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Question Number 54566 by mr W last updated on 06/Feb/19
limn→∞(1+c1+μ1n)n=? with0⩽c⩽1,μ>0
Commented bymaxmathsup by imad last updated on 06/Feb/19
letAn=(1+e1+μ1n)n⇒An=(1+e+μ1n−μ1n1+μ1n)n=(1+e−μ1n1+μ1n)n =enln(1+e−μ1n1+μ1n)butμ1n=eln(μ)n∼1+ln(μ)n⇒e−μ1n∼e−1−ln(μ)nand 1+μ1n=2+ln(μ)n⇒1+e−μ1n1+μ1n∼1+e−12=1+e2⇒An∼enln(1+e2)→+∞(n→+∞) (lookthatln(1+e2)>0)weconsiderμandefixed..
Commented bymr W last updated on 06/Feb/19
thankyousir! c⩽1 thereforeln(1+c2)⩽0 itseemsifc<1,(1+e1+μ1n)n→0 whatifc=1?,i.e.(21+μ1n)n→?
youarerightsirihavecommitedaerrorofcalculuswehave 0<e<1⇒1<1+e<2⇒12<1+e2<1⇒ln(1+e2)<0⇒limn→+∞An=0 forgivemeiwastired...
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