Calculate the right oblique asymptote of the function

[Anakin_99]

Hi guys I was doing this exercise when I noticed that if use taylor developments at first order or second order I find different asymptote lines, in this function I was supposed to find x + 1/2 as asymptote, but how do I understand which taylor order should I use to find the correct asymptote ? Thank you if you can help me and sorry if my english is bad .
I got the same results if I immediatly calculate the asymptote for the general function, depending if I use the first or second order

 

[Steven G]

The 2nd equation starts with what symbol?
The 3rd equation starts with 9?
Why are you computing the limit of y-x? Where is the x coming from?

 

[Anakin_99]

The 2nd equation starts with what symbol?
The 3rd equation starts with 9?
Why are you computing the limit of y-x? Where is the x coming from?

thats the symbols we use in my country for the equation of a generic line y = mx + q
thats a q not a 9 sorry

 

[topsquark]

The first order approximation of [imath]x^2 (e^{1/z} - 1) - x[/imath] at infinity is 1/2 since the x term cancels. The approximation at infinity is
[imath]x^2 (e^{1/z} - 1) - x \to \dfrac{1}{2} + \dfrac{1}{6x} + \text{ ...}[/imath]

The second term isn't linear so isn't going to be acceptable for an asymptote.

-Dan