L Hopital Limit

Mathematical dominion for evaluating certain limits

In calculus,
L’Hôpital’s dominion
or
L’Hospital’s rule
(French:
[lopital], ,
loh-pee-TAHL
), as well known every bit
Bernoulli’s dominion, is a theorem which provides a technique to evaluate limits of indeterminate forms. Application (or repeated application) of the dominion often converts an indeterminate form to an expression that can exist easily evaluated by exchange. The dominion is named after the 17th-century French mathematician Guillaume de fifty’Hôpital. Although the dominion is often attributed to Fifty’Hospital, the theorem was outset introduced to him in 1694 by the Swiss mathematician Johann Bernoulli.

Fifty’Hôpital’due south rule states that for functions
f
and
g
which are differentiable on an open interval
I
except maybe at a point
c
contained in
I, if

$$



lim

x
→


c


f
(
x
)
=

lim

x
→


c


grand
(
x
)
=


 or

±


∞


,


{\textstyle \lim _{10\to c}f(x)=\lim _{x\to c}thou(10)=0{\text{ or }}\pm \infty ,}

and

$$



grand
′

(
ten
)
≠





{\textstyle g'(x)\neq 0}

for all
x
in
I
with

x
≠
c
, and

$$



lim

10
→


c






f
′

(
10
)



g
′

(
ten
)





{\textstyle \lim _{10\to c}{\frac {f'(ten)}{g'(x)}}}

exists, then

$$



lim

x
→


c





f
(
x
)


g
(
ten
)



=

lim

x
→


c






f
′

(
ten
)



g
′

(
x
)



.


{\displaystyle \lim _{10\to c}{\frac {f(x)}{g(x)}}=\lim _{10\to c}{\frac {f'(x)}{1000′(ten)}}.}

The differentiation of the numerator and denominator often simplifies the caliber or converts it to a limit that can exist evaluated straight.

History

[edit]

Guillaume de l’Hôpital (also written l’Hospital^[a]) published this rule in his 1696 book
Analyse des Infiniment Petits cascade l’Intelligence des Lignes Courbes
(literal translation:
Analysis of the Infinitely Pocket-size for the Understanding of Curved Lines), the first textbook on differential calculus.^[i]
[b]
Notwithstanding, information technology is believed that the dominion was discovered by the Swiss mathematician Johann Bernoulli.^[three]
[4]

Full general form

[edit]

The full general form of L’Hôpital’s rule covers many cases. Let

c

and

L

be extended real numbers (i.e., existent numbers, positive infinity, or negative infinity). Allow

I

be an open interval containing

c

(for a two-sided limit) or an open interval with endpoint

c

(for a one-sided limit, or a limit at infinity if

c

is infinite). The real valued functions

f

and

chiliad

are assumed to be differentiable on

I

except possibly at

c
, and additionally

$$



g
′

(
x
)
≠





{\displaystyle thousand'(x)\neq 0}

on

I

except possibly at

c
. It is also assumed that

$$



lim

x
→


c






f
′

(
x
)



chiliad
′

(
10
)



=
L
.


{\textstyle \lim _{x\to c}{\frac {f'(ten)}{g'(x)}}=L.}

Thus the rule applies to situations in which the ratio of the derivatives has a finite or infinite limit, simply non to situations in which that ratio fluctuates permanently equally

10

gets closer and closer to

c
.

If either

or

then

Although we have written

ten
→
c

throughout, the limits may also be i-sided limits (
x
→
c
⁺

or

x
→
c
⁻
), when

c

is a finite endpoint of

I
.

In the second case, the hypothesis that

f

diverges to infinity is not used in the proof (encounter note at the end of the proof section); thus, while the conditions of the rule are ordinarily stated as in a higher place, the second sufficient condition for the rule’s procedure to be valid tin be more than briefly stated as

$$



lim

x
→


c



|

chiliad
(
x
)

|

=
∞


.


{\textstyle \lim _{ten\to c}|g(x)|=\infty .}

The hypothesis that

$$



grand
′

(
10
)
≠





{\displaystyle yard'(x)\neq 0}

appears most commonly in the literature, but some authors sidestep this hypothesis by adding other hypotheses elsewhere. 1 method^[5]
is to ascertain the limit of a function with the boosted requirement that the limiting function is defined everywhere on the relevant interval

I

except possibly at

c
.^[c]
Some other method^[6]
is to require that both

f

and

thousand

be differentiable everywhere on an interval containing

c
.

Cases where theorem cannot be applied (Necessity of conditions)

[edit]

All four weather for L’Hôpital’s rule are necessary:

1. Indeterminancy of form:
   
   
   $$
   
   
   
   lim
   
   x
   →
   
   
   c
   
   
   f
   (
   x
   )
   =
   
   lim
   
   x
   →
   
   
   c
   
   
   g
   (
   10
   )
   =
   
   
   
   {\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(ten)=0}
   
   

   
   
   
   or
   
   

   $$
   
   
   ±
   
   
   ∞
   
   
   
   
   {\displaystyle \pm \infty }
   
   

   
   
    ; and

2. Differentiability of functions:
   
   
   $$
   
   
   f
   (
   10
   )
   
   
   {\displaystyle f(x)}
   
   

   
   
   
   and
   
   

   $$
   
   
   g
   (
   ten
   )
   
   
   {\displaystyle grand(x)}
   
   

   
   
   
   are differentiable on an open interval
   
   

   $$
   
   
   
   
   I
   
   
   
   
   {\displaystyle {\mathcal {I}}}
   
   

   
   
   
   except possibly at a point
   
   

   $$
   
   
   c
   
   
   {\displaystyle c}
   
   

   
   
   
   independent in
   
   

   $$
   
   
   
   
   I
   
   
   
   
   {\displaystyle {\mathcal {I}}}
   
   

   
   
   
   (the same indicate from the limit) ; and

3. Non-zero derivative of denominator:
   
   
   $$
   
   
   
   g
   ′
   
   (
   ten
   )
   ≠
   
   
   
   
   
   {\displaystyle g'(x)\neq 0}
   
   

   
   
   
   for all
   
   

   $$
   
   
   x
   
   
   {\displaystyle x}
   
   

   
   
   
   in
   
   

   $$
   
   
   
   
   I
   
   
   
   
   {\displaystyle {\mathcal {I}}}
   
   

   
   
   
   with
   
   

   $$
   
   
   x
   ≠
   
   
   c
   
   
   {\displaystyle x\neq c}
   
   

   
   
    ; and

4. Existence of limit of the caliber of the derivatives:
   
   
   $$
   
   
   
   lim
   
   10
   →
   
   
   c
   
   
   
   
   
   
   f
   ′
   
   (
   ten
   )
   
   
   
   g
   ′
   
   (
   x
   )
   
   
   
   
   
   {\displaystyle \lim _{ten\to c}{\frac {f'(ten)}{g'(x)}}}
   
   

   
   
   
   exists.

Where ane of the to a higher place atmospheric condition is non satisfied, L’Hôpital’due south rule is not valid in general, and and then information technology cannot always be practical.

Form is non indeterminate

[edit]

The necessity of the first condition tin exist seen by considering the counterexample where the functions are

$$


f
(
10
)
=
x
+
one


{\displaystyle f(x)=x+1}

and

$$


g
(
ten
)
=
ii
ten
+
1


{\displaystyle g(10)=2x+one}

and the limit is

$$


ten
→


i


{\displaystyle x\to 1}

.

The kickoff status is non satisfied for this counterexample because

$$



lim

x
→


1


f
(
x
)
=

lim

x
→


1


(
x
+
one
)
=
(
1
)
+
1
=
2
≠





{\displaystyle \lim _{x\to 1}f(x)=\lim _{ten\to 1}(x+ane)=(1)+i=2\neq 0}

and

$$



lim

x
→


1


1000
(
ten
)
=

lim

x
→


i


(
2
10
+
i
)
=
ii
(
1
)
+
1
=
3
≠





{\displaystyle \lim _{x\to 1}g(x)=\lim _{x\to i}(2x+1)=2(i)+one=iii\neq 0}

. This means that the grade is not indeterminate.

The second and third conditions are satisfied by

$$


f
(
ten
)


{\displaystyle f(x)}

and

$$


g
(
x
)


{\displaystyle g(x)}

. The quaternary condition is also satisfied with

$$



lim

x
→


1






f
′

(
10
)



g
′

(
x
)



=

lim

x
→


1





(
10
+
i

)
′



(
2
x
+
ane

)
′




=

lim

x
→


1




1
2


=


ane
2




{\displaystyle \lim _{x\to 1}{\frac {f'(x)}{m'(ten)}}=\lim _{x\to 1}{\frac {(10+1)’}{(2x+i)’}}=\lim _{ten\to i}{\frac {1}{2}}={\frac {1}{2}}}

.

But, L’Hôpital’s rule fails in this counterexample, since

$$



lim

x
→


1





f
(
x
)


1000
(
x
)



=

lim

x
→


ane





x
+
1


2
ten
+
1



=




lim

ten
→


one


(
x
+
i
)



lim

x
→


1


(
2
x
+
1
)



=


two
3


≠




1
ii


=

lim

10
→


1






f
′

(
x
)



thousand
′

(
x
)





{\displaystyle \lim _{ten\to ane}{\frac {f(x)}{g(10)}}=\lim _{x\to one}{\frac {x+1}{2x+ane}}={\frac {\lim _{x\to 1}(x+1)}{\lim _{x\to ane}(2x+1)}}={\frac {2}{3}}\neq {\frac {1}{ii}}=\lim _{x\to 1}{\frac {f'(x)}{g'(10)}}}

.

Differentiability of functions

[edit]

Differentiability of functions is a requirement because if a function is not differentiable, then the derivative of the functions is non guaranteed to be at each point in

$$




I




{\displaystyle {\mathcal {I}}}

. The fact that

$$




I




{\displaystyle {\mathcal {I}}}

is an open interval is grandfathered in from the hypothesis of the Cauchy Mean Value Theorem. The notable exception of the possibility of the functions being not differentiable at

$$


c


{\displaystyle c}

exists because L’Hôpital’s rule but requires the derivative to exist every bit the function approaches

$$


c


{\displaystyle c}

; the derivative does non demand to be taken at

$$


c


{\displaystyle c}

.

For case, let

f
(
x
)
=


{



sin
⁡


ten
,


x
≠







1
,


x
=









{\displaystyle f(x)={\brainstorm{cases}\sin x,&x\neq 0\\1,&10=0\end{cases}}}

,

$$


g
(
ten
)
=
x


{\displaystyle one thousand(x)=x}

, and

$$


c
=



{\displaystyle c=0}

. In this instance,

$$


f
(
x
)


{\displaystyle f(ten)}

is non differentiable at

$$


c


{\displaystyle c}

. However, since

$$


f
(
x
)


{\displaystyle f(10)}

is differentiable everywhere except

$$


c


{\displaystyle c}

, then

$$



lim

x
→


c



f
′

(
x
)


{\displaystyle \lim _{x\to c}f'(10)}

still exists. Thus, since

$$



lim

x
→


c





f
(
10
)


g
(
x
)



=








{\displaystyle \lim _{x\to c}{\frac {f(ten)}{yard(ten)}}={\frac {0}{0}}}

and

$$



lim

x
→


c






f
′

(
x
)



yard
′

(
x
)





{\displaystyle \lim _{10\to c}{\frac {f'(ten)}{grand'(x)}}}

exists, Fifty’Hôpital’due south dominion still holds.

Derivative of denominator is zero

[edit]

The necessity of the condition that

$$



g
′

(
x
)
≠





{\displaystyle m'(10)\neq 0}

nigh

$$


c


{\displaystyle c}

tin be seen by the following counterexample due to Otto Stolz.^[7]
Let

$$


f
(
x
)
=
x
+
sin
⁡


ten
cos
⁡


x


{\displaystyle f(x)=x+\sin 10\cos x}

and

$$


g
(
x
)
=
f
(
x
)

east

sin
⁡


x


.


{\displaystyle g(x)=f(x)e^{\sin x}.}

Then there is no limit for

$$


f
(
x
)

/

g
(
x
)


{\displaystyle f(x)/g(x)}

equally

$$


x
→


∞


.


{\displaystyle x\to \infty .}

However,

f
′

(
x
)



1000
′

(
x
)






=



2

cos

2


⁡


ten


(
2

cos

2


⁡


x
)

due east

sin
⁡


10


+
(
ten
+
sin
⁡


x
cos
⁡


x
)

east

sin
⁡


ten


cos
⁡


x









=



ii
cos
⁡


x


2
cos
⁡


ten
+
x
+
sin
⁡


x
cos
⁡


x




due east

−


sin
⁡


10


,






{\displaystyle {\begin{aligned}{\frac {f'(x)}{g'(x)}}&={\frac {2\cos ^{ii}x}{(2\cos ^{ii}x)e^{\sin x}+(x+\sin x\cos x)e^{\sin x}\cos 10}}\\&={\frac {2\cos ten}{ii\cos x+x+\sin x\cos 10}}e^{-\sin x},\cease{aligned}}}

which tends to 0 every bit

$$


x
→


∞




{\displaystyle x\to \infty }

. Further examples of this type were found by Ralph P. Boas Jr.^[8]

Limit of derivatives does not exist

[edit]

The requirement that the limit

$$



lim

x
→


c






f
′

(
x
)



g
′

(
10
)





{\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}

exist is essential. Without this condition,

$$



f
′



{\displaystyle f’}

or

$$



g
′



{\displaystyle one thousand’}

may exhibit undamped oscillations as

$$


x


{\displaystyle x}

approaches

$$


c


{\displaystyle c}

, in which case L’Hôpital’s dominion does not apply. For case, if

$$


f
(
ten
)
=
ten
+
sin
⁡


(
x
)


{\displaystyle f(x)=x+\sin(x)}

,

$$


k
(
x
)
=
x


{\displaystyle thousand(x)=x}

and

$$


c
=
±


∞




{\displaystyle c=\pm \infty }

, then

$$






f
′

(
x
)



m
′

(
10
)



=



one
+
cos
⁡


(
x
)

i


;


{\displaystyle {\frac {f'(ten)}{one thousand'(x)}}={\frac {1+\cos(ten)}{1}};}

this expression does non approach a limit as

$$


10


{\displaystyle 10}

goes to

$$


c


{\displaystyle c}

, since the cosine function oscillates between
ane
and
−ane. But working with the original functions,

$$



lim

x
→


∞







f
(
10
)


g
(
ten
)





{\displaystyle \lim _{x\to \infty }{\frac {f(x)}{g(x)}}}

can exist shown to exist:

$$



lim

x
→


∞







f
(
x
)


one thousand
(
x
)



=

lim

x
→


∞





(



x
+
sin
⁡


(
10
)

10


)

=

lim

x
→


∞





(

one
+



sin
⁡


(
x
)

x



)

=
1
+

lim

x
→


∞





(



sin
⁡


(
x
)

x


)

=
1
+

=
1.


{\displaystyle \lim _{x\to \infty }{\frac {f(x)}{g(x)}}=\lim _{x\to \infty }\left({\frac {x+\sin(x)}{10}}\right)=\lim _{ten\to \infty }\left(1+{\frac {\sin(ten)}{x}}\right)=ane+\lim _{x\to \infty }\left({\frac {\sin(x)}{ten}}\right)=i+0=i.}

In a case such as this, all that tin can be concluded is that

$$



lim inf

10
→


c






f
′

(
x
)



grand
′

(
ten
)



≤



lim inf

x
→


c





f
(
x
)


g
(
x
)



≤



lim sup

x
→


c





f
(
10
)


k
(
ten
)



≤



lim sup

x
→


c






f
′

(
x
)



chiliad
′

(
x
)



,


{\displaystyle \liminf _{x\to c}{\frac {f'(x)}{g'(x)}}\leq \liminf _{x\to c}{\frac {f(x)}{g(x)}}\leq \limsup _{ten\to c}{\frac {f(x)}{grand(x)}}\leq \limsup _{x\to c}{\frac {f'(ten)}{g'(x)}},}

so that if the limit of
f/g
exists, then it must lie between the inferior and superior limits of
f′/g′. (In the instance to a higher place, this is true, since 1 indeed lies between 0 and 2.)

Examples

[edit]

• Hither is a basic example involving the exponential part, which involves the indeterminate form
  
  
  /
  
  
  at
  
  10
  = 0:

  
  
  
  
  
  
  
  
  lim
  
  x
  →
  
  
  
  
  
  
  
  
  
  e
  
  ten
  
  
  −
  
  
  1
  
  
  
  x
  
  2
  
  
  +
  10
  
  
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  
  
  
  
  
  
  
  
  d
  
  d
  x
  
  
  
  (
  
  due east
  
  x
  
  
  −
  
  
  1
  )
  
  
  
  
  d
  
  d
  x
  
  
  
  (
  
  x
  
  two
  
  
  +
  ten
  )
  
  
  
  
  
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  
  
  
  
  
  
  east
  
  x
  
  
  
  ii
  ten
  +
  one
  
  
  
  
  
  
  
  
  
  =
  1.
  
  
  
  
  
  
  {\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {east^{x}-1}{x^{2}+10}}&=\lim _{x\to 0}{\frac {{\frac {d}{dx}}(east^{10}-ane)}{{\frac {d}{dx}}(ten^{two}+10)}}\\[4pt]&=\lim _{x\to 0}{\frac {e^{x}}{2x+1}}\\[4pt]&=1.\end{aligned}}}
  
  

  

• This is a more elaborate example involving
  
  
  /
  
  . Applying L’Hôpital’s rule a unmarried time however results in an indeterminate form. In this instance, the limit may exist evaluated by applying the rule iii times:

  
  
  
  
  
  
  
  
  lim
  
  x
  →
  
  
  
  
  
  
  
  
  2
  sin
  ⁡
  
  
  (
  10
  )
  −
  
  
  sin
  ⁡
  
  
  (
  2
  ten
  )
  
  
  x
  −
  
  
  sin
  ⁡
  
  
  (
  x
  )
  
  
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  
  
  
  
  
  
  2
  cos
  ⁡
  
  
  (
  x
  )
  −
  
  
  2
  cos
  ⁡
  
  
  (
  2
  x
  )
  
  
  one
  −
  
  
  cos
  ⁡
  
  
  (
  ten
  )
  
  
  
  
  
  
  
  
  
  =
  
  lim
  
  ten
  →
  
  
  
  
  
  
  
  
  −
  
  
  ii
  sin
  ⁡
  
  
  (
  x
  )
  +
  4
  sin
  ⁡
  
  
  (
  2
  10
  )
  
  
  sin
  ⁡
  
  
  (
  ten
  )
  
  
  
  
  
  
  
  
  
  =
  
  lim
  
  ten
  →
  
  
  
  
  
  
  
  
  −
  
  
  2
  cos
  ⁡
  
  
  (
  x
  )
  +
  8
  cos
  ⁡
  
  
  (
  2
  ten
  )
  
  
  cos
  ⁡
  
  
  (
  10
  )
  
  
  
  
  
  
  
  
  
  =
  
  
  
  −
  
  
  2
  +
  8
  
  1
  
  
  
  
  
  
  
  
  =
  6.
  
  
  
  
  
  
  {\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {2\sin(x)-\sin(2x)}{x-\sin(10)}}&=\lim _{x\to 0}{\frac {ii\cos(x)-2\cos(2x)}{1-\cos(x)}}\\[4pt]&=\lim _{ten\to 0}{\frac {-2\sin(x)+4\sin(2x)}{\sin(10)}}\\[4pt]&=\lim _{x\to 0}{\frac {-2\cos(x)+8\cos(2x)}{\cos(x)}}\\[4pt]&={\frac {-2+eight}{1}}\\[4pt]&=half-dozen.\terminate{aligned}}}
  
  

  

• Hither is an example involving
  
  ∞
  /
  ∞
  :

  $$$$
  
  
  
  lim
  
  10
  →
  
  
  ∞
  
  
  
  
  
  x
  
  n
  
  
  ⋅
  
  
  
  e
  
  −
  
  
  x
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  ten
  
  n
  
  
  
  due east
  
  x
  
  
  
  
  =
  
  lim
  
  10
  →
  
  
  ∞
  
  
  
  
  
  
  
  n
  
  x
  
  due north
  −
  
  
  1
  
  
  
  
  e
  
  ten
  
  
  
  
  =
  n
  ⋅
  
  
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  x
  
  n
  −
  
  
  ane
  
  
  
  e
  
  x
  
  
  
  
  .
  
  
  {\displaystyle \lim _{ten\to \infty }x^{due north}\cdot e^{-x}=\lim _{10\to \infty }{\frac {ten^{north}}{eastward^{10}}}=\lim _{ten\to \infty }{\frac {nx^{due north-1}}{due east^{ten}}}=n\cdot \lim _{ten\to \infty }{\frac {ten^{n-1}}{east^{x}}}.}
  
  

  

  Repeatedly use L’Hôpital’southward rule until the exponent is zero (if
  n
  is an integer) or negative (if
  n
  is fractional) to conclude that the limit is zero.

• Here is an example involving the indeterminate form
  0 · ∞
  (see below), which is rewritten as the grade
  
  ∞
  /
  ∞
  :

  $$$$
  
  
  
  lim
  
  x
  →
  
  
  
  
  
  +
  
  
  
  
  x
  ln
  ⁡
  
  
  10
  =
  
  lim
  
  ten
  →
  
  
  
  
  
  +
  
  
  
  
  
  
  
  ln
  ⁡
  
  
  10
  
  
  i
  x
  
  
  
  =
  
  lim
  
  x
  →
  
  
  
  
  
  +
  
  
  
  
  
  
  
  1
  x
  
  
  −
  
  
  
  
  1
  
  ten
  
  two
  
  
  
  
  
  
  
  =
  
  lim
  
  10
  →
  
  
  
  
  
  +
  
  
  
  
  −
  
  
  x
  =
  0.
  
  
  {\displaystyle \lim _{x\to 0^{+}}x\ln ten=\lim _{x\to 0^{+}}{\frac {\ln x}{\frac {one}{10}}}=\lim _{x\to 0^{+}}{\frac {\frac {one}{x}}{-{\frac {i}{x^{2}}}}}=\lim _{x\to 0^{+}}-10=0.}
  
  

  

• Here is an example involving the mortgage repayment formula and
  
  
  /
  
  . Allow
  
  P
  
  be the master (loan amount),
  
  r
  
  the interest charge per unit per period and
  
  northward
  
  the number of periods. When
  
  r
  
  is cipher, the repayment corporeality per period is
  
  
  $$
  
  
  
  
  P
  n
  
  
  
  
  {\displaystyle {\frac {P}{n}}}
  
  

  
  
  
  (since only principal is being repaid); this is consistent with the formula for non-nothing involvement rates:

  
  
  
  
  
  
  
  
  lim
  
  r
  →
  
  
  
  
  
  
  
  
  P
  r
  (
  1
  +
  r
  
  )
  
  due north
  
  
  
  
  (
  1
  +
  r
  
  )
  
  n
  
  
  −
  
  
  1
  
  
  
  
  
  
  =
  P
  
  lim
  
  r
  →
  
  
  
  
  
  
  
  
  (
  ane
  +
  r
  
  )
  
  n
  
  
  +
  r
  n
  (
  ane
  +
  r
  
  )
  
  n
  −
  
  
  i
  
  
  
  
  n
  (
  1
  +
  r
  
  )
  
  n
  −
  
  
  1
  
  
  
  
  
  
  
  
  
  
  
  =
  
  
  P
  n
  
  
  .
  
  
  
  
  
  
  {\displaystyle {\begin{aligned}\lim _{r\to 0}{\frac {Pr(1+r)^{n}}{(ane+r)^{n}-1}}&=P\lim _{r\to 0}{\frac {(1+r)^{due north}+rn(1+r)^{n-1}}{due north(1+r)^{due north-1}}}\\[4pt]&={\frac {P}{n}}.\end{aligned}}}
  
  

  

• 1 can also use L’Hôpital’due south rule to prove the following theorem. If
  
  f
  
  is twice-differentiable in a neighborhood of
  
  10
  
  and that its second derivative is continuous on this neighbourhood, and then

  
  
  
  
  
  
  
  
  lim
  
  h
  →
  
  
  
  
  
  
  
  
  f
  (
  x
  +
  h
  )
  +
  f
  (
  x
  −
  
  
  h
  )
  −
  
  
  two
  f
  (
  ten
  )
  
  
  h
  
  2
  
  
  
  
  
  
  
  =
  
  lim
  
  h
  →
  
  
  
  
  
  
  
  
  
  f
  ′
  
  (
  ten
  +
  h
  )
  −
  
  
  
  f
  ′
  
  (
  10
  −
  
  
  h
  )
  
  
  ii
  h
  
  
  
  
  
  
  
  
  
  =
  
  lim
  
  h
  →
  
  
  
  
  
  
  
  
  
  f
  ″
  
  (
  10
  +
  h
  )
  +
  
  f
  ″
  
  (
  10
  −
  
  
  h
  )
  
  2
  
  
  
  
  
  
  
  
  =
  
  f
  ″
  
  (
  ten
  )
  .
  
  
  
  
  
  
  {\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {f(x+h)+f(x-h)-2f(ten)}{h^{2}}}&=\lim _{h\to 0}{\frac {f'(ten+h)-f'(10-h)}{2h}}\\[4pt]&=\lim _{h\to 0}{\frac {f”(x+h)+f”(x-h)}{2}}\\[4pt]&=f”(x).\end{aligned}}}
  
  

  

• Sometimes L’Hôpital’s rule is invoked in a catchy style: suppose
  
  f(x) +
  f′(x)
  converges every bit
  
  10
  → ∞
  and that
  
  

  $$
  
  
  
  e
  
  x
  
  
  ⋅
  
  
  f
  (
  x
  )
  
  
  {\displaystyle e^{10}\cdot f(10)}
  
  

  
  
  
  converges to positive or negative infinity. Then:

  $$$$
  
  
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  f
  (
  ten
  )
  =
  
  lim
  
  ten
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  eastward
  
  x
  
  
  ⋅
  
  
  f
  (
  x
  )
  
  
  due east
  
  10
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  e
  
  10
  
  
  
  
  (
  
  
  f
  (
  ten
  )
  +
  
  f
  ′
  
  (
  x
  )
  
  
  )
  
  
  
  
  east
  
  x
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  (
  
  
  f
  (
  10
  )
  +
  
  f
  ′
  
  (
  x
  )
  
  
  )
  
  
  
  
  {\displaystyle \lim _{x\to \infty }f(10)=\lim _{ten\to \infty }{\frac {e^{x}\cdot f(x)}{e^{x}}}=\lim _{10\to \infty }{\frac {east^{x}{\bigl (}f(x)+f'(ten){\bigr )}}{e^{ten}}}=\lim _{x\to \infty }{\bigl (}f(x)+f'(x){\bigr )}}
  
  

  

  and so,
  
  

  $$
  
  
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  f
  (
  x
  )
  
  
  {\textstyle \lim _{10\to \infty }f(x)}
  
  

  
  
  
  exists and
  
  

  $$
  
  
  
  lim
  
  10
  →
  
  
  ∞
  
  
  
  
  
  f
  ′
  
  (
  x
  )
  =
  0.
  
  
  {\textstyle \lim _{x\to \infty }f'(x)=0.}
  
  

  
  
  

  The result remains true without the added hypothesis that
  
  

  $$
  
  
  
  e
  
  x
  
  
  ⋅
  
  
  f
  (
  10
  )
  
  
  {\displaystyle east^{ten}\cdot f(x)}
  
  

  
  
  
  converges to positive or negative infinity, but the justification is so incomplete.

Complications

[edit]

Sometimes L’Hôpital’s dominion does not lead to an answer in a finite number of steps unless some additional steps are applied. Examples include the following:

• Ii applications can lead to a return to the original expression that was to be evaluated:

  $$$$
  
  
  
  lim
  
  ten
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  e
  
  x
  
  
  +
  
  eastward
  
  −
  
  
  x
  
  
  
  
  
  east
  
  10
  
  
  −
  
  
  
  e
  
  −
  
  
  x
  
  
  
  
  
  =
  
  lim
  
  10
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  e
  
  x
  
  
  −
  
  
  
  e
  
  −
  
  
  x
  
  
  
  
  
  e
  
  x
  
  
  +
  
  e
  
  −
  
  
  x
  
  
  
  
  
  =
  
  lim
  
  ten
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  e
  
  ten
  
  
  +
  
  e
  
  −
  
  
  x
  
  
  
  
  
  e
  
  10
  
  
  −
  
  
  
  e
  
  −
  
  
  ten
  
  
  
  
  
  =
  ⋯
  
  
  .
  
  
  {\displaystyle \lim _{10\to \infty }{\frac {e^{ten}+east^{-10}}{e^{ten}-e^{-x}}}=\lim _{ten\to \infty }{\frac {e^{10}-eastward^{-x}}{due east^{x}+east^{-x}}}=\lim _{x\to \infty }{\frac {e^{x}+eastward^{-x}}{e^{x}-e^{-ten}}}=\cdots .}
  
  

  

  This situation can exist dealt with past substituting
  
  

  $$
  
  
  y
  =
  
  e
  
  x
  
  
  
  
  {\displaystyle y=east^{10}}
  
  

  
  
  
  and noting that
  
  y
  
  goes to infinity as
  
  x
  
  goes to infinity; with this commutation, this problem can be solved with a single application of the rule:

  $$$$
  
  
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  due east
  
  10
  
  
  +
  
  e
  
  −
  
  
  10
  
  
  
  
  
  due east
  
  10
  
  
  −
  
  
  
  e
  
  −
  
  
  ten
  
  
  
  
  
  =
  
  lim
  
  y
  →
  
  
  ∞
  
  
  
  
  
  
  
  y
  +
  
  y
  
  −
  
  
  1
  
  
  
  
  y
  −
  
  
  
  y
  
  −
  
  
  1
  
  
  
  
  
  =
  
  lim
  
  y
  →
  
  
  ∞
  
  
  
  
  
  
  
  i
  −
  
  
  
  y
  
  −
  
  
  two
  
  
  
  
  one
  +
  
  y
  
  −
  
  
  two
  
  
  
  
  
  =
  
  
  one
  1
  
  
  =
  1.
  
  
  {\displaystyle \lim _{x\to \infty }{\frac {due east^{x}+e^{-x}}{e^{ten}-e^{-ten}}}=\lim _{y\to \infty }{\frac {y+y^{-ane}}{y-y^{-1}}}=\lim _{y\to \infty }{\frac {1-y^{-ii}}{1+y^{-two}}}={\frac {ane}{one}}=1.}
  
  

  

  Alternatively, the numerator and denominator tin can both be multiplied by
  
  

  $$
  
  
  
  e
  
  x
  
  
  ,
  
  
  {\displaystyle e^{ten},}
  
  

  
  
  
  at which signal L’Hôpital’due south rule tin can immediately be practical successfully:^[nine]

  $$$$
  
  
  
  lim
  
  ten
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  e
  
  x
  
  
  +
  
  e
  
  −
  
  
  x
  
  
  
  
  
  eastward
  
  ten
  
  
  −
  
  
  
  e
  
  −
  
  
  x
  
  
  
  
  
  =
  
  lim
  
  10
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  e
  
  2
  10
  
  
  +
  1
  
  
  
  e
  
  two
  x
  
  
  −
  
  
  1
  
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  2
  
  e
  
  2
  x
  
  
  
  
  2
  
  eastward
  
  2
  x
  
  
  
  
  
  =
  1.
  
  
  {\displaystyle \lim _{ten\to \infty }{\frac {e^{10}+eastward^{-x}}{e^{x}-e^{-ten}}}=\lim _{x\to \infty }{\frac {due east^{2x}+i}{e^{2x}-1}}=\lim _{x\to \infty }{\frac {2e^{2x}}{2e^{2x}}}=ane.}
  
  

  

• An arbitrarily large number of applications may never atomic number 82 to an answer even without repeating:

  $$$$
  
  
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  x
  
  
  ane
  two
  
  
  
  +
  
  10
  
  −
  
  
  
  
  i
  ii
  
  
  
  
  
  
  
  10
  
  
  ane
  2
  
  
  
  −
  
  
  
  ten
  
  −
  
  
  
  
  one
  ii
  
  
  
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  
  1
  2
  
  
  
  x
  
  −
  
  
  
  
  1
  two
  
  
  
  
  −
  
  
  
  
  ane
  2
  
  
  
  x
  
  −
  
  
  
  
  3
  ii
  
  
  
  
  
  
  
  
  1
  2
  
  
  
  x
  
  −
  
  
  
  
  1
  ii
  
  
  
  
  +
  
  
  1
  ii
  
  
  
  10
  
  −
  
  
  
  
  3
  2
  
  
  
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  −
  
  
  
  
  ane
  4
  
  
  
  x
  
  −
  
  
  
  
  3
  ii
  
  
  
  
  +
  
  
  iii
  4
  
  
  
  ten
  
  −
  
  
  
  
  5
  2
  
  
  
  
  
  
  −
  
  
  
  
  1
  4
  
  
  
  10
  
  −
  
  
  
  
  3
  2
  
  
  
  
  −
  
  
  
  
  3
  four
  
  
  
  x
  
  −
  
  
  
  
  5
  two
  
  
  
  
  
  
  
  =
  ⋯
  
  
  .
  
  
  {\displaystyle \lim _{x\to \infty }{\frac {x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}}{ten^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}=\lim _{ten\to \infty }{\frac {{\frac {1}{two}}x^{-{\frac {one}{2}}}-{\frac {ane}{2}}x^{-{\frac {3}{2}}}}{{\frac {i}{2}}x^{-{\frac {1}{2}}}+{\frac {1}{2}}x^{-{\frac {3}{2}}}}}=\lim _{x\to \infty }{\frac {-{\frac {1}{4}}x^{-{\frac {3}{ii}}}+{\frac {3}{4}}x^{-{\frac {5}{2}}}}{-{\frac {1}{4}}x^{-{\frac {3}{2}}}-{\frac {iii}{4}}x^{-{\frac {v}{2}}}}}=\cdots .}
  
  

  

  This state of affairs too tin can be dealt with by a transformation of variables, in this instance
  
  

  $$
  
  
  y
  =
  
  
  x
  
  
  
  
  {\displaystyle y={\sqrt {x}}}
  
  

  
  
  :

  $$$$
  
  
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  x
  
  
  i
  ii
  
  
  
  +
  
  ten
  
  −
  
  
  
  
  i
  2
  
  
  
  
  
  
  
  ten
  
  
  1
  2
  
  
  
  −
  
  
  
  x
  
  −
  
  
  
  
  one
  ii
  
  
  
  
  
  
  
  =
  
  lim
  
  y
  →
  
  
  ∞
  
  
  
  
  
  
  
  y
  +
  
  y
  
  −
  
  
  i
  
  
  
  
  y
  −
  
  
  
  y
  
  −
  
  
  1
  
  
  
  
  
  =
  
  lim
  
  y
  →
  
  
  ∞
  
  
  
  
  
  
  
  1
  −
  
  
  
  y
  
  −
  
  
  two
  
  
  
  
  1
  +
  
  y
  
  −
  
  
  2
  
  
  
  
  
  =
  
  
  one
  1
  
  
  =
  1.
  
  
  {\displaystyle \lim _{x\to \infty }{\frac {10^{\frac {ane}{2}}+x^{-{\frac {ane}{2}}}}{x^{\frac {1}{2}}-x^{-{\frac {1}{2}}}}}=\lim _{y\to \infty }{\frac {y+y^{-1}}{y-y^{-1}}}=\lim _{y\to \infty }{\frac {one-y^{-2}}{1+y^{-two}}}={\frac {ane}{1}}=ane.}
  
  

  

  Again, an culling approach is to multiply numerator and denominator past
  
  

  $$
  
  
  
  x
  
  i
  
  /
  
  two
  
  
  
  
  {\displaystyle 10^{1/2}}
  
  

  
  
  
  before applying L’Hôpital’s rule:

  $$$$
  
  
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  
  x
  
  
  1
  ii
  
  
  
  +
  
  10
  
  −
  
  
  
  
  ane
  2
  
  
  
  
  
  
  
  x
  
  
  1
  two
  
  
  
  −
  
  
  
  x
  
  −
  
  
  
  
  1
  2
  
  
  
  
  
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  
  x
  +
  1
  
  
  x
  −
  
  
  1
  
  
  
  =
  
  lim
  
  x
  →
  
  
  ∞
  
  
  
  
  
  
  1
  i
  
  
  =
  1.
  
  
  {\displaystyle \lim _{10\to \infty }{\frac {ten^{\frac {one}{2}}+x^{-{\frac {1}{2}}}}{x^{\frac {i}{two}}-x^{-{\frac {one}{ii}}}}}=\lim _{ten\to \infty }{\frac {x+1}{ten-1}}=\lim _{x\to \infty }{\frac {one}{ane}}=1.}
  
  

  

A common pitfall is using L’Hôpital’s rule with some circular reasoning to compute a derivative via a difference caliber. For example, consider the job of proving the derivative formula for powers of
10:

$$



lim

h
→








(
x
+
h

)

north


−



ten

north



h


=
n

x

north
−


1


.


{\displaystyle \lim _{h\to 0}{\frac {(x+h)^{n}-10^{n}}{h}}=nx^{n-1}.}

Applying L’Hôpital’s rule and finding the derivatives with respect to

h

of the numerator and the denominator yields

nx

ⁿ⁻¹

as expected. Even so, differentiating the numerator requires the use of the very fact that is being proven. This is an example of begging the question, since one may not assume the fact to be proven during the course of the proof.

Other indeterminate forms

[edit]

Other indeterminate forms, such equally
1^∞
,

,
∞
,
0 · ∞, and
∞ − ∞, tin sometimes be evaluated using L’Hôpital’s dominion. For example, to evaluate a limit involving
∞ − ∞, convert the deviation of two functions to a caliber:

lim

ten
→


ane



(



10

ten
−


1



−




ane

ln
⁡


10




)




=

lim

ten
→


i





ten
⋅


ln
⁡


x
−


x
+
one


(
10
−


1
)
⋅


ln
⁡


x






(
i
)






=

lim

x
→


1





ln
⁡


x





x
−


1

10


+
ln
⁡


10






(
2
)






=

lim

x
→


1





x
⋅


ln
⁡


x


x
−


1
+
10
⋅


ln
⁡


x






(
iii
)






=

lim

x
→


1





1
+
ln
⁡


x


one
+
1
+
ln
⁡


10






(
four
)






=

lim

10
→


1





1
+
ln
⁡


10


two
+
ln
⁡


x









=


1
2


,






{\displaystyle {\begin{aligned}\lim _{x\to 1}\left({\frac {x}{x-i}}-{\frac {1}{\ln ten}}\correct)&=\lim _{10\to 1}{\frac {x\cdot \ln x-x+one}{(x-1)\cdot \ln x}}&\quad (one)\\[6pt]&=\lim _{x\to 1}{\frac {\ln 10}{{\frac {x-i}{x}}+\ln x}}&\quad (2)\\[6pt]&=\lim _{ten\to ane}{\frac {x\cdot \ln x}{10-1+x\cdot \ln ten}}&\quad (3)\\[6pt]&=\lim _{x\to one}{\frac {1+\ln x}{1+ane+\ln ten}}&\quad (4)\\[6pt]&=\lim _{ten\to 1}{\frac {1+\ln x}{ii+\ln x}}\\[6pt]&={\frac {one}{ii}},\cease{aligned}}}

where L’Hôpital’s rule is practical when going from (ane) to (two) and again when going from (3) to (iv).

L’Hôpital’south rule can be used on indeterminate forms involving exponents by using logarithms to “move the exponent downward”. Here is an example involving the indeterminate form

:

$$



lim

x
→





+





x

x


=

lim

x
→





+





eastward

ln
⁡


(

x

ten


)


=

lim

x
→





+





e

x
⋅


ln
⁡


x


=

east


lim

x
→





+




(
10
⋅


ln
⁡


ten
)


.


{\displaystyle \lim _{ten\to 0^{+}}x^{10}=\lim _{10\to 0^{+}}due east^{\ln(ten^{x})}=\lim _{x\to 0^{+}}e^{x\cdot \ln ten}=e^{\lim \limits _{10\to 0^{+}}(10\cdot \ln 10)}.}

It is valid to motion the limit inside the exponential function considering the exponential function is continuous. Now the exponent

$$


x


{\displaystyle x}

has been “moved downwardly”. The limit

$$



lim

ten
→





+




x
⋅


ln
⁡


10


{\displaystyle \lim _{x\to 0^{+}}x\cdot \ln 10}

is of the indeterminate form
0 · ∞, merely every bit shown in an example above, 50’Hôpital’s dominion may be used to make up one’s mind that

$$



lim

10
→





+




x
⋅


ln
⁡


10
=
0.


{\displaystyle \lim _{ten\to 0^{+}}ten\cdot \ln x=0.}

Thus

$$



lim

x
→





+





10

x


=

e




=
1.


{\displaystyle \lim _{ten\to 0^{+}}x^{x}=eastward^{0}=1.}

The following table lists the well-nigh common indeterminate forms, and the transformations for applying l’Hôpital’s rule:

Indeterminate form	Conditions	Transformation to $$ / {\displaystyle 0/0}
/	$$ lim ten → c f ( ten ) = , lim x → c 1000 ( x ) = {\displaystyle \lim _{x\to c}f(10)=0,\ \lim _{x\to c}yard(x)=0\!}	—
$$ ∞ {\displaystyle \infty } / $$ ∞ {\displaystyle \infty }	$$ lim x → c f ( ten ) = ∞ , lim x → c g ( ten ) = ∞ {\displaystyle \lim _{x\to c}f(10)=\infty ,\ \lim _{x\to c}1000(10)=\infty \!}	$$ lim x → c f ( ten ) g ( x ) = lim x → c 1 / g ( x ) 1 / f ( 10 ) {\displaystyle \lim _{10\to c}{\frac {f(x)}{g(x)}}=\lim _{10\to c}{\frac {1/chiliad(x)}{i/f(x)}}\!}
$$ ⋅ ∞ {\displaystyle 0\cdot \infty }	$$ lim x → c f ( x ) = , lim x → c chiliad ( x ) = ∞ {\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}thou(ten)=\infty \!}	$$ lim 10 → c f ( 10 ) grand ( x ) = lim x → c f ( 10 ) 1 / one thousand ( x ) {\displaystyle \lim _{10\to c}f(x)g(x)=\lim _{x\to c}{\frac {f(x)}{one/g(ten)}}\!}
$$ ∞ − ∞ {\displaystyle \infty -\infty }	$$ lim 10 → c f ( x ) = ∞ , lim ten → c chiliad ( x ) = ∞ {\displaystyle \lim _{ten\to c}f(ten)=\infty ,\ \lim _{x\to c}thousand(10)=\infty \!}	$$ lim x → c ( f ( x ) − g ( x ) ) = lim 10 → c 1 / g ( x ) − one / f ( x ) 1 / ( f ( x ) g ( 10 ) ) {\displaystyle \lim _{x\to c}(f(x)-1000(10))=\lim _{x\to c}{\frac {1/g(x)-1/f(x)}{1/(f(x)thou(x))}}\!}
$$ {\displaystyle 0^{0}}	$$ lim x → c f ( x ) = + , lim x → c g ( x ) = {\displaystyle \lim _{x\to c}f(10)=0^{+},\lim _{x\to c}g(x)=0\!}	$$ lim 10 → c f ( x ) grand ( ten ) = exp ⁡ lim x → c m ( x ) one / ln ⁡ f ( x ) {\displaystyle \lim _{x\to c}f(ten)^{g(x)}=\exp \lim _{10\to c}{\frac {g(10)}{1/\ln f(x)}}\!}
$$ one ∞ {\displaystyle one^{\infty }}	$$ lim x → c f ( x ) = 1 , lim ten → c g ( x ) = ∞ {\displaystyle \lim _{x\to c}f(x)=i,\ \lim _{x\to c}chiliad(x)=\infty \!}	$$ lim x → c f ( x ) g ( 10 ) = exp ⁡ lim 10 → c ln ⁡ f ( x ) 1 / chiliad ( x ) {\displaystyle \lim _{x\to c}f(x)^{grand(10)}=\exp \lim _{10\to c}{\frac {\ln f(x)}{1/g(ten)}}\!}
$$ ∞ {\displaystyle \infty ^{0}}	$$ lim x → c f ( x ) = ∞ , lim x → c g ( ten ) = {\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(10)=0\!}	$$ lim x → c f ( 10 ) g ( x ) = exp ⁡ lim 10 → c g ( x ) 1 / ln ⁡ f ( ten ) {\displaystyle \lim _{x\to c}f(ten)^{g(ten)}=\exp \lim _{x\to c}{\frac {g(x)}{one/\ln f(x)}}\!}

Stolz–Cesàro theorem

[edit]

The Stolz–Cesàro theorem is a similar result involving limits of sequences, merely it uses finite difference operators rather than derivatives.

Geometric estimation

[edit]

Consider the curve in the aeroplane whose

x
-coordinate is given by

g(t)
and whose

y
-coordinate is given by

f(t), with both functions continuous, i.due east., the locus of points of the form
[grand(t),
f(t)]. Suppose

f(c) =
g(c) = 0. The limit of the ratio



f(t)
/

m(t)


every bit

t
→
c

is the slope of the tangent to the curve at the indicate
[thousand(c),
f(c)] = [0,0]. The tangent to the curve at the indicate
[chiliad(t),
f(t)]
is given by
[chiliad′(t),
f′(t)]. L’Hôpital’southward rule then states that the slope of the curve when

t
=
c

is the limit of the slope of the tangent to the curve as the curve approaches the origin, provided that this is defined.

Proof of L’Hôpital’s rule

[edit]

Special case

[edit]

The proof of L’Hôpital’s rule is elementary in the instance where

f

and

g

are continuously differentiable at the betoken

c

and where a finite limit is found after the first round of differentiation. It is not a proof of the full general L’Hôpital’s rule because information technology is stricter in its definition, requiring both differentiability and that
c
be a real number. Since many common functions have continuous derivatives (e.k. polynomials, sine and cosine, exponential functions), it is a special case worthy of attention.

Suppose that

f

and

k

are continuously differentiable at a real number

c
, that

$$


f
(
c
)
=
g
(
c
)
=



{\displaystyle f(c)=g(c)=0}

, and that

$$



1000
′

(
c
)
≠





{\displaystyle yard'(c)\neq 0}

. And then

lim

x
→


c





f
(
x
)


yard
(
ten
)



=

lim

x
→


c





f
(
10
)
−





g
(
10
)
−






=

lim

x
→


c





f
(
10
)
−


f
(
c
)


g
(
x
)
−


g
(
c
)







=






lim

x
→


c





(



f
(
10
)
−


f
(
c
)


x
−


c



)


(



g
(
ten
)
−


k
(
c
)


x
−


c



)



=




lim

ten
→


c



(



f
(
ten
)
−


f
(
c
)


x
−


c



)




lim

x
→


c



(



g
(
x
)
−


grand
(
c
)


ten
−


c



)




=




f
′

(
c
)



thousand
′

(
c
)



=

lim

x
→


c






f
′

(
x
)



g
′

(
x
)



.






{\displaystyle {\begin{aligned}&\lim _{x\to c}{\frac {f(x)}{m(x)}}=\lim _{10\to c}{\frac {f(x)-0}{g(ten)-0}}=\lim _{10\to c}{\frac {f(ten)-f(c)}{g(x)-grand(c)}}\\[6pt]={}&\lim _{10\to c}{\frac {\left({\frac {f(x)-f(c)}{10-c}}\right)}{\left({\frac {thousand(ten)-grand(c)}{x-c}}\right)}}={\frac {\lim \limits _{x\to c}\left({\frac {f(x)-f(c)}{x-c}}\right)}{\lim \limits _{x\to c}\left({\frac {yard(ten)-g(c)}{ten-c}}\correct)}}={\frac {f'(c)}{g'(c)}}=\lim _{ten\to c}{\frac {f'(x)}{g'(x)}}.\finish{aligned}}}

This follows from the difference-caliber definition of the derivative. The last equality follows from the continuity of the derivatives at

c
. The limit in the decision is not indeterminate considering

$$



yard
′

(
c
)
≠





{\displaystyle g'(c)\neq 0}

.

The proof of a more general version of L’Hôpital’south rule is given below.

Full general proof

[edit]

The post-obit proof is due to Taylor (1952), where a unified proof for the


/


and

±∞
/
±∞

indeterminate forms is given. Taylor notes that different proofs may be plant in Lettenmeyer (1936) and Wazewski (1949).

Let
f
and
thousand
exist functions satisfying the hypotheses in the General course section. Allow

$$




I




{\displaystyle {\mathcal {I}}}

exist the open interval in the hypothesis with endpoint
c. Considering that

$$



g
′

(
x
)
≠





{\displaystyle m'(x)\neq 0}

on this interval and
k
is continuous,

$$




I




{\displaystyle {\mathcal {I}}}

can be chosen smaller then that
g
is nonzero on

$$




I




{\displaystyle {\mathcal {I}}}

.^[d]

For each
ten
in the interval, ascertain

$$


m
(
x
)
=
inf




f
′

(
ξ


)



g
′

(
ξ


)





{\displaystyle one thousand(ten)=\inf {\frac {f'(\xi )}{g'(\xi )}}}

and

$$


K
(
10
)
=
sup




f
′

(
ξ


)



g
′

(
ξ


)





{\displaystyle M(x)=\sup {\frac {f'(\xi )}{grand'(\xi )}}}

as

$$


ξ




{\displaystyle \eleven }

ranges over all values between
x
and
c. (The symbols inf and sup denote the infimum and supremum.)

From the differentiability of
f
and
g
on

$$




I




{\displaystyle {\mathcal {I}}}

, Cauchy’s mean value theorem ensures that for whatever two distinct points
x
and
y
in

$$




I




{\displaystyle {\mathcal {I}}}

in that location exists a

$$


ξ




{\displaystyle \eleven }

betwixt
ten
and
y
such that

$$





f
(
x
)
−


f
(
y
)


m
(
x
)
−


g
(
y
)



=




f
′

(
ξ


)



yard
′

(
ξ


)





{\displaystyle {\frac {f(x)-f(y)}{yard(10)-g(y)}}={\frac {f'(\xi )}{g'(\xi )}}}

. Consequently,

$$


m
(
x
)
≤





f
(
x
)
−


f
(
y
)


thousand
(
ten
)
−


grand
(
y
)



≤


M
(
x
)


{\displaystyle m(ten)\leq {\frac {f(10)-f(y)}{g(x)-g(y)}}\leq M(ten)}

for all choices of distinct
x
and
y
in the interval. The value
g(x)-thousand(y) is always nonzero for singled-out
x
and
y
in the interval, for if it was not, the hateful value theorem would imply the being of a
p
between
ten
and
y
such that
m’
(p)=0.

The definition of
m(x) and
Yard(x) volition result in an extended real number, and and so it is possible for them to take on the values ±∞. In the post-obit two cases,
g(x) and
One thousand(10) volition establish bounds on the ratio


f

/

thou

.

Instance 1:

$$



lim

ten
→


c


f
(
x
)
=

lim

ten
→


c


g
(
x
)
=



{\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0}

For any
ten
in the interval

$$




I




{\displaystyle {\mathcal {I}}}

, and indicate
y
between
ten
and
c,

$$


1000
(
ten
)
≤





f
(
x
)
−


f
(
y
)


g
(
x
)
−


g
(
y
)



=






f
(
10
)


g
(
ten
)



−





f
(
y
)


g
(
10
)





1
−





g
(
y
)


g
(
ten
)






≤


Grand
(
x
)


{\displaystyle m(x)\leq {\frac {f(10)-f(y)}{g(x)-one thousand(y)}}={\frac {{\frac {f(x)}{thou(ten)}}-{\frac {f(y)}{g(x)}}}{1-{\frac {g(y)}{yard(x)}}}}\leq Thou(10)}

and therefore as
y
approaches
c,

$$





f
(
y
)


chiliad
(
ten
)





{\displaystyle {\frac {f(y)}{g(x)}}}

and

$$





1000
(
y
)


1000
(
x
)





{\displaystyle {\frac {g(y)}{yard(ten)}}}

become zero, and then

$$


m
(
x
)
≤





f
(
x
)


g
(
x
)



≤


M
(
x
)
.


{\displaystyle k(x)\leq {\frac {f(10)}{one thousand(x)}}\leq M(x).}

Instance 2:

$$



lim

x
→


c



|

g
(
10
)

|

=
∞




{\displaystyle \lim _{x\to c}|g(ten)|=\infty }

For every
x
in the interval

$$




I




{\displaystyle {\mathcal {I}}}

, ascertain

$$



Due south

x


=
{
y
∣


y

 is betwixt

ten

 and

c
}


{\displaystyle S_{ten}=\{y\mid y{\text{ is between }}10{\text{ and }}c\}}

. For every point
y
betwixt
10
and
c,

$$


m
(
x
)
≤





f
(
y
)
−


f
(
ten
)


g
(
y
)
−


k
(
ten
)



=






f
(
y
)


g
(
y
)



−





f
(
x
)


g
(
y
)





1
−





1000
(
x
)


g
(
y
)






≤


M
(
x
)
.


{\displaystyle chiliad(ten)\leq {\frac {f(y)-f(10)}{grand(y)-g(x)}}={\frac {{\frac {f(y)}{m(y)}}-{\frac {f(x)}{one thousand(y)}}}{1-{\frac {one thousand(x)}{m(y)}}}}\leq M(x).}

As
y
approaches
c, both

$$





f
(
x
)


g
(
y
)





{\displaystyle {\frac {f(x)}{thousand(y)}}}

and

$$





m
(
x
)


g
(
y
)





{\displaystyle {\frac {k(x)}{g(y)}}}

become zero, and therefore

$$


1000
(
x
)
≤



lim inf

y
∈



S

x







f
(
y
)


grand
(
y
)



≤



lim sup

y
∈



Due south

x







f
(
y
)


g
(
y
)



≤


Yard
(
10
)
.


{\displaystyle m(10)\leq \liminf _{y\in S_{x}}{\frac {f(y)}{grand(y)}}\leq \limsup _{y\in S_{x}}{\frac {f(y)}{k(y)}}\leq G(ten).}

The limit superior and limit inferior are necessary since the existence of the limit of


f

/

grand


has not nevertheless been established.

It is as well the instance that

$$



lim

ten
→


c


m
(
x
)
=

lim

x
→


c


Thou
(
x
)
=

lim

x
→


c






f
′

(
x
)



g
′

(
10
)



=
50
.


{\displaystyle \lim _{ten\to c}one thousand(x)=\lim _{10\to c}Thousand(x)=\lim _{10\to c}{\frac {f'(ten)}{yard'(ten)}}=L.}

^[e]
and

$$



lim

ten
→


c



(


lim inf

y
∈



S

x







f
(
y
)


g
(
y
)




)

=

lim inf

ten
→


c





f
(
10
)


g
(
ten
)





{\displaystyle \lim _{x\to c}\left(\liminf _{y\in S_{x}}{\frac {f(y)}{g(y)}}\right)=\liminf _{x\to c}{\frac {f(10)}{g(x)}}}

and

$$



lim

x
→


c



(


lim sup

y
∈



South

10







f
(
y
)


g
(
y
)




)

=

lim sup

x
→


c





f
(
x
)


k
(
x
)



.


{\displaystyle \lim _{x\to c}\left(\limsup _{y\in S_{x}}{\frac {f(y)}{thou(y)}}\correct)=\limsup _{x\to c}{\frac {f(x)}{g(x)}}.}

In instance 1, the squeeze theorem establishes that

$$



lim

10
→


c





f
(
10
)


1000
(
x
)





{\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}}

exists and is equal to
L. In the instance 2, and the squeeze theorem again asserts that

$$



lim inf

x
→


c





f
(
x
)


one thousand
(
x
)



=

lim sup

x
→


c





f
(
x
)


g
(
x
)



=
50


{\displaystyle \liminf _{x\to c}{\frac {f(x)}{g(x)}}=\limsup _{x\to c}{\frac {f(x)}{g(10)}}=50}

, then the limit

$$



lim

x
→


c





f
(
ten
)


one thousand
(
x
)





{\displaystyle \lim _{x\to c}{\frac {f(x)}{k(10)}}}

exists and is equal to
L. This is the event that was to be proven.

In case 2 the assumption that
f(x) diverges to infinity was not used within the proof. This means that if |g(x)| diverges to infinity equally
x
approaches
c
and both
f
and
yard
satisfy the hypotheses of L’Hôpital’due south rule, then no boosted assumption is needed almost the limit of
f(10): It could even be the case that the limit of
f(x) does not exist. In this case, 50’Hopital’s theorem is really a consequence of Cesàro–Stolz.^[10]

In the example when |g(x)| diverges to infinity as
x
approaches
c
and
f(x) converges to a finite limit at
c, then L’Hôpital’s rule would be applicable, merely not absolutely necessary, since basic limit calculus will testify that the limit of
f(ten)/one thousand(x) as
x
approaches
c
must be zero.

Corollary

[edit]

A uncomplicated but very useful consequence of L’Hopital’s rule is a well-known criterion for differentiability. It states the post-obit: suppose that
f
is continuous at
a, and that

$$



f
′

(
10
)


{\displaystyle f'(x)}

exists for all
x
in some open interval containing
a, except perhaps for

$$


10
=
a


{\displaystyle x=a}

. Suppose, moreover, that

$$



lim

x
→


a



f
′

(
x
)


{\displaystyle \lim _{x\to a}f'(ten)}

exists. And so

$$



f
′

(
a
)


{\displaystyle f'(a)}

also exists and

$$



f
′

(
a
)
=

lim

x
→


a



f
′

(
ten
)
.


{\displaystyle f'(a)=\lim _{10\to a}f'(x).}

In particular,
f’
is also continuous at
a.

Proof

[edit]

Consider the functions

$$


h
(
10
)
=
f
(
10
)
−


f
(
a
)


{\displaystyle h(10)=f(x)-f(a)}

and

$$


1000
(
x
)
=
x
−


a


{\displaystyle g(x)=x-a}

. The continuity of
f
at
a
tells us that

$$



lim

x
→


a


h
(
x
)
=



{\displaystyle \lim _{10\to a}h(10)=0}

. Moreover,

$$



lim

x
→


a


grand
(
x
)
=



{\displaystyle \lim _{x\to a}g(x)=0}

since a polynomial function is always continuous everywhere. Applying L’Hopital’due south rule shows that

$$



f
′

(
a
)
:=

lim

ten
→


a





f
(
x
)
−


f
(
a
)


x
−


a



=

lim

x
→


a





h
(
x
)


chiliad
(
x
)



=

lim

x
→


a



f
′

(
10
)


{\displaystyle f'(a):=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}=\lim _{10\to a}{\frac {h(x)}{1000(x)}}=\lim _{ten\to a}f'(x)}

.

Run across also

[edit]

• L’Hôpital controversy

Notes

[edit]

References

[edit]

Sources

[edit]

• Abhilekh-Chatterjee, Dipak (2005),
  Existent Analysis, PHI Learning Pvt. Ltd, ISBN81-203-2678-iv
  
  
• Krantz, Steven G. (2004),
  A handbook of real variables. With applications to differential equations and Fourier analysis, Boston, MA: Birkhäuser Boston Inc., pp. 14+201, doi:10.1007/978-0-8176-8128-nine, ISBN0-8176-4329-X, MR 2015447
  
• Lettenmeyer, F. (1936), “Über die sogenannte Hospitalsche Regel”,
  Journal für dice reine und angewandte Mathematik,
  1936
  (174): 246–247, doi:x.1515/crll.1936.174.246, S2CID 199546754
  
• Taylor, A. E. (1952), “L’Hospital’due south rule”,
  Amer. Math. Monthly,
  59
  (1): twenty–24, doi:10.2307/2307183, ISSN 0002-9890, JSTOR 2307183, MR 0044602
  
• Wazewski, T. (1949), “Quelques démonstrations uniformes pour tous les cas du théorème de l’Hôpital. Généralisations”,
  Prace Mat.-Fiz.
  (in French),
  47: 117–128, MR 0034430