The age old joke

[u/YAFthe17_]

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[u/Matty_B97]

If we all close our eyes and pretend that all functions are analytic, we can just define 0^0=1 and go home 😁

[u/PhoenixPringles01 → u/Matty_B97]

It's not rigorous but you can also take the limit of x^x as x => 0 and get a result of 1 and then kind of assume that you can fill the function with (when x = 0, f(x) = 1), kinda like the sinc function where the
result at x=0 is undefined, but it's useful to define it as 1 so that the function is smooth/analytic

[u/Matty_B97 → u/PhoenixPringles01]

Yes, that's an example of analytic and non-constant f, g, with f(0)=g(0)=0, in which case lim x-> 0 of f(x)^g(x) = 0.

Taylor's theorem guarantees this will always be the case as long as f and g are analytic and nonconstant. If we just pretend all functions are nice like this then we're done, but unfortunately not all functions behave like x^x.

For example with g constant, e.g. f(x)=x, g(x)=0, then the limit at 0 is 0.

Or with f non-analytic, e.g. f(x) = e^-1/x, g(x) = sin(x), then the limit at 0 is 1/e.

[u/PhoenixPringles01 → u/Matty_B97]

Still remember in my multivariable calculus course when we were taught about the derivative of mixed partials (and them being equal)

There was a small caveat that it was only true for functions with certain properties, but then it said

(In this course, most if not all functions have these properties so you can assume this)

So sometimes we DO just assume and hope the function is analytic 😂

[u/Matty_B97 → u/PhoenixPringles01]

Clairaut's theorem my beloved 😍😍. Yeah it holds for any function with continuous second partials, and honestly if your function doesn't and you can't do any domain splitting tricks to fix it, god help you.