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 } , tai skaičių a {\displaystyle a} vadiname sekos riba . Jei tokio skaičiaus nėra – seka ribos neturi. Kitaip tariant, jeigu egzistuoja toks sekos narys a N {\displaystyle a_{N}} , nuo kurio pradedant, skirtumas tarp visų tolimesnių narių ir kažkokio skaičiaus a {\displaystyle a} yra mažesnis, nei kažkoks iš anksto nustatytas skaičius (jis gali būti kiek norima mažas), tai sakome, kad a {\displaystyle a} yra šios sekos 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 } reiškia ribą , n → ∞ {\displaystyle n\rightarrow \infty } yra simbolinis žymėjimas, kad eilės numeris n {\displaystyle n} tolsta į begalybę , o a n {\displaystyle a_{n}} yra n - tasis, t. y. bendrasis sekos narys.
Dalinės ribos
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Koši kriterijus
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Augustinas Koši suformulavo kriterijų, kurį tenkinančios sekos vadinamos Koši sekomis :
Seka { x n } {\displaystyle {\{x_{n}\}}} yra Koši seka , jei ∑ n = 1 ∞ a n {\displaystyle {\sum _{n=1}^{\infty }a_{n}}} konverguoja tada ir tik tada kai ∀ 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.
Ribų savybės
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Skaičiavimas
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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.}
ir t. t. Dažnai ribos ženklas nerašomas, o rašoma tiesiog, pvz.: 1 ∞ = 0 {\displaystyle {\frac {1}{\infty }}=0} . Toks užrašas suprantamas ne kaip lygybė, o kaip riba.
Ieškodami ribų galime tiesiog įrašyti begalybę vietoj n {\displaystyle n} , tačiau dažniausiai gauname 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 } diverguoja, t. y. ribos neturi. 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}.} Kadangi x → 0 , {\displaystyle x\rightarrow 0,} tai 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,} kai 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}} ir 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.}