Lopitalio taisyklė (Liopitalio taisyklė ) skirta riboms neapibrėžtumo atvejais skaičiuoti, pasiūlyta Gijomo Lopitalio (1661-1704).
Pagrindinė Lopitalio taisyklės esmė yra išvestinės taikymas skaitikliui ir vardikliui atskirai.
I. Neapibrėžtumai
0
0
{\displaystyle {\frac {0}{0}}}
∞
∞
{\displaystyle {\frac {\infty }{\infty }}}
Teorema. Sakykime, kad
1)funkcijos f(x) ir g(x) apibrėžtos ir diferencijuojamos taško x=a aplinkoje ;
2)
lim
x
→
a
f
(
x
)
=
lim
x
→
a
g
(
x
)
=
0
{\displaystyle \lim _{x\to a}f(x)=\lim _{x\to a}g(x)=0}
arba
lim
x
→
a
f
(
x
)
=
lim
x
→
a
g
(
x
)
=
∞
;
{\displaystyle \lim _{x\to a}f(x)=\lim _{x\to a}g(x)=\infty ;}
3) egzistuoja
lim
x
→
a
f
′
(
x
)
g
′
(
x
)
.
{\displaystyle \lim _{x\to a}{\frac {f'(x)}{g'(x)}}.}
Tada
lim
x
→
a
f
(
x
)
g
(
x
)
=
lim
x
→
a
f
′
(
x
)
g
′
(
x
)
.
{\displaystyle \lim _{x\to a}{\frac {f(x)}{g(x)}}=\lim _{x\to a}{\frac {f'(x)}{g'(x)}}.}
0
⋅
∞
{\displaystyle 0\cdot \infty }
Šio tipo neapibrėžtumą galima pakeisti neapibrėžtumu
0
0
{\displaystyle {\frac {0}{0}}}
∞
∞
.
{\displaystyle {\frac {\infty }{\infty }}.}
lim
x
→
a
f
(
x
)
=
0
,
lim
x
→
a
g
(
x
)
=
∞
.
{\displaystyle \lim _{x\to a}f(x)=0,\qquad \lim _{x\to a}g(x)=\infty .}
Kadangi
f
(
x
)
⋅
g
(
x
)
=
f
(
x
)
1
/
g
(
x
)
,
{\displaystyle f(x)\cdot g(x)={\frac {f(x)}{1/g(x)}},}
tai
lim
x
→
a
f
(
x
)
⋅
g
(
x
)
=
lim
x
→
a
f
(
x
)
1
/
g
(
x
)
,
{\displaystyle \lim _{x\to a}f(x)\cdot g(x)=\lim _{x\to a}{\frac {f(x)}{1/g(x)}},}
ir gauname neapibrėžtumą
0
0
.
{\displaystyle {\frac {0}{0}}.}
∞
∞
:
{\displaystyle {\frac {\infty }{\infty }}:}
lim
x
→
a
f
(
x
)
⋅
g
(
x
)
=
lim
x
→
a
g
(
x
)
1
/
f
(
x
)
.
{\displaystyle \lim _{x\to a}f(x)\cdot g(x)=\lim _{x\to a}{\frac {g(x)}{1/f(x)}}.}
III. Neapibrėžtumas
∞
−
∞
{\displaystyle \infty -\infty }
Jį galime pakeisti neapibrėžtumu
0
0
.
{\displaystyle {\frac {0}{0}}.}
lim
x
→
a
f
(
x
)
=
+
∞
{\displaystyle \lim _{x\to a}f(x)=+\infty }
lim
x
→
a
g
(
x
)
=
+
∞
.
{\displaystyle \lim _{x\to a}g(x)=+\infty .}
lim
x
→
a
[
f
(
x
)
−
g
(
x
)
]
=
lim
x
→
a
[
1
1
/
f
(
x
)
−
1
1
/
g
(
x
)
]
=
lim
x
→
a
1
/
g
(
x
)
−
1
/
f
(
x
)
1
/
f
(
x
)
⋅
1
/
g
(
x
)
.
{\displaystyle \lim _{x\to a}[f(x)-g(x)]=\lim _{x\to a}[{\frac {1}{1/f(x)}}-{\frac {1}{1/g(x)}}]=\lim _{x\to a}{\frac {1/g(x)-1/f(x)}{1/f(x)\cdot 1/g(x)}}.}
Gavome neapibrėžtumą
0
0
,
{\displaystyle {\frac {0}{0}},}
1
∞
;
{\displaystyle 1^{\infty };}
0
0
;
{\displaystyle 0^{0};}
∞
0
{\displaystyle \infty ^{0}}
Šio tipo neapibrėžtumai pakeičiami neapibrėžtumu
0
⋅
∞
{\displaystyle 0\cdot \infty }
f
(
x
)
ϕ
(
x
)
=
e
ϕ
(
x
)
ln
f
(
x
)
{\displaystyle f(x)^{\phi (x)}=e^{\phi (x)\ln f(x)}}
lim
x
→
a
f
(
x
)
ϕ
(
x
)
=
lim
x
→
a
e
ϕ
(
x
)
ln
f
(
x
)
=
e
lim
x
→
a
ϕ
(
x
)
ln
f
(
x
)
.
{\displaystyle \lim _{x\to a}f(x)^{\phi (x)}=\lim _{x\to a}e^{\phi (x)\ln f(x)}=e^{\lim _{x\to a}\phi (x)\ln f(x)}.}
Laipsnio rodiklyje turime neapibrėžtumą
0
⋅
∞
.
{\displaystyle 0\cdot \infty .}
Apskaičiuosime
L
=
lim
x
→
∞
ln
x
x
m
{\displaystyle L=\lim _{x\to \infty }{\frac {\ln x}{x^{m}}}}
∞
∞
.
{\displaystyle {\frac {\infty }{\infty }}.}
L
=
lim
x
→
∞
(
ln
x
)
′
(
x
m
)
′
=
lim
x
→
∞
1
/
x
m
x
m
−
1
=
lim
x
→
∞
1
m
x
m
=
0.
{\displaystyle L=\lim _{x\to \infty }{\frac {(\ln x)'}{(x^{m})'}}=\lim _{x\to \infty }{\frac {1/x}{mx^{m-1}}}=\lim _{x\to \infty }{\frac {1}{mx^{m}}}=0.}
Apskaičiuosime
L
=
lim
x
→
+
0
x
m
ln
x
{\displaystyle L=\lim _{x\to +0}x^{m}\ln x}
0
⋅
(
−
∞
)
.
{\displaystyle 0\cdot (-\infty ).}
L
=
lim
x
→
+
0
ln
x
x
−
m
=
lim
x
→
+
0
1
/
x
−
m
x
−
m
−
1
=
lim
x
→
+
0
x
m
−
m
=
0.
{\displaystyle L=\lim _{x\to +0}{\frac {\ln x}{x^{-m}}}=\lim _{x\to +0}{\frac {1/x}{-mx^{-m-1}}}=\lim _{x\to +0}{\frac {x^{m}}{-m}}=0.}
Apskaičiuosime
L
=
lim
x
→
1
(
1
ln
x
−
1
x
−
1
)
.
{\displaystyle L=\lim _{x\to 1}({\frac {1}{\ln x}}-{\frac {1}{x-1}}).}
∞
−
∞
.
{\displaystyle \infty -\infty .}
L
=
lim
x
→
1
x
−
1
−
ln
x
(
x
−
1
)
ln
x
=
lim
x
→
1
1
−
1
/
x
ln
x
+
1
−
1
/
x
=
lim
x
→
1
x
−
1
x
ln
x
+
x
−
1
=
lim
x
→
1
1
ln
x
+
x
/
x
+
1
=
1
2
.
{\displaystyle L=\lim _{x\to 1}{\frac {x-1-\ln x}{(x-1)\ln x}}=\lim _{x\to 1}{\frac {1-1/x}{\ln x+1-1/x}}=\lim _{x\to 1}{\frac {x-1}{x\ln x+x-1}}=\lim _{x\to 1}{\frac {1}{\ln x+x/x+1}}={\frac {1}{2}}.}
L
=
lim
x
→
1
x
−
1
−
ln
x
(
x
−
1
)
ln
x
=
lim
x
→
1
1
−
1
/
x
ln
x
+
1
−
1
/
x
=
lim
x
→
1
1
/
x
2
1
/
x
+
1
/
x
2
=
1
/
1
2
1
/
1
+
1
/
1
2
=
1
2
.
{\displaystyle L=\lim _{x\to 1}{\frac {x-1-\ln x}{(x-1)\ln x}}=\lim _{x\to 1}{\frac {1-1/x}{\ln x+1-1/x}}=\lim _{x\to 1}{\frac {1/x^{2}}{1/x+1/x^{2}}}={\frac {1/1^{2}}{1/1+1/1^{2}}}={\frac {1}{2}}.}
Apskaičiuosime
L
=
lim
x
→
1
(
2
x
2
−
1
−
1
x
−
1
)
.
{\displaystyle L=\lim _{x\to 1}({\frac {2}{x^{2}-1}}-{\frac {1}{x-1}}).}
∞
−
∞
.
{\displaystyle \infty -\infty .}
L
=
lim
x
→
1
2
−
x
−
1
x
2
−
1
=
lim
x
→
1
−
(
x
−
1
)
(
x
−
1
)
(
x
+
1
)
=
−
lim
x
→
1
1
x
+
1
=
−
1
2
.
{\displaystyle L=\lim _{x\to 1}{\frac {2-x-1}{x^{2}-1}}=\lim _{x\to 1}{\frac {-(x-1)}{(x-1)(x+1)}}=-\lim _{x\to 1}{\frac {1}{x+1}}=-{\frac {1}{2}}.}
lim
x
→
1
1
−
x
−
ln
x
1
−
2
x
−
x
2
=
(
0
0
)
=
lim
x
→
1
−
1
+
1
x
−
2
−
2
x
2
2
x
−
x
2
=
lim
x
→
1
−
x
+
1
x
−
1
−
x
2
x
−
x
2
=
−
lim
x
→
1
(
1
−
x
)
2
x
−
x
2
(
1
−
x
)
x
=
−
1.
{\displaystyle \lim _{x\to 1}{\frac {1-x-\ln x}{1-{\sqrt {2x-x^{2}}}}}=({\frac {0}{0}})=\lim _{x\to 1}{\frac {-1+{\frac {1}{x}}}{-{\frac {2-2x}{2{\sqrt {2x-x^{2}}}}}}}=\lim _{x\to 1}{\frac {\frac {-x+1}{x}}{-{\frac {1-x}{\sqrt {2x-x^{2}}}}}}=-\lim _{x\to 1}{\frac {(1-x){\sqrt {2x-x^{2}}}}{(1-x)x}}=-1.}
lim
x
→
0
e
x
⋅
sin
x
−
5
x
4
x
2
+
7
x
=
(
0
0
)
=
lim
x
→
0
e
x
⋅
sin
x
+
e
x
⋅
cos
x
−
5
8
x
+
7
=
0
+
1
−
5
0
+
7
=
−
4
7
.
{\displaystyle \lim _{x\to 0}{\frac {e^{x}\cdot \sin x-5x}{4x^{2}+7x}}=({\frac {0}{0}})=\lim _{x\to 0}{\frac {e^{x}\cdot \sin x+e^{x}\cdot \cos x-5}{8x+7}}={\frac {0+1-5}{0+7}}=-{\frac {4}{7}}.}
lim
x
→
0
e
x
−
e
−
x
ln
(
e
−
x
)
+
x
−
1
=
(
0
0
)
=
lim
x
→
0
e
x
+
e
−
x
−
1
e
−
x
+
1
=
1
+
1
1
−
1
e
=
2
e
−
1
e
=
2
e
e
−
1
.
{\displaystyle \lim _{x\to 0}{\frac {e^{x}-e^{-x}}{\ln(e-x)+x-1}}=({\frac {0}{0}})=\lim _{x\to 0}{\frac {e^{x}+e^{-x}}{-{\frac {1}{e-x}}+1}}={\frac {1+1}{1-{\frac {1}{e}}}}={\frac {2}{\frac {e-1}{e}}}={\frac {2e}{e-1}}.}
lim
x
→
0
1
−
cos
x
x
2
=
(
0
0
)
=
lim
x
→
0
sin
x
2
x
=
1
2
.
{\displaystyle \lim _{x\to 0}{\frac {1-\cos x}{x^{2}}}=({\frac {0}{0}})=\lim _{x\to 0}{\frac {\sin x}{2x}}={\frac {1}{2}}.}
Du kartus pritaikius Lopitalio taisyklę, apskaičiuojama ribinė reikšmė
lim
x
→
0
x
−
sin
x
x
3
=
(
0
0
)
=
lim
x
→
0
1
−
cos
x
3
x
2
=
lim
x
→
0
sin
x
6
x
=
1
6
.
{\displaystyle \lim _{x\to 0}{\frac {x-\sin x}{x^{3}}}=({\frac {0}{0}})=\lim _{x\to 0}{\frac {1-\cos x}{3x^{2}}}=\lim _{x\to 0}{\frac {\sin x}{6x}}={\frac {1}{6}}.}
Tris kartus pritaikius Lopitalio taisyklę, apskaičiuojama ribinė reikšmė
lim
x
→
0
x
4
x
2
+
2
cos
x
−
2
=
(
0
0
)
=
lim
x
→
0
4
x
3
2
x
−
2
sin
x
=
lim
x
→
0
12
x
2
2
−
2
cos
x
=
lim
x
→
0
24
x
2
sin
x
=
12.
{\displaystyle \lim _{x\to 0}{\frac {x^{4}}{x^{2}+2\cos x-2}}=({\frac {0}{0}})=\lim _{x\to 0}{\frac {4x^{3}}{2x-2\sin x}}=\lim _{x\to 0}{\frac {12x^{2}}{2-2\cos x}}=\lim _{x\to 0}{\frac {24x}{2\sin x}}=12.}
lim
x
→
+
0
ln
tan
x
ln
tan
2
x
=
(
−
∞
−
∞
)
=
lim
x
→
+
0
1
tan
x
⋅
1
cos
2
x
1
tan
2
x
⋅
2
cos
2
2
x
=
1
2
lim
x
→
+
0
cos
2
2
x
cos
2
x
lim
x
→
+
0
tan
2
x
tan
x
=
{\displaystyle \lim _{x\to +0}{\frac {\ln \tan x}{\ln \tan 2x}}=({\frac {-\infty }{-\infty }})=\lim _{x\to +0}{\frac {{\frac {1}{\tan x}}\cdot {\frac {1}{\cos ^{2}x}}}{{\frac {1}{\tan 2x}}\cdot {\frac {2}{\cos ^{2}2x}}}}={\frac {1}{2}}\lim _{x\to +0}{\frac {\cos ^{2}2x}{\cos ^{2}x}}\lim _{x\to +0}{\frac {\tan 2x}{\tan x}}=}
=
1
2
⋅
1
⋅
lim
x
→
+
0
2
⋅
1
cos
2
2
x
1
cos
2
x
=
2
2
=
1.
{\displaystyle ={\frac {1}{2}}\cdot 1\cdot \lim _{x\to +0}{\frac {2\cdot {\frac {1}{\cos ^{2}2x}}}{\frac {1}{\cos ^{2}x}}}={\frac {2}{2}}=1.}
lim
x
→
0
(
1
x
−
1
sin
x
)
=
(
∞
−
∞
)
=
lim
x
→
0
sin
x
−
x
x
sin
x
=
lim
x
→
0
(
sin
x
x
−
x
x
)
lim
x
→
0
1
sin
x
=
(
1
−
1
)
/
0
=
0
/
0.
{\displaystyle \lim _{x\to 0}({\frac {1}{x}}-{\frac {1}{\sin x}})=(\infty -\infty )=\lim _{x\to 0}{\frac {\sin x-x}{x\sin x}}=\lim _{x\to 0}({\frac {\sin x}{x}}-{\frac {x}{x}})\lim _{x\to 0}{\frac {1}{\sin x}}=(1-1)/0=0/0.}
lim
x
→
0
(
1
x
−
1
sin
x
)
=
(
∞
−
∞
)
=
lim
x
→
0
sin
x
−
x
x
sin
x
=
lim
x
→
0
cos
x
−
1
sin
x
+
x
cos
x
=
lim
x
→
0
−
sin
x
cos
x
+
cos
x
−
x
sin
x
=
lim
x
→
0
−
cos
x
−
sin
x
−
sin
x
−
(
sin
x
+
x
cos
x
)
=
{\displaystyle \lim _{x\to 0}({\frac {1}{x}}-{\frac {1}{\sin x}})=(\infty -\infty )=\lim _{x\to 0}{\frac {\sin x-x}{x\sin x}}=\lim _{x\to 0}{\frac {\cos x-1}{\sin x+x\cos x}}=\lim _{x\to 0}{\frac {-\sin x}{\cos x+\cos x-x\sin x}}=\lim _{x\to 0}{\frac {-\cos x}{-\sin x-\sin x-(\sin x+x\cos x)}}=}
=
lim
x
→
0
−
cos
x
−
3
sin
x
−
x
cos
x
=
lim
x
→
0
sin
x
−
3
cos
x
−
(
cos
x
−
x
sin
x
)
=
lim
x
→
0
sin
x
−
4
cos
x
+
x
sin
x
=
0
−
4
⋅
1
+
0
⋅
0
=
0
−
4
=
0.
{\displaystyle =\lim _{x\to 0}{\frac {-\cos x}{-3\sin x-x\cos x}}=\lim _{x\to 0}{\frac {\sin x}{-3\cos x-(\cos x-x\sin x)}}=\lim _{x\to 0}{\frac {\sin x}{-4\cos x+x\sin x}}={\frac {0}{-4\cdot 1+0\cdot 0}}={\frac {0}{-4}}=0.}
lim
x
→
a
f
(
x
)
ϕ
(
x
)
=
lim
x
→
a
e
ϕ
(
x
)
ln
f
(
x
)
=
lim
x
→
+
∞
x
1
x
=
(
∞
0
)
=
lim
x
→
+
∞
e
1
x
ln
x
=
e
lim
x
→
+
∞
ln
x
x
=
e
lim
x
→
+
∞
1
x
1
=
e
0
=
1.
{\displaystyle \lim _{x\to a}f(x)^{\phi (x)}=\lim _{x\to a}e^{\phi (x)\ln f(x)}=\lim _{x\to +\infty }x^{\frac {1}{x}}=(\infty ^{0})=\lim _{x\to +\infty }e^{{\frac {1}{x}}\ln x}=e^{\lim _{x\to +\infty }{\frac {\ln x}{x}}}=e^{\lim _{x\to +\infty }{\frac {\frac {1}{x}}{1}}}=e^{0}=1.}
lim
x
→
+
0
x
x
=
(
0
0
)
=
lim
x
→
+
0
e
x
ln
x
=
e
lim
x
→
+
0
ln
x
1
x
=
e
lim
x
→
+
0
1
x
−
1
x
2
=
e
−
lim
x
→
+
0
x
=
e
0
=
1.
{\displaystyle \lim _{x\to +0}x^{x}=(0^{0})=\lim _{x\to +0}e^{x\ln x}=e^{\lim _{x\to +0}{\frac {\ln x}{\frac {1}{x}}}}=e^{\lim _{x\to +0}{\frac {\frac {1}{x}}{-{\frac {1}{x^{2}}}}}}=e^{-\lim _{x\to +0}x}=e^{0}=1.}
lim
x
→
+
0
(
1
+
x
)
1
x
=
(
1
∞
)
=
lim
x
→
+
0
e
1
x
ln
(
1
+
x
)
=
e
lim
x
→
+
0
ln
(
1
+
x
)
x
=
e
lim
x
→
+
0
1
1
+
x
=
e
1
=
e
.
{\displaystyle \lim _{x\to +0}(1+x)^{\frac {1}{x}}=(1^{\infty })=\lim _{x\to +0}e^{{\frac {1}{x}}\ln(1+x)}=e^{\lim _{x\to +0}{\frac {\ln(1+x)}{x}}}=e^{\lim _{x\to +0}{\frac {1}{1+x}}}=e^{1}=e.}
lim
x
→
1
+
0
(
x
−
1
)
ln
x
=
(
0
0
)
=
lim
x
→
1
+
0
e
ln
x
⋅
ln
(
x
−
1
)
=
e
lim
x
→
1
+
0
[
ln
x
⋅
ln
(
x
−
1
)
]
;
{\displaystyle \lim _{x\to 1+0}(x-1)^{\ln x}=(0^{0})=\lim _{x\to 1+0}e^{\ln x\cdot \ln(x-1)}=e^{\lim _{x\to 1+0}[\ln x\cdot \ln(x-1)]};}
lim
x
→
1
+
0
[
ln
x
⋅
ln
(
x
−
1
)
]
=
lim
x
→
1
+
0
ln
(
x
−
1
)
1
ln
x
=
lim
x
→
1
+
0
1
x
−
1
−
1
x
ln
2
x
=
{\displaystyle \lim _{x\to 1+0}[\ln x\cdot \ln(x-1)]=\lim _{x\to 1+0}{\frac {\ln(x-1)}{\frac {1}{\ln x}}}=\lim _{x\to 1+0}{\frac {\frac {1}{x-1}}{-{\frac {1}{x\ln ^{2}x}}}}=}
=
−
lim
x
→
1
x
ln
2
x
x
−
1
=
−
lim
x
→
1
ln
2
x
+
2
x
ln
x
⋅
1
x
1
=
−
lim
x
→
1
(
ln
2
x
+
2
ln
x
)
=
0
;
e
0
=
1.
{\displaystyle =-\lim _{x\to 1}{\frac {x\ln ^{2}x}{x-1}}=-\lim _{x\to 1}{\frac {\ln ^{2}x+2x\ln x\cdot {\frac {1}{x}}}{1}}=-\lim _{x\to 1}(\ln ^{2}x+2\ln x)=0;\;e^{0}=1.}
lim
x
→
∞
x
+
ln
x
2
3
x
+
2
=
(
∞
∞
)
=
lim
x
→
∞
1
+
2
x
x
2
3
=
lim
x
→
∞
1
+
2
x
3
=
1
3
.
{\displaystyle \lim _{x\to \infty }{\frac {x+\ln x^{2}}{3x+2}}=({\frac {\infty }{\infty }})=\lim _{x\to \infty }{\frac {1+{\frac {2x}{x^{2}}}}{3}}=\lim _{x\to \infty }{\frac {1+{\frac {2}{x}}}{3}}={\frac {1}{3}}.}
lim
x
→
0
1
−
sin
(
π
/
2
−
2
x
)
tan
2
(
3
x
)
=
(
0
0
)
=
lim
x
→
0
2
cos
(
π
/
2
−
2
x
)
2
tan
(
3
x
)
⋅
3
/
cos
2
(
3
x
)
=
1
3
lim
x
→
0
cos
2
(
3
x
)
cos
(
π
/
2
−
2
x
)
tan
(
3
x
)
=
{\displaystyle \lim _{x\to 0}{1-\sin(\pi /2-2x) \over \tan ^{2}(3x)}=({\frac {0}{0}})=\lim _{x\to 0}{2\cos(\pi /2-2x) \over 2\tan(3x)\cdot 3/\cos ^{2}(3x)}={1 \over 3}\lim _{x\to 0}{\cos ^{2}(3x)\cos(\pi /2-2x) \over \tan(3x)}=}
=
1
3
lim
x
→
0
−
2
cos
(
3
x
)
sin
(
3
x
)
3
cos
(
π
/
2
−
2
x
)
+
cos
2
(
3
x
)
sin
(
π
/
2
−
2
x
)
2
3
/
cos
2
(
3
x
)
=
1
3
⋅
0
+
2
3
=
2
9
.
{\displaystyle ={1 \over 3}\lim _{x\to 0}{-2\cos(3x)\sin(3x)3\cos(\pi /2-2x)+\cos ^{2}(3x)\sin(\pi /2-2x)2 \over 3/\cos ^{2}(3x)}={1 \over 3}\cdot {0+2 \over 3}={2 \over 9}.}
lim
x
→
0
1
−
sin
(
π
/
2
−
2
x
)
tan
2
(
3
x
)
=
(
0
0
)
=
lim
x
→
0
1
−
cos
(
2
x
)
tan
2
(
3
x
)
=
lim
x
→
0
2
sin
2
(
x
)
tan
2
(
3
x
)
=
2
(
lim
x
→
0
sin
(
x
)
tan
(
3
x
)
)
2
=
{\displaystyle \lim _{x\to 0}{1-\sin(\pi /2-2x) \over \tan ^{2}(3x)}=({\frac {0}{0}})=\lim _{x\to 0}{1-\cos(2x) \over \tan ^{2}(3x)}=\lim _{x\to 0}{2\sin ^{2}(x) \over \tan ^{2}(3x)}=2(\lim _{x\to 0}{\sin(x) \over \tan(3x)})^{2}=}
=
2
(
lim
x
→
0
cos
(
x
)
3
/
cos
2
(
3
x
)
)
2
=
2
⋅
(
1
3
)
2
=
2
9
{\displaystyle =2(\lim _{x\to 0}{\cos(x) \over 3/\cos ^{2}(3x)})^{2}=2\cdot ({1 \over 3})^{2}={2 \over 9}}
lim
x
→
0
+
0
x
ln
x
=
(
0
⋅
(
−
∞
)
)
=
lim
x
→
0
+
0
ln
x
x
−
1
/
2
=
lim
x
→
0
+
0
1
x
(
−
1
2
)
x
−
3
/
2
=
−
2
lim
x
→
0
+
0
x
=
0.
{\displaystyle \lim _{x\to 0+0}{\sqrt {x}}\ln x=(0\cdot (-\infty ))=\lim _{x\to 0+0}{\frac {\ln x}{x^{-1/2}}}=\lim _{x\to 0+0}{\frac {\frac {1}{x}}{(-{\frac {1}{2}})x^{-3/2}}}=-2\lim _{x\to 0+0}{\sqrt {x}}=0.}
Taikėmė II taisyklę.
Pritaikius Lopitalio taisyklę n kartų, apskaičiuojama ribinė reikšmė
lim
x
→
+
∞
x
n
e
x
=
lim
x
→
+
∞
n
x
n
−
1
e
x
=
lim
x
→
+
∞
n
(
n
−
1
)
x
n
−
2
e
x
=
lim
x
→
+
∞
n
(
n
−
1
)
(
n
−
2
)
x
n
−
3
e
x
=
.
.
.
=
lim
x
→
+
∞
n
!
e
x
=
0.
{\displaystyle \lim _{x\to +\infty }{\frac {x^{n}}{e^{x}}}=\lim _{x\to +\infty }{\frac {nx^{n-1}}{e^{x}}}=\lim _{x\to +\infty }{\frac {n(n-1)x^{n-2}}{e^{x}}}=\lim _{x\to +\infty }{\frac {n(n-1)(n-2)x^{n-3}}{e^{x}}}=...=\lim _{x\to +\infty }{\frac {n!}{e^{x}}}=0.}
lim
x
→
0
(
1
+
x
2
)
1
e
x
−
1
−
x
=
(
1
∞
)
.
{\displaystyle \lim _{x\to 0}(1+x^{2})^{\frac {1}{e^{x}-1-x}}=(1^{\infty }).}
y
=
(
1
+
x
2
)
1
e
x
−
1
−
x
.
{\displaystyle y=(1+x^{2})^{\frac {1}{e^{x}-1-x}}.}
ln
y
=
1
e
x
−
1
−
x
⋅
ln
(
1
+
x
2
)
.
{\displaystyle \ln y={\frac {1}{e^{x}-1-x}}\cdot \ln(1+x^{2}).}
Pritaikę Lopitalio taisyklę, gauname
lim
x
→
0
ln
y
=
lim
x
→
0
ln
(
1
+
x
2
)
e
x
−
1
−
x
=
lim
x
→
0
2
x
1
+
x
2
e
x
−
1
=
lim
x
→
0
2
x
(
e
x
−
1
)
(
1
+
x
2
)
=
lim
x
→
0
2
e
x
(
1
+
x
2
)
+
(
e
x
−
1
)
2
x
=
2.
{\displaystyle \lim _{x\to 0}\ln y=\lim _{x\to 0}{\frac {\ln(1+x^{2})}{e^{x}-1-x}}=\lim _{x\to 0}{\frac {\frac {2x}{1+x^{2}}}{e^{x}-1}}=\lim _{x\to 0}{\frac {2x}{(e^{x}-1)(1+x^{2})}}=\lim _{x\to 0}{\frac {2}{e^{x}(1+x^{2})+(e^{x}-1)2x}}=2.}
Iš to aišku, kad
lim
x
→
0
y
=
lim
x
→
0
e
ln
y
=
e
2
.
{\displaystyle \lim _{x\to 0}y=\lim _{x\to 0}e^{\ln y}=e^{2}.}
[ keisti ] Pirmiausia, kad įrodyti Lopitalio taisyklę reikia žinoti Lagranžo formulę ir Koši formulę . Lagranžo formulė yra tokia:
f
(
b
)
−
f
(
a
)
b
−
a
=
f
′
(
ξ
)
;
{\displaystyle {\frac {f(b)-f(a)}{b-a}}=f'(\xi );}
čia
ξ
{\displaystyle \xi }
Koši formulę galima gauti iš Lagranžo formulės . Tarkime, kad intervale [a; b] yra dvi tolydžios funkcijos f(x) ir g(x) (
g
(
x
)
≠
0
{\displaystyle g(x)\neq 0}
ξ
1
{\displaystyle \xi _{1}}
ξ
2
,
{\displaystyle \xi _{2},}
f
(
b
)
−
f
(
a
)
g
(
b
)
−
g
(
a
)
=
f
′
(
ξ
1
)
g
′
(
ξ
2
)
.
{\displaystyle {\frac {f(b)-f(a)}{g(b)-g(a)}}={\frac {f'(\xi _{1})}{g'(\xi _{2})}}.}
Atskirai funkcijai f(x) turime Lagranžo formulę :
f
(
b
)
−
f
(
a
)
b
−
a
=
f
′
(
ξ
1
)
;
{\displaystyle {\frac {f(b)-f(a)}{b-a}}=f'(\xi _{1});}
ir atskirai funkcijai g(x) turime Lagranžo formulę :
g
(
b
)
−
g
(
a
)
b
−
a
=
g
′
(
ξ
2
)
.
{\displaystyle {\frac {g(b)-g(a)}{b-a}}=g'(\xi _{2}).}
Padalinus kairiąsias abiejų funkcijų puses vieną iš kitos ir padalinus abiejų funkcijų dešiniąsias puses gausime Koši formulę :
f
(
b
)
−
f
(
a
)
b
−
a
g
(
b
)
−
g
(
a
)
b
−
a
=
f
′
(
ξ
1
)
g
′
(
ξ
2
)
,
{\displaystyle {\frac {\frac {f(b)-f(a)}{b-a}}{\frac {g(b)-g(a)}{b-a}}}={\frac {f'(\xi _{1})}{g'(\xi _{2})}},}
f
(
b
)
−
f
(
a
)
g
(
b
)
−
g
(
a
)
=
f
′
(
ξ
1
)
g
′
(
ξ
2
)
.
{\displaystyle {\frac {f(b)-f(a)}{g(b)-g(a)}}={\frac {f'(\xi _{1})}{g'(\xi _{2})}}.}
(Iš tikro, Koši formulėje
ξ
1
=
ξ
2
=
ξ
{\displaystyle \xi _{1}=\xi _{2}=\xi }
Lagranžo formulės , bet įrodinėjant Lopitalio taisyklę galima naudotis ir tokia Koši-Paraboloido formule ).
Neapibrėžtumo
0
0
{\displaystyle {\frac {0}{0}}}
aiškinimas . Sakome, kad dviejų funkcijų santykis
f
(
x
)
g
(
x
)
,
{\displaystyle {\frac {f(x)}{g(x)}},}
x
→
a
,
{\displaystyle x\to a,}
0
0
,
{\displaystyle {\frac {0}{0}},}
lim
x
→
a
f
(
x
)
=
lim
x
→
a
g
(
x
)
=
0.
{\displaystyle \lim _{x\to a}f(x)=\lim _{x\to a}g(x)=0.}
Kadangi
f
(
a
)
{\displaystyle f(a)}
g
(
a
)
{\displaystyle g(a)}
x
n
{\displaystyle x_{n}}
a (
x
n
{\displaystyle x_{n}}
a aplinkoje). Intervalas [a;
x
n
{\displaystyle x_{n}}
Koši teoremos sąlygas. Pagal tą teoremą intervalo [a;
x
n
{\displaystyle x_{n}}
ξ
n
{\displaystyle \xi _{n}}
ξ
1
{\displaystyle \xi _{1}}
ξ
2
{\displaystyle \xi _{2}}
f(x) ir g(x) ), kad
f
(
x
n
)
−
f
(
a
)
g
(
x
n
)
−
g
(
a
)
=
f
′
(
ξ
n
)
g
′
(
ξ
n
)
(
8.25
)
{\displaystyle {\frac {f(x_{n})-f(a)}{g(x_{n})-g(a)}}={\frac {f'(\xi _{n})}{g'(\xi _{n})}}\;\;(8.25)}
f
(
x
n
)
−
f
(
a
)
g
(
x
n
)
−
g
(
a
)
=
f
′
(
ξ
1
)
g
′
(
ξ
2
)
.
{\displaystyle {\frac {f(x_{n})-f(a)}{g(x_{n})-g(a)}}={\frac {f'(\xi _{1})}{g'(\xi _{2})}}.}
Atsižvelgę į tai, kad pagal papildomą apibrėžimą
f
(
a
)
=
g
(
a
)
=
0
,
{\displaystyle f(a)=g(a)=0,\;}
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
n
)
g
′
(
ξ
n
)
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{n})}{g'(\xi _{n})}}}
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
1
)
g
′
(
ξ
2
)
.
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{1})}{g'(\xi _{2})}}.}
Dabar tarkime, kad šioje lygybėje
x
n
→
a
{\displaystyle x_{n}\to a}
ξ
n
→
a
{\displaystyle \xi _{n}\to a}
ξ
1
→
a
{\displaystyle \xi _{1}\to a}
ξ
2
→
a
{\displaystyle \xi _{2}\to a}
lim
x
→
a
f
(
x
)
g
(
x
)
=
f
′
(
a
)
g
′
(
a
)
.
{\displaystyle \lim _{x\to a}{\frac {f(x)}{g(x)}}={\frac {f'(a)}{g'(a)}}.}
Neapibrėžtumo
∞
∞
{\displaystyle {\frac {\infty }{\infty }}}
aiškinimas . Sakome, kad dviejų funkcijų santykis
f
(
x
)
g
(
x
)
,
{\displaystyle {\frac {f(x)}{g(x)}},}
x
→
a
,
{\displaystyle x\to a,}
∞
∞
,
{\displaystyle {\frac {\infty }{\infty }},}
lim
x
→
a
f
(
x
)
=
lim
x
→
a
g
(
x
)
=
∞
{\displaystyle \lim _{x\to a}f(x)=\lim _{x\to a}g(x)=\infty \;\;}
∞
{\displaystyle \infty }
+
∞
{\displaystyle +\infty }
−
∞
{\displaystyle -\infty }
Taikydami Koši formulę segmentui (segmentu vadinamas uždaras intervalas aukšojoj matematikoje) [x; a], galime tvirtinti, jog jame yra toks taškas
ξ
,
{\displaystyle \xi ,}
f
(
a
)
−
f
(
x
)
g
(
a
)
−
g
(
x
)
=
f
(
a
)
g
(
a
)
1
−
f
(
x
)
f
(
a
)
1
−
g
(
x
)
g
(
a
)
=
f
′
(
ξ
)
g
′
(
ξ
)
.
{\displaystyle {\frac {f(a)-f(x)}{g(a)-g(x)}}={\frac {f(a)}{g(a)}}{\frac {1-{\frac {f(x)}{f(a)}}}{1-{\frac {g(x)}{g(a)}}}}={\frac {f'(\xi )}{g'(\xi )}}.}
Iš čia
f
(
a
)
g
(
a
)
=
f
′
(
ξ
)
g
′
(
ξ
)
1
−
g
(
x
)
g
(
a
)
1
−
f
(
x
)
f
(
a
)
.
{\displaystyle {\frac {f(a)}{g(a)}}={\frac {f'(\xi )}{g'(\xi )}}{\frac {1-{\frac {g(x)}{g(a)}}}{1-{\frac {f(x)}{f(a)}}}}.}
Kai x artėja prie a , bet niekad nepriartėja (niekad netampa a ), tai
f
(
a
)
g
(
a
)
=
f
′
(
ξ
)
g
′
(
ξ
)
1
−
g
(
x
)
∞
1
−
f
(
x
)
∞
=
f
′
(
ξ
)
g
′
(
ξ
)
1
−
0
1
−
0
=
f
′
(
ξ
)
g
′
(
ξ
)
;
{\displaystyle {\frac {f(a)}{g(a)}}={\frac {f'(\xi )}{g'(\xi )}}{\frac {1-{\frac {g(x)}{\infty }}}{1-{\frac {f(x)}{\infty }}}}={\frac {f'(\xi )}{g'(\xi )}}{\frac {1-0}{1-0}}={\frac {f'(\xi )}{g'(\xi )}};}
be kita ko, kai
x
→
a
,
{\displaystyle x\to a,}
ξ
{\displaystyle \xi }
a , bet kokiu norimu tikslumu. Tokiu budu, gauname apytikslę formulę, bet kokiu norimu tikslumu tašką
ξ
{\displaystyle \xi }
a , bet ne apsoliučiai lygiu a reikšmei. Todėl paskutinę formulę galime užrašyti šitaip:
f
(
a
)
g
(
a
)
=
lim
x
→
a
f
′
(
ξ
)
g
′
(
ξ
)
≈
f
′
(
a
)
g
′
(
a
)
.
{\displaystyle {\frac {f(a)}{g(a)}}=\lim _{x\to a}{\frac {f'(\xi )}{g'(\xi )}}\approx {\frac {f'(a)}{g'(a)}}.}
Profesionalesnis neapibrėžtumo
∞
∞
{\displaystyle {\frac {\infty }{\infty }}}
aiškinimas . Tarkime, kad
lim
x
→
a
f
(
x
)
=
lim
x
→
a
g
(
x
)
=
+
∞
.
{\displaystyle \lim _{x\to a}f(x)=\lim _{x\to a}g(x)=+\infty .}
Sakykime,
x
n
{\displaystyle x_{n}}
x
m
{\displaystyle x_{m}}
a (tad dėl šios priežasties
x
n
−
x
m
{\displaystyle x_{n}-x_{m}}
x
n
>
x
m
.
{\displaystyle x_{n}>x_{m}.}
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
,
{\displaystyle f(x_{m}),}
g
(
x
n
)
{\displaystyle g(x_{n})}
g
(
x
m
)
.
{\displaystyle g(x_{m}).}
Taikydami Koši formulę segmentui
[
x
m
;
x
n
]
,
{\displaystyle [x_{m};\;x_{n}],}
ξ
m
n
,
{\displaystyle \xi _{mn},}
f
(
x
n
)
−
f
(
x
m
)
g
(
x
n
)
−
g
(
x
m
)
=
f
(
x
n
)
g
(
x
n
)
1
−
f
(
x
m
)
f
(
x
n
)
1
−
g
(
x
m
)
g
(
x
n
)
=
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
.
{\displaystyle {\frac {f(x_{n})-f(x_{m})}{g(x_{n})-g(x_{m})}}={\frac {f(x_{n})}{g(x_{n})}}{\frac {1-{\frac {f(x_{m})}{f(x_{n})}}}{1-{\frac {g(x_{m})}{g(x_{n})}}}}={\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}.}
Iš čia
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
1
−
g
(
x
m
)
g
(
x
n
)
1
−
f
(
x
m
)
f
(
x
n
)
.
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}{\frac {1-{\frac {g(x_{m})}{g(x_{n})}}}{1-{\frac {f(x_{m})}{f(x_{n})}}}}.}
Kadangi pagal sąlygą
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
{\displaystyle f(x_{m})}
g
(
x
n
)
{\displaystyle g(x_{n})}
g
(
x
m
)
,
{\displaystyle g(x_{m}),}
f
(
x
n
)
g
(
x
n
)
≈
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
1
−
0.00001
1
−
0.000001
≈
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
.
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}\approx {\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}{\frac {1-0.00001}{1-0.000001}}\approx {\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}.}
Kai
x
n
→
a
,
{\displaystyle x_{n}\to a,}
ξ
m
n
,
{\displaystyle \xi _{mn},}
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
a . Todėl
lim
x
n
→
a
f
(
x
n
)
g
(
x
n
)
≈
lim
ξ
m
n
→
a
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
≈
f
′
(
a
)
g
′
(
a
)
.
{\displaystyle \lim _{x_{n}\to a}{\frac {f(x_{n})}{g(x_{n})}}\approx \lim _{\xi _{mn}\to a}{\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}\approx {\frac {f'(a)}{g'(a)}}.}
Gali kilti natūralus klausimas: kas, jeigu skirtumas tarp
x
n
{\displaystyle x_{n}}
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
x
m
{\displaystyle x_{m}}
a ) ir be to
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
{\displaystyle f(x_{m})}
g
(
x
n
)
{\displaystyle g(x_{n})}
g
(
x
m
)
{\displaystyle g(x_{m})}
Atsakymas yra toks, kad tada formulėje
lim
x
n
→
a
f
(
x
n
)
g
(
x
n
)
≈
lim
ξ
m
n
→
a
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
≈
f
′
(
a
)
g
′
(
a
)
{\displaystyle \lim _{x_{n}\to a}{\frac {f(x_{n})}{g(x_{n})}}\approx \lim _{\xi _{mn}\to a}{\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}\approx {\frac {f'(a)}{g'(a)}}}
lim
x
n
→
a
f
(
x
n
)
g
(
x
n
)
{\displaystyle \lim _{x_{n}\to a}{\frac {f(x_{n})}{g(x_{n})}}}
f
′
(
a
)
g
′
(
a
)
,
{\displaystyle {\frac {f'(a)}{g'(a)}},}
f
(
x
n
)
g
(
x
n
)
≈
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
1
−
0.9
1
−
0.7
≠
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
.
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}\approx {\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}{\frac {1-0.9}{1-0.7}}\neq {\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}.}
Todėl formulė
lim
x
n
→
a
f
(
x
n
)
g
(
x
n
)
≈
lim
x
n
→
a
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
=
f
′
(
a
)
g
′
(
a
)
{\displaystyle \lim _{x_{n}\to a}{\frac {f(x_{n})}{g(x_{n})}}\approx \lim _{x_{n}\to a}{\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}={\frac {f'(a)}{g'(a)}}}
bus teisinga [ir egzistuoja] tik tada, kai tenkinama sąlyga, kad
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
{\displaystyle f(x_{m})}
g
(
x
n
)
{\displaystyle g(x_{n})}
g
(
x
m
)
.
{\displaystyle g(x_{m}).}
Pavyzdys apie tai, kada Lopitalio taisyklė egzistuoja ir kada neegzistuoja. Tarkime turime funkciją
f
(
x
)
=
1
1
−
x
2
.
{\displaystyle f(x)={\frac {1}{1-x^{2}}}.}
a
=
1
{\displaystyle a=1}
lim
x
→
1
f
(
x
)
=
lim
x
→
1
1
1
−
x
2
=
∞
.
{\displaystyle \lim _{x\to 1}f(x)=\lim _{x\to 1}{\frac {1}{1-x^{2}}}=\infty .}
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
;
{\displaystyle f(x_{m});}
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
.
{\displaystyle f(x_{m}).}
Išnagrinėkime pirmąjį atvejį. Imkime
x
m
=
0.9997
{\displaystyle x_{m}=0.9997}
x
n
=
0.9999
{\displaystyle x_{n}=0.9999}
f
(
x
m
)
=
1
1
−
0.9997
2
=
1
1
−
0.99940009
=
1
0.00059991
=
1666.91670417229
;
{\displaystyle f(x_{m})={\frac {1}{1-0.9997^{2}}}={\frac {1}{1-0.99940009}}={\frac {1}{0.00059991}}=1666.91670417229;}
f
(
x
n
)
=
1
1
−
0.9999
2
=
1
1
−
0.99980001
=
1
0.00019999
=
5000.2500125
;
{\displaystyle f(x_{n})={\frac {1}{1-0.9999^{2}}}={\frac {1}{1-0.99980001}}={\frac {1}{0.00019999}}=5000.2500125;}
1
−
f
(
x
m
)
f
(
x
n
)
=
1
−
1666.91670417229
5000.2500125
=
1
−
0.3333666716674
=
0.66663332833.
{\displaystyle 1-{\frac {f(x_{m})}{f(x_{n})}}=1-{\frac {1666.91670417229}{5000.2500125}}=1-0.3333666716674=0.66663332833.}
Išnagrinėkime antrąjį atvejį. Imkime
x
m
=
0.999
{\displaystyle x_{m}=0.999}
x
n
=
0.999999
{\displaystyle x_{n}=0.999999}
f
(
x
m
)
=
1
1
−
0.999
2
=
1
1
−
0.998001
=
1
0.001999
=
500.25012506253
;
{\displaystyle f(x_{m})={\frac {1}{1-0.999^{2}}}={\frac {1}{1-0.998001}}={\frac {1}{0.001999}}=500.25012506253;}
f
(
x
n
)
=
1
1
−
0.999999
2
=
1
1
−
0.999998000001
=
1
0.000001999999
=
500000.250000125
;
{\displaystyle f(x_{n})={\frac {1}{1-0.999999^{2}}}={\frac {1}{1-0.999998000001}}={\frac {1}{0.000001999999}}=500000.250000125;}
1
−
f
(
x
m
)
f
(
x
n
)
=
1
−
500.25012506253
500000.250000125
=
1
−
0.0010004997498749
=
0.998999500250125.
{\displaystyle 1-{\frac {f(x_{m})}{f(x_{n})}}=1-{\frac {500.25012506253}{500000.250000125}}=1-0.0010004997498749=0.998999500250125.}
Matome, kad pirmuoju atveju formulė
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
1
−
g
(
x
m
)
g
(
x
n
)
1
−
f
(
x
m
)
f
(
x
n
)
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}{\frac {1-{\frac {g(x_{m})}{g(x_{n})}}}{1-{\frac {f(x_{m})}{f(x_{n})}}}}}
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
C
;
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}C;}
kur C yra nemažas skaičius (gali gautis, priklausomai nuo g(x) funkcijos, apie 0.3 arba apie 3).
Antruoju atveju formulė
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
1
−
g
(
x
m
)
g
(
x
n
)
1
−
f
(
x
m
)
f
(
x
n
)
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}{\frac {1-{\frac {g(x_{m})}{g(x_{n})}}}{1-{\frac {f(x_{m})}{f(x_{n})}}}}}
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
c
;
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}c;}
kur c yra labai mažas skaičius (priklausomai nuo g(x) funkcijos, c gali gautis lygus apie 0.999 arba apie 1.001).
Pirmuoju atveju Lopitalio taisyklė neegzistuoja. Antruoju atveju Lopitalio taisyklė egzistuoja (nes antruoju atveju Lopitalio taisyklės formulėje nėra jokios gana didelės konstantos C ). Pirmuoju atveju gaunama tokia formulė:
lim
x
n
→
a
f
(
x
n
)
g
(
x
n
)
≈
lim
x
n
→
a
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
C
=
f
′
(
a
)
g
′
(
a
)
C
,
{\displaystyle \lim _{x_{n}\to a}{\frac {f(x_{n})}{g(x_{n})}}\approx \lim _{x_{n}\to a}{\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}C={\frac {f'(a)}{g'(a)}}C,}
kuri nėra Lopitalio taisyklės formulė (dėl gana didelės konstantos C ). Vadinasi, Lopitalio taisyklė neskaičiuoja pagal formulę, kai
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
{\displaystyle f(x_{m})}
g
(
x
n
)
{\displaystyle g(x_{n})}
g
(
x
m
)
,
{\displaystyle g(x_{m}),}
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
a . Kai
a
=
+
∞
,
{\displaystyle a=+\infty ,}
f
(
x
)
=
5
x
{\displaystyle f(x)=5^{x}}
x
n
{\displaystyle x_{n}}
x
m
,
{\displaystyle x_{m},}
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
{\displaystyle f(x_{m})}
g
(
x
n
)
{\displaystyle g(x_{n})}
g
(
x
m
)
{\displaystyle g(x_{m})}
Update 1 . Gali būti (turbūt taip ir yra), kad jeigu riba
lim
x
→
a
f
(
x
)
g
(
x
)
{\displaystyle \lim _{x\to a}{\frac {f(x)}{g(x)}}}
lim
x
→
a
f
(
x
)
=
lim
x
→
a
g
(
x
)
=
∞
{\displaystyle \lim _{x\to a}f(x)=\lim _{x\to a}g(x)=\infty }
∞
{\displaystyle \infty }
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
a , vistiek formulėje
f
(
x
n
)
g
(
x
n
)
=
f
′
(
ξ
m
n
)
g
′
(
ξ
m
n
)
1
−
g
(
x
m
)
g
(
x
n
)
1
−
f
(
x
m
)
f
(
x
n
)
{\displaystyle {\frac {f(x_{n})}{g(x_{n})}}={\frac {f'(\xi _{mn})}{g'(\xi _{mn})}}{\frac {1-{\frac {g(x_{m})}{g(x_{n})}}}{1-{\frac {f(x_{m})}{f(x_{n})}}}}}
šitas dėmuo
1
−
g
(
x
m
)
g
(
x
n
)
1
−
f
(
x
m
)
f
(
x
n
)
{\displaystyle {\frac {1-{\frac {g(x_{m})}{g(x_{n})}}}{1-{\frac {f(x_{m})}{f(x_{n})}}}}}
f
(
x
n
)
{\displaystyle f(x_{n})}
f
(
x
m
)
{\displaystyle f(x_{m})}
g
(
x
n
)
{\displaystyle g(x_{n})}
g
(
x
m
)
{\displaystyle g(x_{m})}
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
a ) ar skirtumai dideli ( tarp
f
(
x
m
)
{\displaystyle f(x_{m})}
f
(
x
n
)
{\displaystyle f(x_{n})}
g
(
x
m
)
{\displaystyle g(x_{m})}
g
(
x
n
)
{\displaystyle g(x_{n})}
x
m
{\displaystyle x_{m}}
x
n
{\displaystyle x_{n}}
a ). 
![{\displaystyle \lim _{x\to a}[f(x)-g(x)]=\lim _{x\to a}[{\frac {1}{1/f(x)}}-{\frac {1}{1/g(x)}}]=\lim _{x\to a}{\frac {1/g(x)-1/f(x)}{1/f(x)\cdot 1/g(x)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54c99dee7620e5f7ae927eb32eddfc83cd99cb03)
![{\displaystyle \lim _{x\to 1+0}(x-1)^{\ln x}=(0^{0})=\lim _{x\to 1+0}e^{\ln x\cdot \ln(x-1)}=e^{\lim _{x\to 1+0}[\ln x\cdot \ln(x-1)]};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4758a2e2e74ca4871cc3b443aaed24a65db14b11)
![{\displaystyle \lim _{x\to 1+0}[\ln x\cdot \ln(x-1)]=\lim _{x\to 1+0}{\frac {\ln(x-1)}{\frac {1}{\ln x}}}=\lim _{x\to 1+0}{\frac {\frac {1}{x-1}}{-{\frac {1}{x\ln ^{2}x}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f20854d87dc8bb4c32f929696476c8165481c223)
![{\displaystyle [x_{m};\;x_{n}],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f4d1409cf48335f2754a0bcd889432caa5f0909)
