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[math] 0^0 = 1 [/math] and here is my quick proof:[eqn] \displaystyle \lim_{x \rightarrow 0} x^x = \lim_{x \rightarrow 0} e^{x\ln{(x)}} = e^{\lim_{x \rightarrow 0} x\ln{(x)}} = \Bigg( \lim_{x \rightarrow 0} x \Bigg)\Bigg(\lim_{x \rightarrow 0} x\ln{(x)} \Bigg) [/eqn]so by squeeze theorem:[eqn] \displaystyle \Bigg( \lim_{x \rightarrow 0} x \Bigg)\Bigg(\lim_{x \rightarrow 0} x\ln{(x)} \Bigg) = 0 \implies \lim_{x \rightarrow 0} x^x = e^0 = 1 [/eqn]Q.E.D :)

Your proof shows that [eqn]\lim_{x \rightarrow 0} x^x = 1[/eqn]. That part is not not disputed, I think. But that does not mean that 0^0 = 1.It is, mind you. But it doesn't follow from this limit.

Math is the shittiest meme to ever come out of humanity

>>10254963Okay, but where's your proof that exponentiation is continuous.

>>10254974how come it doesnt??>>10255016why do I need to proof that??

>>10255024>how come it doesnt??Why would it?

>>10255024x^x approaches 1 on the real axis, but not anywhere on the complex plane

>>10254974>He doesn't understand limits

>>10255052Do tell.

>>10255055im not this anon>>10255052[math] \displaystyle \lim_{x \to c} f(x) = a [/math] doesnt imply that [math] f(c) = a [/math]thats why even though sin(x)/x approaches 1 as x approaches 0, sin(0)/0 isnt equal to one, its undefined.

>>10255061Indeed.

>>10255061>partial sum definition of infinite series ignores thisand that's why its the worst way to define infinite series

>>10254963Isn't a^n just 1×a×...×a n times?So a^0 = 1 by defnAlso 0^0 is then 1 without ambiguity.Just like 0! = 1 by def

>>10254963Wolframed

>>10255098Not quite.[math]a^0=1[/math] and [math]a^n=aa^{n-1}[/math] define exponentiation.

>>10255103[math]\frac{a^n}{a^n}=a^{n-n}=a^0=1[/math]

>>10255407Yeah, but if you just define [math]a^ba^c=a^{b+c}[/math] you still have to define [math]a^1=a[/math], which is really short but still gives longer proofs.

>>10254963here is a definition of the exponent:a(n,0)=1a(n,k+1)=n*a(n,k)> every natural number is 0 or of the form k+1 for some k> this always halts for any input because the second argument decreases> the proof that a(0,0)=1 is obviousThis is why sometimes 0^0 = 1. Attempting to prove this with calculus is beyond retarded.

>>10255437>its 1 by definitionfaggot

>>10255024are you retarded? by definition the function doesn't have to be defined in the exact point.

>>10254974based and redpilled answer. Here is a real proof that 0^0=1. def: for cardinal numbers A, B, we define A^B as the number of functions from a set of size A to a set of size B. Since there is precisely one function from the empty set to itself, 0^0=1

>>10254963whoa dude! You're a genius!Doing a similar thing you can prove that:[math]\frac{1}{0}=infinity[/math]

>>10255632ah yes, its obvious that 0^0 is 1 if you only work with the nonnegative integers