# Thread: 0 Divided by 0: X and Y Axis

1. Looking at this: , represents the dividend and is just the input.

When is 1, we have our familiar rational function.

As increases, the two curves seen expand. And inversely, as decreases the curves get closer and closer to the x and y axis.

What happens when is simply 0? We'd have the function Okay, so the function is 0 everywhere, except for . Raising the question, what is 0 divided by 0?

Some say 0, some say 1, or undefined, or indeterminate. I think most of us would probably agree on an indeterminate form. so . We know 0 times 0 is 0. We know 1, 2, 3, pi, -1, and every real number times zero is 0.

So the value of 0/0 is indeterminate, so how would I show that on the graph?

Seeing how 1/x, 0.5/x, 0.1/x etc... gets closer and closer to the x and y axis, we'd eventually see lines coincidental to the x and y axis. Right?

It makes sense since indeterminate forms mean the value could be anything, so every value for the domain 0 would be represented, a vertical line.

( I started a thread on a similar topic on "f(x)=x^x (tetrative functions)" )  2.

3. This function is not defined for x=0; you cannot really show it on the graph, but technically the graph has no value at the point x=0.
For n->0 the functions approach the x-axis; for n=0 you get f(x)=0 for all x except 0, where the function is not defined. On the other hand there is no combination for n/x where the graph is equal to or parallel to the y-axis. You can approach the y-axis asymptotically by taking x->infinity, but the axis itself is never reached or intersected. The function diverges for both x->0 and n->infinity.  4. If 0/0 was undefined, the function would asymptotically approach x=0. A discontinuation. But, if I were to interpret it as an indeterminate form... The possible values of x are all real numbers (hence the indeterminacy) therefore 0/0 in this case brings On the interval f(0)=(-∞, +∞)

Therefore, the representation for this, where x=0, I would think to put a vertical line.

If I hadn't thought about the 1/x I wouldn't have come up with this, thinking that 0/x would remain in continuum on f(x)=0 for the entire interval.

But again, in , as n approaches zero, the curves approach a point of coindendentiality with the x and y axis.  5. An indeterminate form is not the same as the original function, they are disctinct mathematical constructs; the form does not allow you to assign a value for f(0), because a unique value would be required to solve the form. This is not the case, however, as all values for x are possible solutions to the indeterminate form.
If you plot the graph of your function for an arbitrary n in the interval of, say, x=-10...10, you will easily see why f(0) is not defined. A good overview of this topic can be found here :

Division by zero - Wikipedia, the free encyclopedia

A more in-depths treatment of the topic of division by zero can be found here :
I. Herstein: Topics in Algebra. ISBN 0-471-02371-X  Bookmarks
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