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Question Number 33865 by 33 last updated on 26/Apr/18
evaluatelimx→∞π(aπ)xx!
Commented by abdo imad last updated on 26/Apr/18
wehaveforx∈V(+∞)x!∼xxe−x2πx(stirlingformulageneralised)⇒A(x)=π(aπ)xx!∼π(aπ)xxxe−x2πx=πex(ax)xx−x−12.12π=π2exaxx−12wemusthavea>0anda≠1π⇒A(x)∼π2exln(ae)e−12ln(x)=π2exlna+x−12ln(x)⇒A(x)∼π2ex(lna+1−12lnxx)→+∞(x→+∞)
limA(x)=+∞ifln(a)+1>0⇔a>1eif0<a<eln(a)+1<0⇒limx→+∞A(x)=0
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