Finding Oblique Asympote by polidiv

[BananaJoe]

If you are looking for obli. asympt. for a function where the x^n+1/x^n or any, and you do a polynom division and you get a perfect match wihout any Rests. Does that mean the function doesn't have any obliique asymptotes?

 

[mfb]

That depends on the definition of asymptote - do you call a function "asymptote" if it is identical to the function you consider?
As in, is g(x)=x+1 an asymptote to f(x)=x+1?

 

[Mark44]

If you are looking for obli. asympt. for a function where the x^n+1/x^n or any, and you do a polynom division and you get a perfect match wihout any Rests. Does that mean the function doesn't have any obliique asymptotes?

I'm having a hard time understanding what you're asking.

"where the x^n+1/x^n or any" - what does this mean?
Also, the expression you wrote probably isn't what you meant. What you wrote is this:
xⁿ + 1/xⁿ

I can't tell if you meant $\frac{x^{n + 1}}{x^n}$ or $\frac{x^n + 1}{x^n}$. Suitably placed parentheses would be a great help.

"perfect match wihout any Rests" - ??