
x
x
x
x
g
x
f
x
x
x
.
Therefore, the first rule of L’Hôpital applies, and we have
( )
( )
(
)
2
1
1
ln
lim
lim
1
1
1
−
=
−
−
+
=
+
→
→
x
x
x
x
x
g
x
f
x
x
x
.
Example 5. Find the limit of a matrix function
( )
(
)
+
=
x
x
x
x
x
e
x
x
Arshx
x
arctgx
x
x
x
A
1
1
1
1
1
1
sin
2
2
2
.
Since
( )
( )
(
)
x
a
x
A
ij
a
x
a
x
→
→
=
lim
lim
, where are
( )
x
a
ij
the elements of the functional matrix
( )
x
A
, then we calculate the limit of this matrix element by element. We have
z
x
x
e
x
x
=
→
2
1
0
sin
lim
, where
2
0
sin
ln
lim
x
x
x
z
x
→
=
.
Applying L’Hôpital’s rule
6
1
6
sin
lim
2
sin
cos
lim
sin
cos
sin
2
lim
2
0
3
0
2
0
−
=
=
−
=
−
=
→
→
→
x
x
x
x
x
x
x
x
x
x
x
x
x
x
z
x
x
x
.
Similarly, we obtain for all other elements:
z
x
x
e
x
arctgx
=
→
2
1
0
lim
,
(
)
=
+
−
=
−
+
=
=
→
→
→
3
2
0
3
2
0
2
0
2
1
lim
2
1
lim
ln
lim
x
arctgx
x
x
x
arctgx
x
x
arctgx
x
x
x
arctgx
z
x
x
x
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(
)
3
1
6
2
lim
2
1
lim
2
0
3
2
0
−
=
=
+
−
=
→
→
x
arctgx
x
x
arctgx
x
x
x
x
.
z
x
x
e
x
Arshx
=
→
2
1
0
lim
,
( )
=
−
=
−
+
=
=
→
→
→
3
0
3
2
0
2
0
lim
2
1
2
1
lim
ln
lim
x
x
u
x
x
Arshx
x
x
x
Arshx
x
x
Arshx
z
x
x
x
6
1
3
lim
2
0
−
=
=
→
x
Arshx
x
.
(Here we have introduced the notation
( )
Arshx
x
x
u
+
=
2
1
.)
(
)
z
x
x
x
e
e
x
=
+
→
1
1
0
1
lim
,
(
)
(
)
(
)
2
1
2
1
1
lim
2
1
1
1
lim
1
ln
lim
1
ln
lim
2
0
0
2
0
1
0
=
+
−
=
−
+
=
−
+
=
+
=
→
→
→
→
x
x
x
x
x
x
x
e
x
z
x
x
x
x
x
.
So, finally we have
( )
=
−
−
−
−
→
2
1
6
1
3
1
6
1
0
lim
e
e
e
e
x
A
x
.
Example 6.
Find
+
−
=
→
x
x
x
e
chx
tgx
x
x
x
e
x
x
z
cos
det
1
1
sin
det
lim
2
0
.
These determinants as functions of a variable satisfy all the conditions of L’Hôspital’s rule
in some neighborhood of the point
0
=
x
. Therefore, applying the rule, we get
=
+
−
+
+
−
=
+
−
=
−
→
→
x
x
x
x
x
x
x
x
e
shx
tgx
x
x
e
chx
x
x
x
x
x
e
x
x
x
e
x
e
chx
tgx
x
x
x
e
x
x
z
cos
det
cos
sin
cos
det
2
sin
det
1
1
cos
1
det
lim
cos
det
1
1
sin
det
lim
2
2
0
2
0
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1
1
1
0
0
det
1
0
1
1
det
0
1
0
0
det
1
0
1
1
det
lim
0
=
+
+
=
→
x
.
Example 7.
Find the asymptote of the curve
(
)
x
x
x
x
y
+
=
+
1
1
,
0
x
.
The oblique asymptote equation has the form
b
kx
y
+
=
. Using the equation of the curve,
we find
k
and
b
:
(
)
e
x
x
x
k
x
x
x
x
x
1
1
1
1
lim
1
lim
=
+
=
+
=
+
→
+
→
,
(
)
=
+
−
=
−
+
=
−
+
=
+
→
+
→
+
+
→
x
x
x
x
x
x
x
x
e
x
e
e
x
x
e
x
x
x
b
1
1
lim
1
1
1
1
1
lim
1
lim
2
1
( )
( )
( )
( )
=
+
−
+
+
=
+
−
=
+
→
+
→
2
1
0
2
1
0
2
1
ln
1
1
1
lim
1
1
lim
1
t
t
t
t
t
e
t
t
e
e
t
t
t
t
( ) ( )
( )
( )
e
t
t
t
e
t
t
t
t
t
e
t
t
2
1
3
2
1
ln
lim
1
1
1
ln
1
lim
1
2
0
2
2
0
2
=
+
+
−
=
+
+
+
−
−
=
+
→
+
→
.
Thus, we obtain the asymptote equation
e
e
x
y
2
1
+
=
.
Example 8.
Investigate for differentiability at a point at point
0
=
x
the function
( )
=
−
−
=
0
if
,
2
1
0
if
,
1
1
1
x
x
e
x
x
f
x
.
To investigate the differentiability of a function at a point
0
=
x
means to establish the
existence of a finite limit
( )
x
e
x
f
x
x
2
1
1
1
1
lim
0
'
0
−
−
−
=
→
( )
1
.
We will search for the limit of
( )
1
according to L’Hôpital’s rule, for which we must make
sure that the numerator in
( )
1
tends to zero as
0
→
x
. A test using L’Hôspital’s rule shows that
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(
)
(
)
(
)
0
2
lim
2
1
1
1
2
1
lim
1
2
2
2
lim
2
1
1
1
1
lim
0
0
0
0
=
+
−
=
−
+
−
−
=
−
−
−
−
=
−
−
−
→
→
→
→
x
x
x
x
x
x
x
x
x
x
x
x
x
x
xe
e
xe
x
e
xe
e
e
x
xe
x
e
e
x
.
So, in the formula
( )
1
we have an uncertainty of the form
0
0
. Applying the L’Hôpital rule
to
( )
1
three times, we get
(
)
(
)
=
+
−
−
−
−
=
−
−
−
−
=
−
−
−
→
→
→
x
x
x
x
x
x
x
x
x
x
x
x
e
x
e
x
xe
e
e
e
x
xe
x
e
x
e
x
2
0
2
0
0
2
1
4
1
2
lim
1
2
2
2
lim
2
1
1
1
1
lim
(
) (
)
(
)
(
)
12
1
2
12
12
1
lim
2
8
1
4
lim
2
0
2
0
−
=
+
+
+
−
=
+
+
−
−
=
→
→
x
x
x
x
x
x
x
e
x
x
x
e
x
x
e
e
x
xe
,
( )
12
1
0
'
−
=
f
.
Example 9.
( )
+
=
+
→
x
x
x
x
thx
arctgx
x
x
tg
w
,
2
,
1
2
lim
1
.
To find the limit of a vector function, we calculate the limits of each of its components.
Since the components are exponential expressions, we apply the representation
u
v
v
e
u
ln
=
,
0
u
, and, having reduced the corresponding uncertainties to the form
0
0
, we use the L’Hôpital rule.
We have
(
)
(
)
,
1
1
2
lim
1
2
sin
1
2
lim
2
1
2
1
2
2
sin
lim
2
1
2
1
2
cos
1
2
lim
1
2
1
2
2
1
=
=
=
=
+
+
+
+
+
+
+
+
+
→
→
−
−
+
→
−
−
+
→
x
x
x
x
x
x
x
x
x
x
x
x
tg
x
x
x
x
x
e
e
e
x
x
tg
0
1
2
1
lim
=
+
=
→
x
x
,
z
arctgx
x
x
x
e
e
arctgx
x
=
=
+
→
+
→
2
ln
lim
2
lim
, где
2
1
1
1
1
lim
2
2
−
=
−
+
=
+
→
x
arctgx
x
z
x
,
( )
( )
z
thx
x
x
x
e
e
thx
x
=
=
+
→
+
→
ln
lim
lim
,
ISSN:
2181-3906
2024
International scientific journal
«MODERN
SCIENCE
АND RESEARCH»
VOLUME 3 / ISSUE 4 / UIF:8.2 / MODERNSCIENCE.UZ
387
( )
0
lim
2
lim
2
1
1
1
lim
ln
lim
2
2
2
=
−
=
−
=
−
=
=
+
→
+
→
+
→
+
→
chx
x
shx
x
x
x
ch
thx
thx
x
z
x
x
x
x
.
Consequently,
=
−
1
,
,
1
2
e
w
.
Example 10.
=
−
−
−
+
→
1
100
1
0
,
lim
2
x
x
x
x
x
x
e
w
,
Since for vector - functions
( )
=
=
−
+
→
−
−
+
→
−
−
−
1
0
100
1
0
1
100
1
lim
,
lim
,
2
2
x
x
x
x
x
x
x
x
x
x
e
x
x
e
w
,
then we find the limits of each of the components separately. We have
0
lim
!
50
lim
lim
50
100
1
0
2
=
=
=
−
+
→
+
→
−
−
+
→
y
y
y
y
x
x
e
e
y
x
e
.
(here the second L'LHôspital's rule is applied 50 times).
For the second component, we first apply the representation
u
v
v
e
u
ln
=
,
0
u
, and
perform some transformation so that we can use the L’Hôpital rule:
( )
ab
x
x
e
x
x
x
x
x
e
e
x
x
x
x
=
=
−
+
→
−
+
→
ln
1
ln
0
1
0
ln
2
lim
lim
.
(Here we have used the continuity of the function
( )
x
e
x
f
=
and the product limit
theorem). To find
1
2
0
2
0
ln
lim
ln
lim
−
+
→
+
→
=
=
x
x
x
x
a
x
x
, we use the second, and to find
x
x
e
b
x
x
x
ln
1
lim
ln
0
−
=
+
→
, the first rule of L’Hôpital. We have
0
1
2
lim
ln
2
lim
1
ln
2
lim
ln
lim
2
0
1
0
2
0
1
2
0
=
−
−
=
−
=
−
=
=
+
→
−
+
→
−
+
→
−
+
→
x
x
x
x
x
x
x
x
x
a
x
x
x
x
.
1
1
lim
ln
1
lim
0
ln
0
=
−
=
−
=
−
→
+
→
t
e
x
x
e
b
t
t
x
x
x
.
Therefore, finally
( )
1
,
0
=
w
.
CONCLUSION
ISSN:
2181-3906
2024
International scientific journal
«MODERN
SCIENCE
АND RESEARCH»
VOLUME 3 / ISSUE 4 / UIF:8.2 / MODERNSCIENCE.UZ
388
L’Hôpital’s rule often (though not always) allows you to reveal uncertainties of the form
0
0
or
without much thought - if after the first differentiation you again get uncertainty, it does
not matter - you can differentiate again, and so on until we get some specific limit.
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