Oblique Asymtotes (1 Viewer)

[DNETTZ]

Right, was looking through for solving oblique asymtotes, and i was wondering how you do it through limits.
eg;
x+2
-----
x-1

Looks like this ;x+2/x-1 - Wolfram|Alpha
I dont like the long division method, it's dodgy. And it doesnt work here. This is fully simplified save (x+2)(X-1)^-1
Asymptotes: Comparing Graphs

So can someone tell me how?

 

[ohexploitable]

But... there aren't any oblique asymptotes, there is a horizontal one at y=1 and a vertical one at x=1...

 

[kcqn93]

your input is wrong for the wolfram LOL

and yeh there isnt an oblique...

 

[hscishard]

What the hell.
http://community.boredofstudies.org/12/mathematics/245590/unusual-graph-please-help.html

Where are you guys getting these oblique asymptotes from?

 

[adomad]

practise on this one:

 

[nonsenseTM]

if you meant (x+2)/(x-1) [add the bracket when using wolframalpha] , then it has

a vert. aymt. at x=1 cos x-1 can't be 0, consider limits-from 1+ve side (like 1.01+2/1.01-1) , it's +ve infinity , consider the other side of 1 (like 0.99+2/0.99-1), it's -ve infinity

a hor. asyt. at y=1 cos as x approaches inf. it's a reallly large no. divided by something a bit smaller than it (like 1000+2/1000-1), so it should be just more than 1
as x --> -ve infinity, x-1 becomes really large -ve no. which is to divide the slightly smaller x+2 (say -1000+2/-1000-1), y--> a bit less than 1

so it's like y=1/x shifted up 1 unit and right 1 unit

 

[nonsenseTM]

if you wanna pracitce oblique asymtotes, try the one from adomad

it's best done by division which isn't quite "dodgy" and x^2+2/x+1=x+1+3/x+1

but don't worry too much about it , you're given things like x^2+2/x+1=x+1+3/x+1 and you only need to analyse the limits logically

 

[DNETTZ]

Yeah sorry about the misquote onto wolfram.
I got the limits method from wikipedia instead for forming a limit in y=mx+n which is what I was trying to do.
so its good now.