# B Difference between these functions .

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1. Aug 29, 2016

### Nader AbdlGhani

What's the difference between $$f(x)=3$$ and $$f(x)=3x^0$$ ? and why Limit of the second function when $$x\rightarrow0$$ exists ? and is the second function continuous at $$x=0$$ ?

2. Aug 29, 2016

### micromass

Staff Emeritus
You can't present a function by just giving a rule, like you do. You need to give the domain and the codomain too. So what are they?

Also, what is your convention for $0^0$? Yes, mathematicians disagree on what it should be.

3. Aug 29, 2016

### Staff: Mentor

Assuming R->R:
00 does not have a single common definition, so f(0) needs a separate definition in the second case. If you define f(0)=3, then both functions are identical, and the second function is continuous, otherwise it is not.

4. Aug 29, 2016

### Staff: Mentor

5. Aug 29, 2016

### micromass

Staff Emeritus
6. Aug 29, 2016

### Staff: Mentor

Maybe it's not. But I also think it's natural, simply for the reason that taking something to a power is something multiplicative. And therefore it makes sense to define it as the unit 1.

7. Aug 29, 2016

### micromass

Staff Emeritus
Sure, there are plenty of reasons why it should be $1$. The best are set theoretic and category theoretic. But there is no consensus.

8. Aug 29, 2016

### Staff: Mentor

At least its a very binary decision and mathematicians can wear the shirt:

"There are 10 different answers to the equation y = 0^0"

9. Aug 29, 2016

### Staff: Mentor

Binary? What about defining 00 as the limit $$\lim_{x \to 0} x^{1/\log(x)} = e$$? You can also get every other number of course.

10. Aug 29, 2016

### micromass

Staff Emeritus
Yes, the function $y^x$ has an essential singularity around $(0,0)$.

11. Aug 29, 2016

### Staff: Mentor

It all reduces to 1's and 0's somehow. :-)

That's a good example, I didn't see it before. It appears then that there's a whole class of possibilities too with other variations.

12. Aug 29, 2016

### Staff: Mentor

The vast, boring, senseless, stupid, Sisyphus-like, unprofitable and endless job to rewrite formerly beautiful and nice formulas like, e.g.
$$\exp(x) = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$
is justification enough for the choice of $1$.

13. Aug 29, 2016

### micromass

Staff Emeritus
The thing is though that $0^0 = 1$ in this case only when the exponent is already seen as an integer. If the exponent is seen as a real number then it's better to leave $0^0$ undefined. So we have this weird context-dependent rule: $y^x = 1$ if $x$ is an integer and undefined when it can take on a continuous range of variables. Such a definition would be the most interesting one, but have no idea how to formalize that in a neat way. Maybe some kind of typed logic or something.

14. Aug 29, 2016

### micromass

Staff Emeritus
Computer languanges can deal with this very well if you take as input $0^0$. Then it would find the type of the exponent. If the exponent is an INT, then it's $1$, if it's a double then it's undefined. But how to do this in mathematical context?

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