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Asymptote- vertical and diagonal

Mateusz Janicki
Posted 5 years ago

I would like to calculate the vertical and diagonal asymptote f (x) = (x ^ 2 + x + 3) / (x-1). Will somebody help?
POSTED BY: Mateusz Janicki
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Capora Allin
Capora Allin
Posted 5 years ago

For the diagonal one you could also do:

g[x_]:= (x ^ 2 + x + 3)

p[x_]:=(x-1)

PolynomialQuotient[g[x],p[x],x]

Or

f[x_]:= (x ^ 2 + x + 3) / (x-1);

n=Numerator[f[x]];
d=Denominator[f[x]];

PolynomialQuotient[n,d,x]
POSTED BY: Capora Allin

For the diagonal one you could also do:

PolynomialRemainder[x^2+x+3, x-1, x]
POSTED BY: Stephen Perencevich

You can also try the repository function CurveAnalysis which you can find here: https://resources.wolframcloud.com/FunctionRepository/resources/CurveAnalysis
POSTED BY: Artin Hatzikioseyian

The vertical asymptote can be calculated with the definition of limit:

Reduce[ForAll[n,
  Exists[\[Delta], \[Delta] > 0,
   ForAll[x, 0 < Abs[x - x0] < \[Delta],
    Abs[(x^2 + x + 3)/(x - 1)] > n]]],
 x0, Reals]

The diagonal one with Series:

Series[(x^2 + x + 3)/(x - 1),
 {x, Infinity, 0}]
POSTED BY: Gianluca Gorni
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