# B Difference between these functions .

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 $$x=0$$ ?

#### micromass

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.

#### mfb

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.

#### fresh_42

Mentor
2018 Award
I disagree very much with the mathematician's explanation. They're like saying that $0^0 = 1$ is a consensus among mathematicians, it's not.
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.

#### micromass

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

#### jedishrfu

Mentor
Sure, there are plenty of reasons why it should be $1$. The best are set theoretic and category theoretic. But there is no consensus.
At least its a very binary decision and mathematicians can wear the shirt:

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

#### mfb

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

• jedishrfu

#### micromass

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

#### jedishrfu

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.
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.

#### fresh_42

Mentor
2018 Award
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 $1$.

#### micromass

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 $1$.
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.

#### micromass

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?