# Finding asymptotes of |x|/x

1. Jan 21, 2014

### syd9231

1. The problem statement, all variables and given/known data
List all the asymptotes of
f(x) = |x| / x

2. Relevant equations

3. The attempt at a solution
This is a problem on a limits test in a Calculus AB class. I tried vertical asymptotes and horizontal asymptotes by setting the numerator and denominator equal to 0 but only getting x=0

|x|=0

x=0
Unfortunately that wasn't the answer to the question, but I plugged the equation in the calculator and it did appear to have a vertical asymptote at x=0 and two horizontal asymptotes at x=1 and x=-1
However I don't know how to find that algebraically.
I did some research and saw an example where they found the limit of a function as it went to positive and negative infinity in order to find the asymptotes. I tried that:

lim |x|/x = 1
x→∞

lim |x|/x = -1
x→-∞

At the end of this I was left with
x=0 (from my very first attempt)
x=1
x=-1

Questions
1. Is setting finding the limit of the function as it goes to infinity a solid way of finding asymptotes?
2. a. Is my final answer correct?
b. If not, Could someone explain the mistakes/false-reasoning I made or lead me in the right direction?

2. Jan 21, 2014

### Dick

It doesn't have a vertical asymptote at x=0. What are the limits from the right and left? And you are right about the behavior at infinity, but doesn't that make the asymptote lines y=1 and y=(-1) (not x=1 and x=(-1))?

3. Jan 21, 2014

### syd9231

There is no vertical asymptote at x=0 because
the limit as x goes to 0 from the left is -1
the limit as x goes to 0 from the right is 1
Yes?

but there is two horizontal asymptotes at
y=1
y=-1
because of the limit as x goes to ∞ and -∞?

Does everything look good now?

4. Jan 21, 2014

### Dick

Yes. Vertical asymptotes are where a one sided limit goes to infinity. Doesn't happen here.

5. Jan 21, 2014

### Staff: Mentor

I don't think I would trust that calculator.

The function here is simple enough that you can sketch a graph in about the same time it would take to enter it into the calculator.

f(x) = 1 if x > 0, and
f(x) = -1 if x < 0
It is undefined at x = 0 (jump discontinuity).
The graph consists of two horizontal half-lines.