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Question Number 54730 by yorrick23ralph@gmail.com last updated on 09/Feb/19
prove lnx≤x
provelnxx
Commented by Abdo msup. last updated on 09/Feb/19
first x>0  let  f(x)=x−ln(x) ⇒  f^′ (x)=1−(1/x) =((x−1)/x) and f^′ (x)≥0 ⇔x≥1  lim_(x→0^+ )    f(x)=+∞  and lim_(x→+∞) f(x)=  lim_(x→+∞)  x(1−((lnx)/x))=lim_(+∞) x=+∞  variation of f(x)  x         0               1               +∞  f^′ (x)            −             +  f(x)    +∞ dec  1     inc    +∞  we see that f(x)≥0  ∀ x>0 ⇒ln(x)≤x  ∀x>0
firstx>0letf(x)=xln(x)f(x)=11x=x1xandf(x)0x1limx0+f(x)=+andlimx+f(x)=limx+x(1lnxx)=lim+x=+variationoff(x)x01+f(x)+f(x)+dec1inc+weseethatf(x)0x>0ln(x)xx>0
Commented by yorrick23ralph@gmail.com last updated on 09/Feb/19
thanks sir
thankssir
Commented by mr W last updated on 10/Feb/19
x<e^x   ⇒ln x<x  (not ln x≤x)
x<exlnx<x(notlnxx)

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