# Function with an asymptote at y=-1 and y=1 [closed]

I'm looking for a function that has two asymptotes parallel to the x-axis. Preferably it should also only cross the x-axis at (0,0) and be built without using any trigonometric functions. Mind you, if that's not possible then so be it. However, that is my preference.

The purpose of this is that when the function is placed into the floor function and (manipulated a little) it will generate a function that is 0 at 0, 1 whenever the input is positive and -1 when the input is negative. Such a curve is the best bet I can come up with towards guessing a means to build such a construct.

I think I have seen the function somewhere at some point or a variation. I just cannot remember what it was.

Thank you

-The Great Duck

(Also, while I suppose it is probably obvious (can't place it in floor), no piece wise please)

## closed as off-topic by The Great Duck, user91500, Claude Leibovici, Behrouz Maleki, Adam HughesDec 16 '16 at 15:34

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• Hint: Have you seen the graph of the arctangent? – GEdgar Feb 25 '16 at 1:00
• Um.. if you are talking about the integral or derivative of arctan i understand. However, I am specifically avoiding trig functions for this subject. – The Great Duck Feb 25 '16 at 1:01
• The thing that makes it hard is that it cannot be a rational function. – GEdgar Feb 25 '16 at 1:02
• Actually it is. See the answer below. It worked perfectly. – The Great Duck Feb 25 '16 at 1:15
• No, I was referring to the square root answer, but thank you. Alternate formulas will give me room to play around with it. – The Great Duck Feb 25 '16 at 1:32

You can be creative with this and make up your own funtions. For example consider,

$$f(x) = \left\{ \begin{array}{ll} x & \quad -1\leq x \leq 1 \\ 1 & \quad x \geq 1\\ -1 & \quad x \leq 1 \end{array} \right.$$

• @TheGreatDuck: This is a function will all the prescribed notions you mentioned in your question. It's even more simple to understand. Its the identity up to $[\pm1,\pm1]$ and trivially satisfies the definition for horizontal asymptote. – Fundamental Feb 25 '16 at 1:16
• @TheGreatDuck: I don't want to argue. You should really try to give constructive responses and not be rude to users trying to help you. Most of us are ph.d student in mathematics, postdocs or professors. It is not my intention to lead you astray. I think your understand of functions (maps) is a bit loose. – Fundamental Feb 25 '16 at 1:35
• I'm sorry if it came across rude. I wasn't attempting to mean it that way. It's just that to me that feels like the answer was an attempt to be condescending or some kind of attempt at a smart alek answer. I thought it was apparent by the way I phrased it that I was wanting a function with a single equation, not several. It wouldn't be very intuitive to take a function with a single equation and place a piece wise equation into it. That would result in everything after that being piece wise, which is highly undesirable. Hence, why I thought the OP was trying to be coy about their answer. – The Great Duck Feb 25 '16 at 1:46
• @TheGreatDuck You do realize that every function can be written as a piecewise function, right? That a piecewise function is not somehow "less" of a function than any other one? To call the answer stupid or condescending or smart aleck is ridiculous, and you should carefully reconsider your engagement here. – user296602 Feb 25 '16 at 1:50
• "That a piecewise function is not somehow "less" of a function than any other one?" No I never said that. I said that doing a composition and work with a piece wise function would be highly un-intuitive. I.E. the piece wise function would then need to be expressed with a single equality or I'm just working in circles essentially. I'm attempting to represent a function with floor which I described. If I was wanting piece wise I would just say {x = 1 for x>=0, x=-1 for x<0} and be done. I never meant to insult or anger. – The Great Duck Feb 25 '16 at 2:04

$$f(x)=\frac{x}{\sqrt{x^2+1}}$$

It should be apparent that this works because leading coefficient test provides asymptotes for $y=\pm 1$, and plugging in $0$ for $x$ yields $0$.

Edit: I found another one if you prefer it:

$$f(x)=\frac{1-e^{-x}}{1+e^{-x}}$$

• thank you. this will be a great help. – The Great Duck Feb 25 '16 at 1:02
• @TheGreatDuck My pleasure. – Lanier Freeman Feb 25 '16 at 1:05