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March 30

0^0 (cont'd!)

I have read the previous discussions on the definition of $0^{0}$ . The controversy arises solely because the limit $\lim _{(x,y)\rightarrow (0,0)}x^{y}$ does not exist... but does it matter that it doesn't exist? What's wrong with simply defining $0^{0}=1$ and acknowledging that the function $x^{y}$ is discontinuous at $(0,0)$ ? 101.119.129.156 (talk) 13:40, 30 March 2025 (UTC)[reply]

Maybe it is because mathematicians like to be purists. Although typical real-world problems consider $0^{0}=1$ , there are theoretical systems where the limit exists but has a different value. 101.119.129.156 (talk) 14:00, 30 March 2025 (UTC)[reply]

Continuity matters because operations on real numbers require it. Consider the expression π√2. You can't say "Add π to itself √2 times" because that's nonsense. Instead you have to build up the definition from integers to rationals, then from rationals to real numbers. In detail, first define r⋅n as r added to itself n times; this can be done inductively: r⋅0 = 0, r⋅(n+1) = r⋅n + r. Then define r⋅(a/b) when a and b are integers, as the solution to p⋅b=r⋅a. Finally define x⋅y as the limit of (a_i/b_i)(c_i/d_i) where a_i/b_i has limit x and c_i/d_i has limit y. But without knowing that r⋅s is a continuous function of r and s you can't guarantee that your limiting value of (a_i/b_i)(c_i/d_i) doesn't depend on which sequences a_i/b_i and c_i/d_i you're using. Multiplication is continuous so there is no such problem extending the definition from rationals to reals. But exponentiation is not continuous so there is a problem. You have to restrict the domain of the operation so that this issue does not arise. And this has been done to extend the definition as much as possible, though this becomes awkward to state concisely. If you restrict r to positive values then r^s is continuous and can be extended to x^y for real x and y as long as x is positive. If n is a non-negative integer then rⁿ is continuous for all r, so xⁿ can be defined for all real x. Because r^s is not a continuous function in the neighborhood of r=0, s=0, the expression 0⁰ is problematic when considered as the case r=0, s=0 of the expression r^s. It's not that mathematicians like to be purists, but that they like to have expressions mean something definite and not be a matter of opinion. For more detail, the relevant article is Zero to the power of zero. --RDBury (talk) 20:22, 30 March 2025 (UTC)[reply]

Here is a concrete example. Take the problem of determining

\lim _{x\downarrow 0}f(x)^{g(x)}

and consider the rule

\lim _{x\downarrow 0}f(x)^{g(x)}=F^{G},~{\text{where}}~F=\lim _{x\downarrow 0}f(x)~{\text{and}}~G=\lim _{x\downarrow 0}g(x).

This seems reasonable enough. Now apply this to

f(x)=\exp \left(-{\frac {1}{x}}\right),~g(x)=x.

In this case

F=G=0,

so the rule suggests that the answer is

0^{0}.

But the actual limit is

{\frac {1}{e}}\approx 0.37\,.

When the exponent is restricted to the domain of the natural numbers, the notion of it having a limit does not apply, so then there is no ground for considering

0^{0}

an indeterminate form, and defining

x^{0}=1

without restriction is perfectly reasonable. With that convention, defining a Taylor series by

\textstyle {\sum _{i=0}^{\infty }}a_{i}x^{i}

means the same as, but is more convenient than,

a_{0}+\textstyle {\sum _{i=1}^{\infty }}a_{i}x^{i}

‑‑Lambiam 21:04, 30 March 2025 (UTC)[reply]

March 31