Funkcija gali vienu metu turėti vertikalias, horizontalias ir pasvirąsias asimptotes, pvz., y = x | x | / x + 1 / x . {\displaystyle y=x^{|x|/x}+1/x.} {\displaystyle y=x^{|x|/x}+1/x.}
Kreivė gali kirsti savo asimptotę be galo daug kartų.
  • Rasime kreivės y = x 2 + 1 x {\displaystyle y={\frac {x^{2}+1}{x}}} {\displaystyle y={\frac {x^{2}+1}{x}}} asimptotes.

Funkcija x 2 + 1 x {\displaystyle {\frac {x^{2}+1}{x}}} {\displaystyle {\frac {x^{2}+1}{x}}} neapibrėžta tik kai x=0, lim x → 0 x 2 + 1 x = ∞ , {\displaystyle \lim _{x\to 0}{\frac {x^{2}+1}{x}}=\infty ,} {\displaystyle \lim _{x\to 0}{\frac {x^{2}+1}{x}}=\infty ,} taigi jos grafikas turi vertikaliąją asimptotę x=0. Ieškosime pasvirųjų ir horizontaliųjų asimptočių. Kadangi

lim x → ∞ f ( x ) x = lim x → ∞ x 2 + 1 x 2 = 1 ( = k ≠ 0 ) {\displaystyle \lim _{x\to \infty }{\frac {f(x)}{x}}=\lim _{x\to \infty }{\frac {x^{2}+1}{x^{2}}}=1(=k\not =0)} {\displaystyle \lim _{x\to \infty }{\frac {f(x)}{x}}=\lim _{x\to \infty }{\frac {x^{2}+1}{x^{2}}}=1(=k\not =0)}

tai horizontalių asimptočių nėra. Kadangi

lim x → ∞ [ f ( x ) − k x ] = lim x → ∞ ( x 2 + 1 x − x ) = 0 = b , {\displaystyle \lim _{x\to \infty }[f(x)-kx]=\lim _{x\to \infty }\left({\frac {x^{2}+1}{x}}-x\right)=0=b,} {\displaystyle \lim _{x\to \infty }[f(x)-kx]=\lim _{x\to \infty }\left({\frac {x^{2}+1}{x}}-x\right)=0=b,}

tai tiesė y=x yra pasviroji asimptotė abiem kreivės šakoms ir kai x → + ∞ , {\displaystyle x\rightarrow +\infty ,} {\displaystyle x\rightarrow +\infty ,} ir kai x → − ∞ . {\displaystyle x\rightarrow -\infty .} {\displaystyle x\rightarrow -\infty .}

  • Rasime kreivės y = x 3 1 − x 2 {\displaystyle y={\frac {x^{3}}{1-x^{2}}}} {\displaystyle y={\frac {x^{3}}{1-x^{2}}}} asimptotes.

Kreivė turi dvi vertikaliasias asimptotes x = ± 1 {\displaystyle x=\pm 1} {\displaystyle x=\pm 1}, kadangi

lim x → ± 1 x 3 1 − x 2 = ∞ . {\displaystyle \lim _{x\to \pm 1}{\frac {x^{3}}{1-x^{2}}}=\infty .} {\displaystyle \lim _{x\to \pm 1}{\frac {x^{3}}{1-x^{2}}}=\infty .}

Kadangi

lim x → ± ∞ f ( x ) x = lim x → ± ∞ x 3 x ( 1 − x 2 ) = − 1 ( = k ) , {\displaystyle \lim _{x\to \pm \infty }{\frac {f(x)}{x}}=\lim _{x\to \pm \infty }{\frac {x^{3}}{x(1-x^{2})}}=-1(=k),} {\displaystyle \lim _{x\to \pm \infty }{\frac {f(x)}{x}}=\lim _{x\to \pm \infty }{\frac {x^{3}}{x(1-x^{2})}}=-1(=k),}
lim x → ± ∞ [ f ( x ) − k x ] = lim x → ± ∞ ( x 3 1 − x 2 + x ) = lim x → ± ∞ x 1 − x 2 = 0 ( = b ) , {\displaystyle \lim _{x\to \pm \infty }[f(x)-kx]=\lim _{x\to \pm \infty }\left({\frac {x^{3}}{1-x^{2}}}+x\right)=\lim _{x\to \pm \infty }{\frac {x}{1-x^{2}}}=0(=b),} {\displaystyle \lim _{x\to \pm \infty }[f(x)-kx]=\lim _{x\to \pm \infty }\left({\frac {x^{3}}{1-x^{2}}}+x\right)=\lim _{x\to \pm \infty }{\frac {x}{1-x^{2}}}=0(=b),}

tai tiesė y=-x yra pasviroji asimptotė.


  • Raskime funkcijos y = − x 2 − 3 x + 2 x + 7 {\displaystyle y={\frac {-x^{2}-3x+2}{x+7}}} {\displaystyle y={\frac {-x^{2}-3x+2}{x+7}}} asimptotes.

Vertikalioji asimptotė - tiesė x=-7, nes lim x → − 7 ± 0 y = ± ∞ . {\displaystyle \lim _{x\to -7\pm 0}y=\pm \infty .} {\displaystyle \lim _{x\to -7\pm 0}y=\pm \infty .}

Apskaičiuosime koeficientus:

k = lim x → ± ∞ − x 2 − 3 x + 2 ( x + 7 ) ⋅ x = lim x → ± ∞ x 2 ( − x 2 / x 2 − 3 x / x 2 + 2 / x 2 ) x 2 ( x 2 / x 2 + 7 x / x 2 ) = − 1 − 0 + 0 1 + 0 = − 1 , {\displaystyle k=\lim _{x\to \pm \infty }{\frac {-x^{2}-3x+2}{(x+7)\cdot x}}=\lim _{x\to \pm \infty }{\frac {x^{2}(-x^{2}/x^{2}-3x/x^{2}+2/x^{2})}{x^{2}(x^{2}/x^{2}+7x/x^{2})}}={\frac {-1-0+0}{1+0}}=-1,} {\displaystyle k=\lim _{x\to \pm \infty }{\frac {-x^{2}-3x+2}{(x+7)\cdot x}}=\lim _{x\to \pm \infty }{\frac {x^{2}(-x^{2}/x^{2}-3x/x^{2}+2/x^{2})}{x^{2}(x^{2}/x^{2}+7x/x^{2})}}={\frac {-1-0+0}{1+0}}=-1,}

b = lim x → ± ∞ ( − x 2 − 3 x + 2 x + 7 − ( − 1 ) ⋅ x ) = lim x → ± ∞ 4 x + 2 x + 7 = lim x → ± ∞ x ( 4 x / x + 2 / x ) x ( x / x + 7 / x ) = lim x → ± ∞ 4 + 2 / x 1 + 7 / x = 4. {\displaystyle b=\lim _{x\to \pm \infty }\left({\frac {-x^{2}-3x+2}{x+7}}-(-1)\cdot x\right)=\lim _{x\to \pm \infty }{\frac {4x+2}{x+7}}=\lim _{x\to \pm \infty }{\frac {x(4x/x+2/x)}{x(x/x+7/x)}}=\lim _{x\to \pm \infty }{\frac {4+2/x}{1+7/x}}=4.} {\displaystyle b=\lim _{x\to \pm \infty }\left({\frac {-x^{2}-3x+2}{x+7}}-(-1)\cdot x\right)=\lim _{x\to \pm \infty }{\frac {4x+2}{x+7}}=\lim _{x\to \pm \infty }{\frac {x(4x/x+2/x)}{x(x/x+7/x)}}=\lim _{x\to \pm \infty }{\frac {4+2/x}{1+7/x}}=4.} Todėl pasvirosios asimptotės lygtis tokia: y = − x + 4. {\displaystyle y=-x+4.} {\displaystyle y=-x+4.}


  • Raskime kreivės y = 5 x x − 3 {\displaystyle y={\frac {5x}{x-3}}} {\displaystyle y={\frac {5x}{x-3}}} asimptotes.

Kadangi lim x → 3 5 x x − 3 = ∞ , {\displaystyle \lim _{x\to 3}{\frac {5x}{x-3}}=\infty ,} {\displaystyle \lim _{x\to 3}{\frac {5x}{x-3}}=\infty ,} tai tiesė x=3 yra vertikalioji asimptotė. Kadangi lim x → ∞ 5 x x − 3 = 5 , {\displaystyle \lim _{x\to \infty }{\frac {5x}{x-3}}=5,} {\displaystyle \lim _{x\to \infty }{\frac {5x}{x-3}}=5,} tai tiesė y=5 yra horizontalioji asimptotė. Kadangi lim x → ∞ f ( x ) x = lim x → ∞ 5 x − 3 = 0 , {\displaystyle \lim _{x\to \infty }{\frac {f(x)}{x}}=\lim _{x\to \infty }{\frac {5}{x-3}}=0,} {\displaystyle \lim _{x\to \infty }{\frac {f(x)}{x}}=\lim _{x\to \infty }{\frac {5}{x-3}}=0,} tai pasvirųjų asimptočių nėra.


  • Rasime kreivės y = x 2 − 6 x + 3 x − 3 {\displaystyle y={\frac {x^{2}-6x+3}{x-3}}} {\displaystyle y={\frac {x^{2}-6x+3}{x-3}}} asimptotes.

Kadangi lim x → 3 x 2 − 6 x + 3 x − 3 = ∞ , {\displaystyle \lim _{x\to 3}{\frac {x^{2}-6x+3}{x-3}}=\infty ,} {\displaystyle \lim _{x\to 3}{\frac {x^{2}-6x+3}{x-3}}=\infty ,} tai x=3 yra vertikalioji asimptotė. Kadangi lim x → ∞ x 2 − 6 x + 3 x − 3 = ∞ , {\displaystyle \lim _{x\to \infty }{\frac {x^{2}-6x+3}{x-3}}=\infty ,} {\displaystyle \lim _{x\to \infty }{\frac {x^{2}-6x+3}{x-3}}=\infty ,} tai horizontaliųjų asimptočių nėra. Raskime pasvirosios asimptotės koeficientus k ir b:

k = lim x → ∞ x 2 − 6 x + 3 x ( x − 3 ) = 1 , {\displaystyle k=\lim _{x\to \infty }{\frac {x^{2}-6x+3}{x(x-3)}}=1,} {\displaystyle k=\lim _{x\to \infty }{\frac {x^{2}-6x+3}{x(x-3)}}=1,}
b = lim x → ∞ x 2 − 6 x + 3 x − 3 − x = lim x → ∞ x 2 − 6 x + 3 − x 2 + 3 x x − 3 = lim x → ∞ − 3 x + 3 x − 3 = − 3. {\displaystyle b=\lim _{x\to \infty }{\frac {x^{2}-6x+3}{x-3}}-x=\lim _{x\to \infty }{\frac {x^{2}-6x+3-x^{2}+3x}{x-3}}=\lim _{x\to \infty }{\frac {-3x+3}{x-3}}=-3.} {\displaystyle b=\lim _{x\to \infty }{\frac {x^{2}-6x+3}{x-3}}-x=\lim _{x\to \infty }{\frac {x^{2}-6x+3-x^{2}+3x}{x-3}}=\lim _{x\to \infty }{\frac {-3x+3}{x-3}}=-3.}

Pasviroji asimptotė yra y = x − 3. {\displaystyle y=x-3.} {\displaystyle y=x-3.}


  • Raskime funkcijos y = 2 x 3 x 2 − 4 {\displaystyle y={\frac {2x^{3}}{x^{2}-4}}} {\displaystyle y={\frac {2x^{3}}{x^{2}-4}}} asimptotes.

Tiesės x = ± 2 {\displaystyle x=\pm 2} {\displaystyle x=\pm 2} yra vertikaliosios asimptotės, nes

lim x → ± 2 2 x 3 x 2 − 4 = ∞ . {\displaystyle \lim _{x\to \pm 2}{\frac {2x^{3}}{x^{2}-4}}=\infty .} {\displaystyle \lim _{x\to \pm 2}{\frac {2x^{3}}{x^{2}-4}}=\infty .}

Kadangi lim x → ∞ 2 x 3 x 2 − 4 = ∞ , {\displaystyle \lim _{x\to \infty }{\frac {2x^{3}}{x^{2}-4}}=\infty ,} {\displaystyle \lim _{x\to \infty }{\frac {2x^{3}}{x^{2}-4}}=\infty ,} tai horizontaliųjų asimptočių nėra. Kadangi

k = lim x → ± ∞ y x = lim x → ± ∞ 2 x 2 x 2 − 4 = 2 , b = lim x → ± ∞ ( y − 2 x ) = lim x → ± ∞ 8 x x 2 − 4 = 0 , {\displaystyle k=\lim _{x\to \pm \infty }{\frac {y}{x}}=\lim _{x\to \pm \infty }{\frac {2x^{2}}{x^{2}-4}}=2,\qquad b=\lim _{x\to \pm \infty }(y-2x)=\lim _{x\to \pm \infty }{\frac {8x}{x^{2}-4}}=0,} {\displaystyle k=\lim _{x\to \pm \infty }{\frac {y}{x}}=\lim _{x\to \pm \infty }{\frac {2x^{2}}{x^{2}-4}}=2,\qquad b=\lim _{x\to \pm \infty }(y-2x)=\lim _{x\to \pm \infty }{\frac {8x}{x^{2}-4}}=0,}

tai tiesė y = 2 x {\displaystyle y=2x} {\displaystyle y=2x} yra pasviroji asimptotė.


  • Rasime kreivės y = x 3 2 ( x + 1 ) 2 {\displaystyle y={\frac {x^{3}}{2(x+1)^{2}}}} {\displaystyle y={\frac {x^{3}}{2(x+1)^{2}}}} asimptotes.

Kadangi lim x → − 1 ± 0 y = − ∞ , {\displaystyle \lim _{x\to -1\pm 0}y=-\infty ,} {\displaystyle \lim _{x\to -1\pm 0}y=-\infty ,} tai tiesė x = − 1 {\displaystyle x=-1} {\displaystyle x=-1} yra vertikalioji asimptotė. Kadangi lim x → ± ∞ y = ± ∞ , {\displaystyle \lim _{x\to \pm \infty }y=\pm \infty ,} {\displaystyle \lim _{x\to \pm \infty }y=\pm \infty ,} tai horizontaliųjų asimptočių nėra.

k = lim x → ± ∞ f ( x ) x = lim x → ± ∞ x 2 2 ( x + 1 ) 2 = 1 2 , {\displaystyle k=\lim _{x\to \pm \infty }{\frac {f(x)}{x}}=\lim _{x\to \pm \infty }{\frac {x^{2}}{2(x+1)^{2}}}={\frac {1}{2}},} {\displaystyle k=\lim _{x\to \pm \infty }{\frac {f(x)}{x}}=\lim _{x\to \pm \infty }{\frac {x^{2}}{2(x+1)^{2}}}={\frac {1}{2}},}
b = lim x → ± ∞ [ f ( x ) − k x ] = lim x → ± ∞ [ x 3 2 ( x + 1 ) 2 − 1 2 x ] = 1 2 lim x → ± ∞ x 3 − x ( x 2 + 2 x + 1 ) ( x + 1 ) 2 = {\displaystyle b=\lim _{x\to \pm \infty }[f(x)-kx]=\lim _{x\to \pm \infty }\left[{\frac {x^{3}}{2(x+1)^{2}}}-{\frac {1}{2}}x\right]={\frac {1}{2}}\lim _{x\to \pm \infty }{\frac {x^{3}-x(x^{2}+2x+1)}{(x+1)^{2}}}=} {\displaystyle b=\lim _{x\to \pm \infty }[f(x)-kx]=\lim _{x\to \pm \infty }\left[{\frac {x^{3}}{2(x+1)^{2}}}-{\frac {1}{2}}x\right]={\frac {1}{2}}\lim _{x\to \pm \infty }{\frac {x^{3}-x(x^{2}+2x+1)}{(x+1)^{2}}}=}
= 1 2 lim x → ± ∞ − 2 x 2 − x ( x + 1 ) 2 = − 1. {\displaystyle ={\frac {1}{2}}\lim _{x\to \pm \infty }{\frac {-2x^{2}-x}{(x+1)^{2}}}=-1.} {\displaystyle ={\frac {1}{2}}\lim _{x\to \pm \infty }{\frac {-2x^{2}-x}{(x+1)^{2}}}=-1.}

Vadinasi, kreivė turi pasvirąją asimptotę y = 1 2 x − 1. {\displaystyle y={\frac {1}{2}}x-1.} {\displaystyle y={\frac {1}{2}}x-1.}


  • Raskime kreivės y = x 2 x 2 − 1 {\displaystyle y={\frac {x^{2}}{\sqrt {x^{2}-1}}}} {\displaystyle y={\frac {x^{2}}{\sqrt {x^{2}-1}}}} asimptotes.

lim x → ± 1 ± 0 f ( x ) = ∞ , {\displaystyle \lim _{x\to \pm 1\pm 0}f(x)=\infty ,} {\displaystyle \lim _{x\to \pm 1\pm 0}f(x)=\infty ,} todėl tiesės x = − 1 {\displaystyle x=-1} {\displaystyle x=-1} ir x = 1 {\displaystyle x=1} {\displaystyle x=1} yra vertikaliosios asimptotės. Kadangi

k = lim x → + ∞ f ( x ) x = lim x → + ∞ x x 2 − 1 = 1 , {\displaystyle k=\lim _{x\to +\infty }{\frac {f(x)}{x}}=\lim _{x\to +\infty }{\frac {x}{\sqrt {x^{2}-1}}}=1,} {\displaystyle k=\lim _{x\to +\infty }{\frac {f(x)}{x}}=\lim _{x\to +\infty }{\frac {x}{\sqrt {x^{2}-1}}}=1,}
b = lim x → + ∞ [ f ( x ) − k x ] = lim x → + ∞ ( x 2 x 2 − 1 − x ) = lim x → + ∞ x ( x − x 2 − 1 ) x 2 − 1 = {\displaystyle b=\lim _{x\to +\infty }[f(x)-kx]=\lim _{x\to +\infty }\left({\frac {x^{2}}{\sqrt {x^{2}-1}}}-x\right)=\lim _{x\to +\infty }{\frac {x(x-{\sqrt {x^{2}-1}})}{\sqrt {x^{2}-1}}}=} {\displaystyle b=\lim _{x\to +\infty }[f(x)-kx]=\lim _{x\to +\infty }\left({\frac {x^{2}}{\sqrt {x^{2}-1}}}-x\right)=\lim _{x\to +\infty }{\frac {x(x-{\sqrt {x^{2}-1}})}{\sqrt {x^{2}-1}}}=}
= lim x → + ∞ x ( x 2 − ( x 2 − 1 ) ) x 2 − 1 ( x + x 2 − 1 ) = lim x → + ∞ 1 x 2 − 1 ( 1 + 1 − 1 / x 2 ) = 0 , {\displaystyle =\lim _{x\to +\infty }{\frac {x(x^{2}-(x^{2}-1))}{{\sqrt {x^{2}-1}}(x+{\sqrt {x^{2}-1}})}}=\lim _{x\to +\infty }{\frac {1}{{\sqrt {x^{2}-1}}(1+{\sqrt {1-1/x^{2}}})}}=0,} {\displaystyle =\lim _{x\to +\infty }{\frac {x(x^{2}-(x^{2}-1))}{{\sqrt {x^{2}-1}}(x+{\sqrt {x^{2}-1}})}}=\lim _{x\to +\infty }{\frac {1}{{\sqrt {x^{2}-1}}(1+{\sqrt {1-1/x^{2}}})}}=0,}

tai tiesė y = x {\displaystyle y=x} {\displaystyle y=x} yra pasviroji asimptotė. Be to

k = lim x → − ∞ f ( x ) x = lim x → − ∞ x x 2 − 1 = − 1 ; {\displaystyle k=\lim _{x\to -\infty }{\frac {f(x)}{x}}=\lim _{x\to -\infty }{\frac {x}{\sqrt {x^{2}-1}}}=-1;} {\displaystyle k=\lim _{x\to -\infty }{\frac {f(x)}{x}}=\lim _{x\to -\infty }{\frac {x}{\sqrt {x^{2}-1}}}=-1;}
b = lim x → − ∞ [ f ( x ) − k x ] = lim x → − ∞ ( x 2 x 2 − 1 − ( − 1 ) ⋅ x ) = lim x → − ∞ x ( x + x 2 − 1 ) x 2 − 1 = {\displaystyle b=\lim _{x\to -\infty }[f(x)-kx]=\lim _{x\to -\infty }\left({\frac {x^{2}}{\sqrt {x^{2}-1}}}-(-1)\cdot x\right)=\lim _{x\to -\infty }{\frac {x(x+{\sqrt {x^{2}-1}})}{\sqrt {x^{2}-1}}}=} {\displaystyle b=\lim _{x\to -\infty }[f(x)-kx]=\lim _{x\to -\infty }\left({\frac {x^{2}}{\sqrt {x^{2}-1}}}-(-1)\cdot x\right)=\lim _{x\to -\infty }{\frac {x(x+{\sqrt {x^{2}-1}})}{\sqrt {x^{2}-1}}}=}
= lim x → − ∞ x x 2 − 1 ( x − x 2 − 1 ) = lim x → − ∞ − | x | x 2 − 1 ( − | x | − x 2 − 1 ) = {\displaystyle =\lim _{x\to -\infty }{\frac {x}{{\sqrt {x^{2}-1}}(x-{\sqrt {x^{2}-1}})}}=\lim _{x\to -\infty }{\frac {-|x|}{{\sqrt {x^{2}-1}}(-|x|-{\sqrt {x^{2}-1}})}}=} {\displaystyle =\lim _{x\to -\infty }{\frac {x}{{\sqrt {x^{2}-1}}(x-{\sqrt {x^{2}-1}})}}=\lim _{x\to -\infty }{\frac {-|x|}{{\sqrt {x^{2}-1}}(-|x|-{\sqrt {x^{2}-1}})}}=}

= lim x → − ∞ | x | x 2 − 1 ( | x | + x 2 − 1 ) = lim x → − ∞ 1 x 2 / x 2 − 1 / x 2 ( | x | + x 2 − 1 ) = 0 , {\displaystyle =\lim _{x\to -\infty }{\frac {|x|}{{\sqrt {x^{2}-1}}(|x|+{\sqrt {x^{2}-1}})}}=\lim _{x\to -\infty }{\frac {1}{{\sqrt {x^{2}/x^{2}-1/x^{2}}}(|x|+{\sqrt {x^{2}-1}})}}=0,} {\displaystyle =\lim _{x\to -\infty }{\frac {|x|}{{\sqrt {x^{2}-1}}(|x|+{\sqrt {x^{2}-1}})}}=\lim _{x\to -\infty }{\frac {1}{{\sqrt {x^{2}/x^{2}-1/x^{2}}}(|x|+{\sqrt {x^{2}-1}})}}=0,} todėl ir tiesė y = − x {\displaystyle y=-x} {\displaystyle y=-x} yra pasviroji asimptotė.