Asymptote of x^3 - x^5 / ( x^2 + 1) and similar curves

  • #1
Playing with some numerical simulations, I plotted this in Wolfram Cloud / Mathematica:
##x^3-\frac{x^5}{x^2+2}##


1575855561058.png

I had naively expected it to approach ##x^3−x^3=0##, but that isn't the case. It approaches 2x.
I can now vaguely understand that the two terms need not cancel at infinity, but I'd like to get a better handle on this.

[1] How to break this down intuitively and estimate the qualitative nature of the asymptote "by inspection"?

[2] How can we obtain the asymptote ##2x## theoretically?

[3] Can we find the equation of the asymptote using Mathematica etc?

[Moderator's note: Moved from a technical forum and thus no template.]
 
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Answers and Replies

  • #2
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##x^3 -\dfrac{x^5}{x^3+2} \sim x^3 -x^{5-3} = x^3 -x^2 \sim x^3## although it looks as if you used another scaling.
 
  • #3
Sorry, the laTex formula was wrong. I've corrected it. It has ##x^2## in the denominator.
Code:
Plot[{x^3-x^5/(x^2+2),2*x},{x,0,20},ImageSize->600,AxesStyle->20,PlotStyle->{Blue,Red},PlotRange->All]
Corrected the title as well.
 
  • #4
14,152
11,454
Then use good old basic fraction algebra! Search a common denominator, expand the first fraction, add them, calculate the new numerator, divide again and see what the leading term is. Where is the problem?
 
Last edited:
  • #6
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##x^3-\frac{x^5}{x^2+2} = \frac{x^3(x^2 + 2) - x^5}{x^2 + 2} = \frac{x^5 + 2x^3 - x^5}{x^2 + 2} = \frac {2x^3}{x^2 + 2}##
If you carry out the polynomial division, you get 2x plus a proper rational expression.

Whenever you have a rational function, as in the third expression above, where the degree of the numerator is one more than the degree of the denominator, there will be a slant asymptote. In this case, the slant asymptote is the line y = 2x, which is what you're seeing in the Wolfram graph.
 
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