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>> No.10254963 [View]
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10254963

[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 :)



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