Sekos riba

Skaičių sekos riba vadinama vertė, prie kurios artėja sekos narių vertės, tolstant į begalybę. Pavyzdžiui, turime seką:

${\displaystyle \lbrace \;{\frac {1}{1}},{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},\dots ,{\frac {1}{n}},\dots \;\rbrace }$

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 ${\displaystyle \forall \varepsilon >0\;\exists N(\varepsilon )\in \mathbb {N} :n>N\Rightarrow |a_{n}-a|<\varepsilon }$, tai skaičių ${\displaystyle a}$ vadiname sekos riba. Jei tokio skaičiaus nėra – seka ribos neturi.

Kitaip tariant, jeigu egzistuoja toks sekos narys ${\displaystyle a_{N}}$, nuo kurio pradedant, skirtumas tarp visų tolimesnių narių ir kažkokio skaičiaus ${\displaystyle a}$ yra mažesnis, nei kažkoks iš anksto nustatytas skaičius (jis gali būti kiek norima mažas), tai sakome, kad ${\displaystyle a}$ yra šios sekos riba. Iš esmės šis apibrėžimas atitinka mūsų natūralų suvokimą apie sekos ribą.

Seka, turinti baigtinę ribą, vadinama konverguojančia, neturinti ribos laikoma diverguojančia.

Sekos ribą žymime:

${\displaystyle \lim _{n\rightarrow \infty }a_{n}=L}$

Čia ${\displaystyle \lim }$ reiškia ribą, ${\displaystyle n\rightarrow \infty }$ yra simbolinis žymėjimas, kad eilės numeris ${\displaystyle n}$ tolsta į begalybę, o ${\displaystyle a_{n}}$ yra n - tasis, t. y. bendrasis sekos narys.

Dalinės ribos

Jei seka {${\displaystyle x_{n}}$} turi konverguojantį posekį {${\displaystyle x_{n}k}$}, šio posekio riba vadinama daline riba. Didžiausia sekos {${\displaystyle x_{n}}$} dalinė riba vadinama sekos viršutiniąja riba (žymima ${\displaystyle \limsup _{n\to \infty }x_{n}}$). Mažiausia sekos dalinė riba – apatinioji riba ${\displaystyle (\liminf _{n\to \infty }x_{n})}$.

Pavyzdžiui, seka ${\displaystyle x_{n}=\lbrace (-1)^{n}\rbrace }$ neturi ribos, tačiau turi du konverguojančius posekius:

• ${\displaystyle x_{2n}=\lbrace (-1)^{2n}\rbrace \to 1}$ ir
• ${\displaystyle x_{2n+1}=\lbrace (-1)^{2n+1}\rbrace \to -1}$

Koši kriterijus

Ogiustenas Lui Koši suformulavo kriterijų, kurį tenkinančios sekos vadinamos Koši sekomis:

Seka ${\displaystyle {\{x_{n}\}}}$ yra Koši seka, jei ${\displaystyle {\sum _{n=1}^{\infty }a_{n}}}$ konverguoja tada ir tik tada kai ${\displaystyle {\forall \varepsilon >0,\exists N\in {\boldsymbol {N}},\forall n>m>N:|\sum _{k=m+1}^{n}a_{k}|<\varepsilon }}$.

Koši kriterijus yra būtina ir pakankama sekos konvergavimo sąlyga – visos konverguojančios sekos yra Koši sekos ir atvirkščiai. [Čia paimtas koši kriterijus ne sekoms, o skaičių eilutėm, reikia pataisyti.]

Ribų savybės

Tegul ${\displaystyle \lim _{n\to \infty }x_{n}=L_{1}}$ ir ${\displaystyle \lim _{n\to \infty }y_{n}=L_{2}}$, tada galime atlikti tokius veiksmus:

• ${\displaystyle \lim _{n\to \infty }(x_{n}+y_{n})=L_{1}+L_{2}}$
• ${\displaystyle \lim _{n\to \infty }(x_{n}y_{n})=L_{1}L_{2}}$
• ${\displaystyle \lim _{n\to \infty }{\frac {x_{n}}{y_{n}}}={\frac {L_{1}}{L_{2}}}}$

arba

• ${\displaystyle \lim _{x\to a}[f(x)\pm g(x)]=\lim _{x\to a}f(x)\pm \lim _{x\to a}g(x)}$
• ${\displaystyle \lim _{x\to a}[f(x)\cdot g(x)]=\lim _{x\to a}f(x)\cdot \lim _{x\to a}g(x)}$
• ${\displaystyle \lim _{x\to a}{\frac {f(x)}{g(x)}}={\frac {\lim _{x\to a}f(x)}{\lim _{x\to a}g(x)}}}$

Skaičiavimas

Skaičiuodami ribas pasiremiame jų savybėmis ir keliomis elementariausiomis ribomis:

• ${\displaystyle \lim _{n\rightarrow \infty }n=\infty }$
• ${\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n}}=0}$
• ${\displaystyle \lim _{n\rightarrow \infty }a^{n}=\infty }$
• ${\displaystyle \lim _{n\rightarrow \infty }a^{\frac {1}{n}}=1}$
• ${\displaystyle \lim _{n\rightarrow \infty }n^{\frac {1}{n}}=1}$

• ${\displaystyle \lim _{x\to 0}{\frac {\sin x}{x}}=1,}$
• ${\displaystyle \lim _{x\to 0}{\frac {\sin(kx)}{x}}=k,}$
• ${\displaystyle \lim _{x\to 0}{\frac {\tan x}{x}}=\lim _{x\to 0}({\frac {\sin x}{x}}\cdot {\frac {1}{\cos x}})=1,}$
• ${\displaystyle \lim _{x\to 0}{\frac {\arcsin x}{x}}=1,}$
• ${\displaystyle \lim _{x\to 0}{\frac {\arctan x}{x}}=1,}$
• ${\displaystyle \lim _{x\to 0}{\frac {\ln(1+x)}{x}}=\lim _{x\to 0}[\ln(1+x)^{\frac {1}{x}}]=\ln[\lim _{x\to 0}(1+x)^{\frac {1}{x}}]=\ln e=1,}$
• ${\displaystyle \lim _{x\to 0}{\frac {e^{x}-1}{x}}=1.}$

ir t. t. Dažnai ribos ženklas nerašomas, o rašoma tiesiog, pvz.: ${\displaystyle {\frac {1}{\infty }}=0}$. Toks užrašas suprantamas ne kaip lygybė, o kaip riba.

Ieškodami ribų galime tiesiog įrašyti begalybę vietoj ${\displaystyle n}$, tačiau dažniausiai gauname , kurį ir reikia pašalinti, pvz.:

• ${\displaystyle \lim _{n\rightarrow \infty }{\frac {2n-1}{n}}={\frac {\infty }{\infty }}=\lim _{n\rightarrow \infty }{\frac {2-{\frac {1}{n}}}{1}}=2-{\frac {1}{\infty }}=2.}$

• ${\displaystyle \lim _{x\to 0}{\frac {\sin 3x}{\sin 7x}}=\lim _{x\to 0}{\frac {{\frac {\sin 3x}{3x}}\cdot 3x}{{\frac {\sin 7x}{7x}}\cdot 7x}}=\lim _{x\to 0}{\frac {3x}{7x}}={\frac {3}{7}}.}$

Skaičius e

Nepaprastai svarbi matematikoje yra tokia riba:

${\displaystyle \lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}\equiv {\mathsf {e}}.}$

${\displaystyle \lim _{x\to 0}(1+x)^{\frac {1}{x}}=e.}$

Ši vertė, vadinama skaičiumi e, yra viena svarbiausių matematinių konstantų.

Pavyzdžiai

• ${\displaystyle \lim _{n\rightarrow \infty }\left(1+{\frac {1}{n(n+2)}}\right)^{n}=\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n(n+2)}}\right)^{n(n+2)\left({\frac {n}{n(n+2)}}\right)}=\lim _{n\rightarrow \infty }{\mathsf {e}}^{\frac {n}{n(n+2)}}={\mathsf {e}}^{\frac {1}{\infty }}=1.}$

• ${\displaystyle \lim _{n\rightarrow \infty }{\frac {n^{2}+4n-5}{n^{2}-1}}=\lim _{n\rightarrow \infty }{\frac {n^{2}\left(1+{\frac {4}{n}}-{\frac {5}{n^{2}}}\right)}{n^{2}\left(1-{\frac {1}{n^{2}}}\right)}}={\frac {1+{\frac {4}{\infty }}-{\frac {5}{\infty }}}{1-{\frac {1}{\infty }}}}=1.}$

• Seka ${\displaystyle \lbrace \;-1,1,-1,1,\dots ,(-1)^{n},\dots \;\rbrace }$ diverguoja, t. y. ribos neturi.

• ${\displaystyle \lim _{x\to \infty }{\frac {x^{3}-4x^{2}+7x-3}{x^{2}+2x-11}}=\lim _{x\to \infty }{\frac {\frac {x^{3}-4x^{2}+7x-3}{x^{3}}}{\frac {x^{2}+2x-11}{x^{3}}}}=\lim _{x\to \infty }{\frac {1-{\frac {4}{x}}+{\frac {7}{x^{2}}}-{\frac {3}{x^{3}}}}{{\frac {1}{x}}+{\frac {2}{x^{2}}}-{\frac {11}{x^{3}}}}}=\infty .}$

• ${\displaystyle \lim _{x\to 0}{\frac {3x^{2}-2x}{2x^{2}-5x}}=({\frac {0}{0}})=\lim _{x\to 0}{\frac {x(3x-2)}{x(2x-5)}}={\frac {-2}{-5}}={\frac {2}{5}}.}$

• ${\displaystyle \lim _{x\to \infty }{\frac {3x^{2}-1}{5x^{2}+2x}}=\lim _{x\to \infty }{\frac {3-1/x^{2}}{5+2/x}}={\frac {3-0}{5+0}}={\frac {3}{5}}.}$

• ${\displaystyle a\lim _{x\to -{\frac {3}{2}}}{\frac {4x^{2}-9}{2x+3}}=\lim _{x\to -{\frac {3}{2}}}{\frac {(2x-3)(2x+3)}{2x+3}}=\lim _{x\to -{\frac {3}{2}}}(2x-3)=-3-3=-6.{\breve {}}}$

• ${\displaystyle \lim _{x\to 1}{\frac {x-1}{1-x^{2}}}=\lim _{x\to 1}{\frac {x-1}{(1-x)(1+x)}}=\lim _{x\to 1}{\frac {x-1}{-(x-1)(1+x)}}=\lim _{x\to 1}{\frac {1}{-(1+x)}}=-{\frac {1}{2}}.}$

• ${\displaystyle \lim _{x\to 2}({\frac {1}{x-2}}-{\frac {4}{x^{2}-4}})=\lim _{x\to 2}{\frac {x+2-4}{x^{2}-4}}=\lim _{x\to 2}{\frac {x-2}{(x-2)(x+2)}}=\lim _{x\to 2}{\frac {1}{x+2}}={\frac {1}{4}}.}$

• ${\displaystyle \lim _{x\to 0}{\frac {1-{\sqrt {1-x^{2}}}}{x}}=\lim _{x\to 0}{\frac {(1-{\sqrt {1-x^{2}}})(1+{\sqrt {1-x^{2}}})}{x(1+{\sqrt {1-x^{2}}})}}=\lim _{x\to 0}{\frac {1-(1-x^{2})}{x(1+{\sqrt {1-x^{2}}})}}=}$

${\displaystyle =\lim _{x\to 0}{\frac {x}{1+{\sqrt {1-x^{2}}}}}={\frac {0}{2}}=0.}$

• ${\displaystyle \lim _{x\to 2}{\frac {2x^{2}-8}{x-2}}=\lim _{x\to 2}{\frac {2(x-2)(x+2)}{x-2}}=\lim _{x\to 2}(2x+4)=8.}$

• ${\displaystyle \lim _{x\to 0}{\frac {x+\sin 3x}{x}}=\lim _{x\to 0}{\frac {3x}{3x}}\cdot ({\frac {x}{x}}+{\frac {\sin 3x}{x}})=\lim _{x\to 0}(1+{\frac {3x\sin 3x}{3x\cdot x}})=\lim _{x\to 0}(1+{\frac {3x}{x}})=4.}$

• ${\displaystyle \lim _{x\to \infty }({\frac {x+1}{x}})^{\frac {x}{3}}=\lim _{x\to \infty }((1+{\frac {1}{x}})^{x})^{\frac {1}{3}}=e^{\frac {1}{3}}.}$

• ${\displaystyle \lim _{x\to \infty }({\frac {2x+1}{2x+3}})^{\frac {3x-2}{5}}=\lim _{x\to \infty }({\frac {2x+3-2}{2x+3}})^{\frac {3x-2}{5}}=\lim _{x\to \infty }((1+{\frac {-2}{2x+3}})^{\frac {2x+3}{-2}})^{{\frac {-2}{2x+3}}\cdot {\frac {3x-2}{5}}}=}$

${\displaystyle =e^{-{\frac {2}{5}}\lim _{x\to \infty }{\frac {3x-2}{2x+3}}}=e^{-{\frac {2}{5}}\cdot {\frac {3}{2}}}=e^{-{\frac {3}{5}}}.}$

• ${\displaystyle \lim _{x\to 3}{\frac {x^{2}-9}{{\sqrt {x+1}}-2}}=\lim _{x\to 3}{\frac {(x-3)(x+3)({\sqrt {x+1}}+2)}{({\sqrt {x+1}}-2)({\sqrt {x+1}}+2)}}=\lim _{x\to 3}{\frac {(x-3)(x+3)({\sqrt {x+1}}+2)}{x+1-4}}=}$

${\displaystyle =\lim _{x\to 3}[(x+3)({\sqrt {x+1}}+2)]=6\cdot 4=24.}$

• ${\displaystyle \lim _{x\to 0}{\frac {(1+x)^{\frac {1}{3}}-1}{{\sqrt {x+1}}-1}}=\lim _{z\to 1}{\frac {z^{\frac {6}{3}}-1}{{\sqrt {z^{6}}}-1}}=\lim _{z\to 1}{\frac {(z-1)(z+1)}{(z-1)(z^{2}+z+1)}}=\lim _{z\to 1}{\frac {z+1}{z^{2}+z+1}}={\frac {2}{3}},}$

kur keičiame kintamąjį: ${\displaystyle 1+x=z^{6}.}$ Kadangi ${\displaystyle x\rightarrow 0,}$ tai ${\displaystyle z\rightarrow 1.}$

• ${\displaystyle \lim _{x\to \pi }{\frac {\sin ^{2}x}{1+\cos ^{3}x}}=\lim _{x\to \pi }{\frac {1-\cos ^{2}x}{(1+\cos x)(1-\cos x+\cos ^{2}x)}}=\lim _{x\to \pi }{\frac {(1-\cos x)(1+\cos x)}{(1+\cos x)(1-\cos x+\cos ^{2}x)}}=}$

${\displaystyle =\lim _{x\to \pi }{\frac {1-\cos x}{1-\cos x+\cos ^{2}x}}={\frac {1-(-1)}{1-(-1)+(-1)^{2}}}={\frac {2}{3}}.}$

• ${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{1-\cos x}}=\lim _{x\to 0}{\frac {x^{2}}{2\sin ^{2}(x/2)}}=2\lim _{x\to 0}({\frac {x/2}{\sin(x/2)}})^{2}=2.}$

• ${\displaystyle \lim _{x\to \infty }(1+{\frac {1}{x^{2}}})^{x}=\lim _{x\to \infty }[(1+{\frac {1}{x^{2}}})^{x^{2}}]^{\frac {1}{x}}=e^{0}=1.}$

• ${\displaystyle \lim _{x\to 0}{\frac {\log _{a}(1+x)}{x}}=\lim _{x\to 0}[\log _{a}(1+x)^{\frac {1}{x}}]=\log _{a}e.}$

• ${\displaystyle \lim _{x\to 2}{\frac {x^{3}-8}{x-2}}=\lim _{x\to 2}{\frac {(x-2)(x^{2}+2x+4)}{x-2}}=\lim _{x\to 2}(x^{2}+2x+4)=12.}$

• ${\displaystyle \lim _{x\to 3}{\frac {x^{2}+x-12}{2x^{2}-9x+9}}=\lim _{x\to 3}{\frac {(x-3)(x+4)}{2(x-3)(x-1.5)}}=\lim _{x\to 3}{\frac {x+4}{2x-3}}={\frac {7}{3}}.}$

• ${\displaystyle \lim _{x\to 0}{\frac {\tan x-\sin x}{x^{3}}}=\lim _{x\to 0}{\frac {{\frac {\sin x}{\cos x}}-\sin x}{x^{3}}}=\lim _{x\to 0}{\frac {\frac {\sin x(1-\cos x)}{\cos x}}{x^{3}}}=\lim _{x\to 0}({\frac {\sin x}{x}}\cdot {\frac {1-\cos x}{x^{2}\cdot \cos x}})=}$

${\displaystyle =\lim _{x\to 0}{\frac {\sin x}{x}}\cdot \lim _{x\to 0}{\frac {1}{\cos x}}\cdot \lim _{x\to 0}{\frac {1-\cos x}{x^{2}}}=\lim _{x\to 0}{\frac {2\sin ^{2}{\frac {x}{2}}}{x^{2}}}={\frac {1}{2}}\lim _{x\to 0}({\frac {\sin {\frac {x}{2}}}{\frac {x}{2}}})^{2}={\frac {1}{2}}.}$

• ${\displaystyle \lim _{x\to \infty }({\frac {x^{3}}{3x^{2}-4}}-{\frac {x^{2}}{3x+2}})=\lim _{x\to \infty }{\frac {x^{3}(3x+2)-x^{2}(3x^{2}-4)}{(3x^{2}-4)(3x+2)}}=\lim _{x\to \infty }{\frac {2x^{3}+4x^{2}}{9x^{3}+6x^{2}-12x-8}}=}$

${\displaystyle =\lim _{x\to \infty }{\frac {2+{\frac {4}{x}}}{9+{\frac {6}{x}}-{\frac {12}{x^{2}}}-{\frac {8}{x^{3}}}}}={\frac {2}{9}}.}$

• ${\displaystyle \lim _{x\to 1}{\frac {2x-2}{(26+x)^{\frac {1}{3}}-3}}=\lim _{t\to 3}{\frac {2(t^{3}-27)}{t-3}}=\lim _{t\to 3}{\frac {2(t-3)(t^{2}+3t+9)}{t-3}}=2\lim _{t\to 3}(t^{2}+3t+9)=54,}$

kur ${\displaystyle 26+x=t^{3};}$ ${\displaystyle x=t^{3}-26;}$ ${\displaystyle t\rightarrow 3,}$ kai ${\displaystyle x\rightarrow 1.}$

• Rasime ribą ${\displaystyle \lim _{x\to {\frac {1}{2}}}{\frac {8x^{3}-1}{6x^{2}+3x-3}}=({\frac {0}{0}})}$

Skaitiklis išskaidomas pagal formulę ${\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})=8x^{3}-1=(2x-1)(4x^{2}+2x+1)}$

Vardiklis gali būti išskaidomas surandant jo sprendinius ${\displaystyle x_{1}}$ ir ${\displaystyle x_{2}}$:

${\displaystyle 6x^{2}+3x-3=0}$

${\displaystyle D=b^{2}-4ac=3^{2}-4\cdot 6\cdot (-3)=9+72=81}$

${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {D}}}{2a}}={\frac {-3\pm {\sqrt {81}}}{2\cdot 6}}={\frac {-3\pm 9}{12}}=-1;{\frac {1}{2}}.}$

Kvadratinė lygtis yra išskaidoma ${\displaystyle ax^{2}+bx+c=a(x-x_{1})(x-x_{2})=6x^{2}+3x-3=6(x-(-1))(x-{\frac {1}{2}}).}$

${\displaystyle \lim _{x\to {\frac {1}{2}}}{\frac {(2x-1)(4x^{2}+2x+1)}{6(x+1)(x-{\frac {1}{2}})}}=\lim _{x\to {\frac {1}{2}}}{\frac {(2x-1)(4x^{2}+2x+1)}{3(x+1)(2x-1)}}=\lim _{x\to {\frac {1}{2}}}{\frac {4x^{2}+2x+1}{3x+3}}={\frac {3}{4.5}}={\frac {2}{3}}.}$

• ${\displaystyle \lim _{x\to \infty }(1-{\frac {1}{x}})^{x}=\lim _{z\to -\infty }[(1+{\frac {1}{z}})^{z}]^{-1}=e^{-1}={\frac {1}{e}}.}$

• ${\displaystyle \lim _{x\to 6}{\frac {{\sqrt {x+3}}-3}{x-6}}=\lim _{x\to 6}{\frac {x+3-9}{(x-6)({\sqrt {x+3}}+3)}}=\lim _{x\to 6}{\frac {1}{{\sqrt {x+3}}+3}}={\frac {1}{6}}.}$

• ${\displaystyle \lim _{x\to 8}{\frac {x-8}{x^{\frac {1}{3}}-2}}=\lim _{x\to 8}{\frac {(x^{\frac {1}{3}}-2)(x^{\frac {2}{3}}+2x^{\frac {1}{3}}+2^{2})}{x^{\frac {1}{3}}-2}}=\lim _{x\to 8}(x^{\frac {2}{3}}+2x^{\frac {1}{3}}+4)=12.}$

• ${\displaystyle \lim _{x\to \infty }({\frac {x-1}{x}})^{5x}=\lim _{x\to \infty }(1-{\frac {1}{x}})^{5x}=\lim _{z\to -\infty }((1+{\frac {1}{z}})^{z})^{-5}=e^{-5}.}$

• ${\displaystyle \lim _{x\to 0}(1+\tan x)^{\frac {1}{\sin x}}=\lim _{x\to 0}((1+\tan x)^{\frac {\cos x}{\sin x}})^{\frac {1}{\cos x}}=e.}$

• ${\displaystyle \lim _{x\to \infty }({\frac {x^{2}+5}{3x^{2}+1}})^{x^{2}}=\lim _{x\to \infty }({\frac {1+{\frac {5}{x^{2}}}}{3+{\frac {1}{x^{2}}}}})^{x^{2}}=\lim _{x\to \infty }({\frac {1}{3}})^{x^{2}}=0.}$

• ${\displaystyle \lim _{x\to 0}\cos {\frac {\pi \cdot x}{\sin x}}=\lim _{x\to 0}\cos {\frac {\pi \cdot x}{x{\frac {\sin x}{x}}}}=\lim _{x\to 0}\cos {\frac {\pi }{\frac {\sin x}{x}}}=\lim _{x\to 0}\cos \pi =-1.}$

Šaltiniai

1. Birutė Gražulevičienė. Mokyklinės matematikos žinynas. – Vilnius: Leidybos centras, 1997. – 21 p. ISBN 9986-03-264-4