Skaičių sekos riba vadinama vertė, prie kurios artėja sekos narių vertės, tolstant į begalybę . Pavyzdžiui, turime seką:
{
1
1
,
1
2
,
1
3
,
1
4
,
1
5
,
…
,
1
n
,
…
}
{\displaystyle \lbrace \;{\frac {1}{1}},{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},\dots ,{\frac {1}{n}},\dots \;\rbrace }
Jokio sekos nario vertė nėra lygi nuliui, tačiau, kuo narys tolimesnis sekoje, tuo jo vertė artimesnė nuliui. Intuityviai suvokiame, kad sekos nariai artėja į nulį.
Tačiau toks apibrėžimas nėra tikslus ir netinkamas naudoti matematikoje . Griežtesnis apibrėžimas yra toks:
Jei
∀
ε
>
0
∃
N
(
ε
)
∈
N
:
n
>
N
⇒
|
a
n
−
a
|
<
ε
{\displaystyle \forall \varepsilon >0\;\exists N(\varepsilon )\in \mathbb {N} :n>N\Rightarrow |a_{n}-a|<\varepsilon }
a
{\displaystyle a}
sekos riba . Jei tokio skaičiaus nėra – seka ribos neturi. Kitaip tariant, jeigu egzistuoja toks sekos narys
a
N
{\displaystyle a_{N}}
a
{\displaystyle a}
a
{\displaystyle a}
riba . Iš esmės šis apibrėžimas atitinka mūsų natūralų suvokimą apie sekos ribą .
Jei seka turi ribą, tai sakome, kad seka konverguoja , kitu atveju – diverguoja .
Sekos ribą žymime:
lim
n
→
∞
a
n
=
L
{\displaystyle \lim _{n\rightarrow \infty }a_{n}=L}
Čia
lim
{\displaystyle \lim }
ribą ,
n
→
∞
{\displaystyle n\rightarrow \infty }
n
{\displaystyle n}
begalybę , o
a
n
{\displaystyle a_{n}}
Augustinas Koši suformulavo kriterijų, kurį tenkinančios sekos vadinamos Koši sekomis :
Seka
{
x
n
}
{\displaystyle {\{x_{n}\}}}
Koši seka , jei
∑
n
=
1
∞
a
n
{\displaystyle {\sum _{n=1}^{\infty }a_{n}}}
∀
ε
>
0
,
∃
N
∈
N
,
∀
n
>
m
>
N
:
|
∑
k
=
m
+
1
n
a
k
|
<
ε
{\displaystyle {\forall \varepsilon >0,\exists N\in {\boldsymbol {N}},\forall n>m>N:|\sum _{k=m+1}^{n}a_{k}|<\varepsilon }}
Koši kriterijus yra būtina ir pakankama sekos konvergavimo sąlyga – visos konverguojančios sekos yra Koši sekos ir atvirkščiai.
Skaičiuodami ribas pasiremiame jų savybėmis ir keliomis elementariausiomis ribomis:
lim
n
→
∞
n
=
∞
{\displaystyle \lim _{n\rightarrow \infty }n=\infty }
lim
n
→
∞
1
n
=
0
{\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n}}=0}
lim
n
→
∞
a
n
=
∞
{\displaystyle \lim _{n\rightarrow \infty }a^{n}=\infty }
lim
n
→
∞
a
1
n
=
1
{\displaystyle \lim _{n\rightarrow \infty }a^{\frac {1}{n}}=1}
lim
n
→
∞
n
1
n
=
1
{\displaystyle \lim _{n\rightarrow \infty }n^{\frac {1}{n}}=1}
lim
x
→
0
sin
x
x
=
1
,
{\displaystyle \lim _{x\to 0}{\frac {\sin x}{x}}=1,}
lim
x
→
0
sin
(
k
x
)
x
=
k
,
{\displaystyle \lim _{x\to 0}{\frac {\sin(kx)}{x}}=k,}
lim
x
→
0
tan
x
x
=
lim
x
→
0
(
sin
x
x
⋅
1
cos
x
)
=
1
,
{\displaystyle \lim _{x\to 0}{\frac {\tan x}{x}}=\lim _{x\to 0}({\frac {\sin x}{x}}\cdot {\frac {1}{\cos x}})=1,}
lim
x
→
0
arcsin
x
x
=
1
,
{\displaystyle \lim _{x\to 0}{\frac {\arcsin x}{x}}=1,}
lim
x
→
0
arctan
x
x
=
1
,
{\displaystyle \lim _{x\to 0}{\frac {\arctan x}{x}}=1,}
lim
x
→
0
ln
(
1
+
x
)
x
=
lim
x
→
0
[
ln
(
1
+
x
)
1
x
]
=
ln
[
lim
x
→
0
(
1
+
x
)
1
x
]
=
ln
e
=
1
,
{\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=\lim _{x\to 0}[\ln(1+x)^{\frac {1}{x}}]=\ln[\lim _{x\to 0}(1+x)^{\frac {1}{x}}]=\ln e=1,}
lim
x
→
0
e
x
−
1
x
=
1.
{\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}=1.}
1
∞
=
0
{\displaystyle {\frac {1}{\infty }}=0}
Ieškodami ribų galime tiesiog įrašyti begalybę vietoj
n
{\displaystyle n}
neapibrėžtumą , kurį ir reikia pašalinti, pvz.:
lim
n
→
∞
2
n
−
1
n
=
∞
∞
=
lim
n
→
∞
2
−
1
n
1
=
2
−
1
∞
=
2.
{\displaystyle \lim _{n\rightarrow \infty }{\frac {2n-1}{n}}={\frac {\infty }{\infty }}=\lim _{n\rightarrow \infty }{\frac {2-{\frac {1}{n}}}{1}}=2-{\frac {1}{\infty }}=2.}
lim
x
→
0
sin
3
x
sin
7
x
=
lim
x
→
0
sin
3
x
3
x
⋅
3
x
sin
7
x
7
x
⋅
7
x
=
lim
x
→
0
3
x
7
x
=
3
7
.
{\displaystyle \lim _{x\to 0}{\frac {\sin 3x}{\sin 7x}}=\lim _{x\to 0}{\frac {{\frac {\sin 3x}{3x}}\cdot 3x}{{\frac {\sin 7x}{7x}}\cdot 7x}}=\lim _{x\to 0}{\frac {3x}{7x}}={\frac {3}{7}}.}
lim
n
→
∞
(
1
+
1
n
(
n
+
2
)
)
n
=
lim
n
→
∞
(
1
+
1
n
(
n
+
2
)
)
n
(
n
+
2
)
(
n
n
(
n
+
2
)
)
=
lim
n
→
∞
e
n
n
(
n
+
2
)
=
e
1
∞
=
1.
{\displaystyle \lim _{n\rightarrow \infty }\left(1+{\frac {1}{n(n+2)}}\right)^{n}=\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n(n+2)}}\right)^{n(n+2)\left({\frac {n}{n(n+2)}}\right)}=\lim _{n\rightarrow \infty }{\mathsf {e}}^{\frac {n}{n(n+2)}}={\mathsf {e}}^{\frac {1}{\infty }}=1.}
lim
n
→
∞
n
2
+
4
n
−
5
n
2
−
1
=
lim
n
→
∞
n
2
(
1
+
4
n
−
5
n
2
)
n
2
(
1
−
1
n
2
)
=
1
+
4
∞
−
5
∞
1
−
1
∞
=
1.
{\displaystyle \lim _{n\rightarrow \infty }{\frac {n^{2}+4n-5}{n^{2}-1}}=\lim _{n\rightarrow \infty }{\frac {n^{2}\left(1+{\frac {4}{n}}-{\frac {5}{n^{2}}}\right)}{n^{2}\left(1-{\frac {1}{n^{2}}}\right)}}={\frac {1+{\frac {4}{\infty }}-{\frac {5}{\infty }}}{1-{\frac {1}{\infty }}}}=1.}
Seka
{
−
1
,
1
,
−
1
,
1
,
…
,
(
−
1
)
n
,
…
}
{\displaystyle \lbrace \;-1,1,-1,1,\dots ,(-1)^{n},\dots \;\rbrace }
lim
x
→
∞
x
3
−
4
x
2
+
7
x
−
3
x
2
+
2
x
−
11
=
lim
x
→
∞
x
3
−
4
x
2
+
7
x
−
3
x
3
x
2
+
2
x
−
11
x
3
=
lim
x
→
∞
1
−
4
x
+
7
x
2
−
3
x
3
1
x
+
2
x
2
−
11
x
3
=
∞
.
{\displaystyle \lim _{x\to \infty }{\frac {x^{3}-4x^{2}+7x-3}{x^{2}+2x-11}}=\lim _{x\to \infty }{\frac {\frac {x^{3}-4x^{2}+7x-3}{x^{3}}}{\frac {x^{2}+2x-11}{x^{3}}}}=\lim _{x\to \infty }{\frac {1-{\frac {4}{x}}+{\frac {7}{x^{2}}}-{\frac {3}{x^{3}}}}{{\frac {1}{x}}+{\frac {2}{x^{2}}}-{\frac {11}{x^{3}}}}}=\infty .}
lim
x
→
0
3
x
2
−
2
x
2
x
2
−
5
x
=
(
0
0
)
=
lim
x
→
0
x
(
3
x
−
2
)
x
(
2
x
−
5
)
=
−
2
−
5
=
2
5
.
{\displaystyle \lim _{x\to 0}{\frac {3x^{2}-2x}{2x^{2}-5x}}=({\frac {0}{0}})=\lim _{x\to 0}{\frac {x(3x-2)}{x(2x-5)}}={\frac {-2}{-5}}={\frac {2}{5}}.}
lim
x
→
∞
3
x
2
−
1
5
x
2
+
2
x
=
lim
x
→
∞
3
−
1
/
x
2
5
+
2
/
x
=
3
−
0
5
+
0
=
3
5
.
{\displaystyle \lim _{x\to \infty }{\frac {3x^{2}-1}{5x^{2}+2x}}=\lim _{x\to \infty }{\frac {3-1/x^{2}}{5+2/x}}={\frac {3-0}{5+0}}={\frac {3}{5}}.}
lim
x
→
−
3
2
4
x
2
−
9
2
x
+
3
=
lim
x
→
−
3
2
(
2
x
−
3
)
(
2
x
+
3
)
2
x
+
3
=
lim
x
→
−
3
2
(
2
x
−
3
)
=
−
2
3
2
−
3
=
−
6.
{\displaystyle \lim _{x\to -{\frac {3}{2}}}{\frac {4x^{2}-9}{2x+3}}=\lim _{x\to -{\frac {3}{2}}}{\frac {(2x-3)(2x+3)}{2x+3}}=\lim _{x\to -{\frac {3}{2}}}(2x-3)=-2{\frac {3}{2}}-3=-6.}
lim
x
→
1
x
−
1
1
−
x
2
=
lim
x
→
1
x
−
1
(
1
−
x
)
(
1
+
x
)
=
lim
x
→
1
x
−
1
−
(
x
−
1
)
(
1
+
x
)
=
lim
x
→
1
1
−
(
1
+
x
)
=
−
1
2
.
{\displaystyle \lim _{x\to 1}{\frac {x-1}{1-x^{2}}}=\lim _{x\to 1}{\frac {x-1}{(1-x)(1+x)}}=\lim _{x\to 1}{\frac {x-1}{-(x-1)(1+x)}}=\lim _{x\to 1}{\frac {1}{-(1+x)}}=-{\frac {1}{2}}.}
lim
x
→
2
(
1
x
−
2
−
4
x
2
−
4
)
=
lim
x
→
2
x
+
2
−
4
x
2
−
4
=
lim
x
→
2
x
−
2
(
x
−
2
)
(
x
+
2
)
=
lim
x
→
2
1
x
+
2
=
1
4
.
{\displaystyle \lim _{x\to 2}({\frac {1}{x-2}}-{\frac {4}{x^{2}-4}})=\lim _{x\to 2}{\frac {x+2-4}{x^{2}-4}}=\lim _{x\to 2}{\frac {x-2}{(x-2)(x+2)}}=\lim _{x\to 2}{\frac {1}{x+2}}={\frac {1}{4}}.}
lim
x
→
0
1
−
1
−
x
2
x
=
lim
x
→
0
(
1
−
1
−
x
2
)
(
1
+
1
−
x
2
)
x
(
1
+
1
−
x
2
)
=
lim
x
→
0
1
−
(
1
−
x
2
)
x
(
1
+
1
−
x
2
)
=
{\displaystyle \lim _{x\to 0}{\frac {1-{\sqrt {1-x^{2}}}}{x}}=\lim _{x\to 0}{\frac {(1-{\sqrt {1-x^{2}}})(1+{\sqrt {1-x^{2}}})}{x(1+{\sqrt {1-x^{2}}})}}=\lim _{x\to 0}{\frac {1-(1-x^{2})}{x(1+{\sqrt {1-x^{2}}})}}=}
=
lim
x
→
0
x
1
+
1
−
x
2
=
0
2
=
0.
{\displaystyle =\lim _{x\to 0}{\frac {x}{1+{\sqrt {1-x^{2}}}}}={\frac {0}{2}}=0.}
lim
x
→
2
2
x
2
−
8
x
−
2
=
lim
x
→
2
2
(
x
−
2
)
(
x
+
2
)
x
−
2
=
lim
x
→
2
(
2
x
+
4
)
=
8.
{\displaystyle \lim _{x\to 2}{\frac {2x^{2}-8}{x-2}}=\lim _{x\to 2}{\frac {2(x-2)(x+2)}{x-2}}=\lim _{x\to 2}(2x+4)=8.}
lim
x
→
0
x
+
sin
3
x
x
=
lim
x
→
0
3
x
3
x
⋅
(
x
x
+
sin
3
x
x
)
=
lim
x
→
0
(
1
+
3
x
sin
3
x
3
x
⋅
x
)
=
lim
x
→
0
(
1
+
3
x
x
)
=
4.
{\displaystyle \lim _{x\to 0}{\frac {x+\sin 3x}{x}}=\lim _{x\to 0}{\frac {3x}{3x}}\cdot ({\frac {x}{x}}+{\frac {\sin 3x}{x}})=\lim _{x\to 0}(1+{\frac {3x\sin 3x}{3x\cdot x}})=\lim _{x\to 0}(1+{\frac {3x}{x}})=4.}
lim
x
→
∞
(
x
+
1
x
)
x
3
=
lim
x
→
∞
(
(
1
+
1
x
)
x
)
1
3
=
e
1
3
.
{\displaystyle \lim _{x\to \infty }({\frac {x+1}{x}})^{\frac {x}{3}}=\lim _{x\to \infty }((1+{\frac {1}{x}})^{x})^{\frac {1}{3}}=e^{\frac {1}{3}}.}
lim
x
→
∞
(
2
x
+
1
2
x
+
3
)
3
x
−
2
5
=
lim
x
→
∞
(
2
x
+
3
−
2
2
x
+
3
)
3
x
−
2
5
=
lim
x
→
∞
(
(
1
+
−
2
2
x
+
3
)
2
x
+
3
−
2
)
−
2
2
x
+
3
⋅
3
x
−
2
5
=
{\displaystyle \lim _{x\to \infty }({\frac {2x+1}{2x+3}})^{\frac {3x-2}{5}}=\lim _{x\to \infty }({\frac {2x+3-2}{2x+3}})^{\frac {3x-2}{5}}=\lim _{x\to \infty }((1+{\frac {-2}{2x+3}})^{\frac {2x+3}{-2}})^{{\frac {-2}{2x+3}}\cdot {\frac {3x-2}{5}}}=}
=
e
−
2
5
lim
x
→
∞
3
x
−
2
2
x
+
3
=
e
−
2
5
⋅
3
2
=
e
−
3
5
.
{\displaystyle =e^{-{\frac {2}{5}}\lim _{x\to \infty }{\frac {3x-2}{2x+3}}}=e^{-{\frac {2}{5}}\cdot {\frac {3}{2}}}=e^{-{\frac {3}{5}}}.}
lim
x
→
3
x
2
−
9
x
+
1
−
2
=
lim
x
→
3
(
x
−
3
)
(
x
+
3
)
(
x
+
1
+
2
)
(
x
+
1
−
2
)
(
x
+
1
+
2
)
=
lim
x
→
3
(
x
−
3
)
(
x
+
3
)
(
x
+
1
+
2
)
x
+
1
−
4
=
{\displaystyle \lim _{x\to 3}{\frac {x^{2}-9}{{\sqrt {x+1}}-2}}=\lim _{x\to 3}{\frac {(x-3)(x+3)({\sqrt {x+1}}+2)}{({\sqrt {x+1}}-2)({\sqrt {x+1}}+2)}}=\lim _{x\to 3}{\frac {(x-3)(x+3)({\sqrt {x+1}}+2)}{x+1-4}}=}
=
lim
x
→
3
[
(
x
+
3
)
(
x
+
1
+
2
)
]
=
6
⋅
4
=
24.
{\displaystyle =\lim _{x\to 3}[(x+3)({\sqrt {x+1}}+2)]=6\cdot 4=24.}
lim
x
→
0
(
1
+
x
)
1
3
−
1
x
+
1
−
1
=
lim
z
→
1
z
6
3
−
1
z
6
−
1
=
lim
z
→
1
(
z
−
1
)
(
z
+
1
)
(
z
−
1
)
(
z
2
+
z
+
1
)
=
lim
z
→
1
z
+
1
z
2
+
z
+
1
=
2
3
,
{\displaystyle \lim _{x\to 0}{\frac {(1+x)^{\frac {1}{3}}-1}{{\sqrt {x+1}}-1}}=\lim _{z\to 1}{\frac {z^{\frac {6}{3}}-1}{{\sqrt {z^{6}}}-1}}=\lim _{z\to 1}{\frac {(z-1)(z+1)}{(z-1)(z^{2}+z+1)}}=\lim _{z\to 1}{\frac {z+1}{z^{2}+z+1}}={\frac {2}{3}},}
kur keičiame kintamąjį:
1
+
x
=
z
6
.
{\displaystyle 1+x=z^{6}.}
x
→
0
,
{\displaystyle x\rightarrow 0,}
z
→
1.
{\displaystyle z\rightarrow 1.}
lim
x
→
π
sin
2
x
1
+
cos
3
x
=
lim
x
→
π
1
−
cos
2
x
(
1
+
cos
x
)
(
1
−
cos
x
+
cos
2
x
)
=
lim
x
→
π
(
1
−
cos
x
)
(
1
+
cos
x
)
(
1
+
cos
x
)
(
1
−
cos
x
+
cos
2
x
)
=
{\displaystyle \lim _{x\to \pi }{\frac {\sin ^{2}x}{1+\cos ^{3}x}}=\lim _{x\to \pi }{\frac {1-\cos ^{2}x}{(1+\cos x)(1-\cos x+\cos ^{2}x)}}=\lim _{x\to \pi }{\frac {(1-\cos x)(1+\cos x)}{(1+\cos x)(1-\cos x+\cos ^{2}x)}}=}
=
lim
x
→
π
1
−
cos
x
1
−
cos
x
+
cos
2
x
=
1
−
(
−
1
)
1
−
(
−
1
)
+
(
−
1
)
2
=
2
3
.
{\displaystyle =\lim _{x\to \pi }{\frac {1-\cos x}{1-\cos x+\cos ^{2}x}}={\frac {1-(-1)}{1-(-1)+(-1)^{2}}}={\frac {2}{3}}.}
lim
x
→
0
x
2
1
−
cos
x
=
lim
x
→
0
x
2
2
sin
2
(
x
/
2
)
=
2
lim
x
→
0
(
x
/
2
sin
(
x
/
2
)
)
2
=
2.
{\displaystyle \lim _{x\to 0}{\frac {x^{2}}{1-\cos x}}=\lim _{x\to 0}{\frac {x^{2}}{2\sin ^{2}(x/2)}}=2\lim _{x\to 0}({\frac {x/2}{\sin(x/2)}})^{2}=2.}
lim
x
→
∞
(
1
+
1
x
2
)
x
=
lim
x
→
∞
[
(
1
+
1
x
2
)
x
2
]
1
x
=
e
0
=
1.
{\displaystyle \lim _{x\to \infty }(1+{\frac {1}{x^{2}}})^{x}=\lim _{x\to \infty }[(1+{\frac {1}{x^{2}}})^{x^{2}}]^{\frac {1}{x}}=e^{0}=1.}
lim
x
→
0
log
a
(
1
+
x
)
x
=
lim
x
→
0
[
log
a
(
1
+
x
)
1
x
]
=
log
a
e
.
{\displaystyle \lim _{x\to 0}{\frac {\log _{a}(1+x)}{x}}=\lim _{x\to 0}[\log _{a}(1+x)^{\frac {1}{x}}]=\log _{a}e.}
lim
x
→
2
x
3
−
8
x
−
2
=
lim
x
→
2
(
x
−
2
)
(
x
2
+
2
x
+
4
)
x
−
2
=
lim
x
→
2
(
x
2
+
2
x
+
4
)
=
12.
{\displaystyle \lim _{x\to 2}{\frac {x^{3}-8}{x-2}}=\lim _{x\to 2}{\frac {(x-2)(x^{2}+2x+4)}{x-2}}=\lim _{x\to 2}(x^{2}+2x+4)=12.}
lim
x
→
3
x
2
+
x
−
12
2
x
2
−
9
x
+
9
=
lim
x
→
3
(
x
−
3
)
(
x
+
4
)
2
(
x
−
3
)
(
x
−
1.5
)
=
lim
x
→
3
x
+
4
2
x
−
3
=
7
3
.
{\displaystyle \lim _{x\to 3}{\frac {x^{2}+x-12}{2x^{2}-9x+9}}=\lim _{x\to 3}{\frac {(x-3)(x+4)}{2(x-3)(x-1.5)}}=\lim _{x\to 3}{\frac {x+4}{2x-3}}={\frac {7}{3}}.}
lim
x
→
0
tan
x
−
sin
x
x
3
=
lim
x
→
0
sin
x
cos
x
−
sin
x
x
3
=
lim
x
→
0
sin
x
(
1
−
cos
x
)
cos
x
x
3
=
lim
x
→
0
(
sin
x
x
⋅
1
−
cos
x
x
2
⋅
cos
x
)
=
{\displaystyle \lim _{x\to 0}{\frac {\tan x-\sin x}{x^{3}}}=\lim _{x\to 0}{\frac {{\frac {\sin x}{\cos x}}-\sin x}{x^{3}}}=\lim _{x\to 0}{\frac {\frac {\sin x(1-\cos x)}{\cos x}}{x^{3}}}=\lim _{x\to 0}({\frac {\sin x}{x}}\cdot {\frac {1-\cos x}{x^{2}\cdot \cos x}})=}
=
lim
x
→
0
sin
x
x
⋅
lim
x
→
0
1
cos
x
⋅
lim
x
→
0
1
−
cos
x
x
2
=
lim
x
→
0
2
sin
2
x
2
x
2
=
1
2
lim
x
→
0
(
sin
x
2
x
2
)
2
=
1
2
.
{\displaystyle =\lim _{x\to 0}{\frac {\sin x}{x}}\cdot \lim _{x\to 0}{\frac {1}{\cos x}}\cdot \lim _{x\to 0}{\frac {1-\cos x}{x^{2}}}=\lim _{x\to 0}{\frac {2\sin ^{2}{\frac {x}{2}}}{x^{2}}}={\frac {1}{2}}\lim _{x\to 0}({\frac {\sin {\frac {x}{2}}}{\frac {x}{2}}})^{2}={\frac {1}{2}}.}
lim
x
→
∞
(
x
3
3
x
2
−
4
−
x
2
3
x
+
2
)
=
lim
x
→
∞
x
3
(
3
x
+
2
)
−
x
2
(
3
x
2
−
4
)
(
3
x
2
−
4
)
(
3
x
+
2
)
=
lim
x
→
∞
2
x
3
+
4
x
2
9
x
3
+
6
x
2
−
12
x
−
8
=
{\displaystyle \lim _{x\to \infty }({\frac {x^{3}}{3x^{2}-4}}-{\frac {x^{2}}{3x+2}})=\lim _{x\to \infty }{\frac {x^{3}(3x+2)-x^{2}(3x^{2}-4)}{(3x^{2}-4)(3x+2)}}=\lim _{x\to \infty }{\frac {2x^{3}+4x^{2}}{9x^{3}+6x^{2}-12x-8}}=}
=
lim
x
→
∞
2
+
4
x
9
+
6
x
−
12
x
2
−
8
x
3
=
2
9
.
{\displaystyle =\lim _{x\to \infty }{\frac {2+{\frac {4}{x}}}{9+{\frac {6}{x}}-{\frac {12}{x^{2}}}-{\frac {8}{x^{3}}}}}={\frac {2}{9}}.}
lim
x
→
1
2
x
−
2
(
26
+
x
)
1
3
−
3
=
lim
t
→
3
2
(
t
3
−
27
)
t
−
3
=
lim
t
→
3
2
(
t
−
3
)
(
t
2
+
3
t
+
9
)
t
−
3
=
2
lim
t
→
3
(
t
2
+
3
t
+
9
)
=
54
,
{\displaystyle \lim _{x\to 1}{\frac {2x-2}{(26+x)^{\frac {1}{3}}-3}}=\lim _{t\to 3}{\frac {2(t^{3}-27)}{t-3}}=\lim _{t\to 3}{\frac {2(t-3)(t^{2}+3t+9)}{t-3}}=2\lim _{t\to 3}(t^{2}+3t+9)=54,}
kur
26
+
x
=
t
3
;
{\displaystyle 26+x=t^{3};}
x
=
t
3
−
26
;
{\displaystyle x=t^{3}-26;}
t
→
3
,
{\displaystyle t\rightarrow 3,}
x
→
1.
{\displaystyle x\rightarrow 1.}
Rasime ribą
lim
x
→
1
2
8
x
3
−
1
6
x
2
+
3
x
−
3
=
(
0
0
)
{\displaystyle \lim _{x\to {\frac {1}{2}}}{\frac {8x^{3}-1}{6x^{2}+3x-3}}=({\frac {0}{0}})}
Skaitiklis išskaidomas pagal formulę
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
a
b
+
b
2
)
=
8
x
3
−
1
=
(
2
x
−
1
)
(
4
x
2
+
2
x
+
1
)
{\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})=8x^{3}-1=(2x-1)(4x^{2}+2x+1)}
Vardiklis gali būti išskaidomas surandant jo sprendinius
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
6
x
2
+
3
x
−
3
=
0
{\displaystyle 6x^{2}+3x-3=0}
D
=
b
2
−
4
a
c
=
3
2
−
4
⋅
6
⋅
(
−
3
)
=
9
+
72
=
81
{\displaystyle D=b^{2}-4ac=3^{2}-4\cdot 6\cdot (-3)=9+72=81}
x
1
,
2
=
−
b
±
D
2
a
=
−
3
±
81
2
⋅
6
=
−
3
±
9
12
=
−
1
;
1
2
.
{\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {D}}}{2a}}={\frac {-3\pm {\sqrt {81}}}{2\cdot 6}}={\frac {-3\pm 9}{12}}=-1;{\frac {1}{2}}.}
Kvadratinė lygtis yra išskaidoma
a
x
2
+
b
x
+
c
=
a
(
x
−
x
1
)
(
x
−
x
2
)
=
6
x
2
+
3
x
−
3
=
6
(
x
−
(
−
1
)
)
(
x
−
1
2
)
.
{\displaystyle ax^{2}+bx+c=a(x-x_{1})(x-x_{2})=6x^{2}+3x-3=6(x-(-1))(x-{\frac {1}{2}}).}
lim
x
→
1
2
(
2
x
−
1
)
(
4
x
2
+
2
x
+
1
)
6
(
x
+
1
)
(
x
−
1
2
)
=
lim
x
→
1
2
(
2
x
−
1
)
(
4
x
2
+
2
x
+
1
)
3
(
x
+
1
)
(
2
x
−
1
)
=
lim
x
→
1
2
4
x
2
+
2
x
+
1
3
x
+
3
=
3
4.5
=
2
3
.
{\displaystyle \lim _{x\to {\frac {1}{2}}}{\frac {(2x-1)(4x^{2}+2x+1)}{6(x+1)(x-{\frac {1}{2}})}}=\lim _{x\to {\frac {1}{2}}}{\frac {(2x-1)(4x^{2}+2x+1)}{3(x+1)(2x-1)}}=\lim _{x\to {\frac {1}{2}}}{\frac {4x^{2}+2x+1}{3x+3}}={\frac {3}{4.5}}={\frac {2}{3}}.}
lim
x
→
∞
(
1
−
1
x
)
x
=
lim
z
→
−
∞
[
(
1
+
1
z
)
z
]
−
1
=
e
−
1
=
1
e
.
{\displaystyle \lim _{x\to \infty }(1-{\frac {1}{x}})^{x}=\lim _{z\to -\infty }[(1+{\frac {1}{z}})^{z}]^{-1}=e^{-1}={\frac {1}{e}}.}
lim
x
→
6
x
+
3
−
3
x
−
6
=
lim
x
→
6
x
+
3
−
9
(
x
−
6
)
(
x
+
3
+
3
)
=
lim
x
→
6
1
x
+
3
+
3
=
1
6
.
{\displaystyle \lim _{x\to 6}{\frac {{\sqrt {x+3}}-3}{x-6}}=\lim _{x\to 6}{\frac {x+3-9}{(x-6)({\sqrt {x+3}}+3)}}=\lim _{x\to 6}{\frac {1}{{\sqrt {x+3}}+3}}={\frac {1}{6}}.}
lim
x
→
8
x
−
8
x
1
3
−
2
=
lim
x
→
8
(
x
1
3
−
2
)
(
x
2
3
+
2
x
1
3
+
2
2
)
x
1
3
−
2
=
lim
x
→
8
(
x
2
3
+
2
x
1
3
+
4
)
=
12.
{\displaystyle \lim _{x\to 8}{\frac {x-8}{x^{\frac {1}{3}}-2}}=\lim _{x\to 8}{\frac {(x^{\frac {1}{3}}-2)(x^{\frac {2}{3}}+2x^{\frac {1}{3}}+2^{2})}{x^{\frac {1}{3}}-2}}=\lim _{x\to 8}(x^{\frac {2}{3}}+2x^{\frac {1}{3}}+4)=12.}
lim
x
→
∞
(
x
−
1
x
)
5
x
=
lim
x
→
∞
(
1
−
1
x
)
5
x
=
lim
z
→
−
∞
(
(
1
+
1
z
)
z
)
−
5
=
e
−
5
.
{\displaystyle \lim _{x\to \infty }({\frac {x-1}{x}})^{5x}=\lim _{x\to \infty }(1-{\frac {1}{x}})^{5x}=\lim _{z\to -\infty }((1+{\frac {1}{z}})^{z})^{-5}=e^{-5}.}
lim
x
→
0
(
1
+
tan
x
)
1
sin
x
=
lim
x
→
0
(
(
1
+
tan
x
)
cos
x
sin
x
)
1
cos
x
=
e
.
{\displaystyle \lim _{x\to 0}(1+\tan x)^{\frac {1}{\sin x}}=\lim _{x\to 0}((1+\tan x)^{\frac {\cos x}{\sin x}})^{\frac {1}{\cos x}}=e.}
lim
x
→
∞
(
x
2
+
5
3
x
2
+
1
)
x
2
=
lim
x
→
∞
(
1
+
5
x
2
3
+
1
x
2
)
x
2
=
lim
x
→
∞
(
1
3
)
x
2
=
0.
{\displaystyle \lim _{x\to \infty }({\frac {x^{2}+5}{3x^{2}+1}})^{x^{2}}=\lim _{x\to \infty }({\frac {1+{\frac {5}{x^{2}}}}{3+{\frac {1}{x^{2}}}}})^{x^{2}}=\lim _{x\to \infty }({\frac {1}{3}})^{x^{2}}=0.}
lim
x
→
0
cos
π
⋅
x
sin
x
=
lim
x
→
0
cos
π
⋅
x
x
sin
x
x
=
lim
x
→
0
cos
π
sin
x
x
=
lim
x
→
0
cos
π
=
−
1.
{\displaystyle \lim _{x\to 0}\cos {\frac {\pi \cdot x}{\sin x}}=\lim _{x\to 0}\cos {\frac {\pi \cdot x}{x{\frac {\sin x}{x}}}}=\lim _{x\to 0}\cos {\frac {\pi }{\frac {\sin x}{x}}}=\lim _{x\to 0}\cos \pi =-1.}
![{\displaystyle \lim _{x\to a}[f(x)\pm g(x)]=\lim _{x\to a}f(x)\pm \lim _{x\to a}g(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/891d882bacf88ecb35eda0a086ad2ec18be80eca)
![{\displaystyle \lim _{x\to a}[f(x)\cdot g(x)]=\lim _{x\to a}f(x)\cdot \lim _{x\to a}g(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ab395c7036419cacee5d3406322074e91d170cb)
![{\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=\lim _{x\to 0}[\ln(1+x)^{\frac {1}{x}}]=\ln[\lim _{x\to 0}(1+x)^{\frac {1}{x}}]=\ln e=1,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40ffa9184d283b1d7219bd6cda1f1699f25bb75f)
![{\displaystyle =\lim _{x\to 3}[(x+3)({\sqrt {x+1}}+2)]=6\cdot 4=24.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae915797228b2da2b266e27e360d1716a4f94bf5)
![{\displaystyle \lim _{x\to \infty }(1+{\frac {1}{x^{2}}})^{x}=\lim _{x\to \infty }[(1+{\frac {1}{x^{2}}})^{x^{2}}]^{\frac {1}{x}}=e^{0}=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9de2c3bf33f1f3c41a4ea6de716f6e81edffeb39)
![{\displaystyle \lim _{x\to 0}{\frac {\log _{a}(1+x)}{x}}=\lim _{x\to 0}[\log _{a}(1+x)^{\frac {1}{x}}]=\log _{a}e.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38c70172281ca52fa6124effa9dbe5f2b7903b12)
![{\displaystyle \lim _{x\to \infty }(1-{\frac {1}{x}})^{x}=\lim _{z\to -\infty }[(1+{\frac {1}{z}})^{z}]^{-1}=e^{-1}={\frac {1}{e}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccc2cba9c7b216981f9528fe7ef2a2926915d52e)
