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Question Number 54730 by yorrick23ralph@gmail.com last updated on 09/Feb/19
prove lnx≤x
$${prove}\:{lnx}\leqslant{x} \\ $$
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
$${first}\:{x}>\mathrm{0}\:\:{let}\:\:{f}\left({x}\right)={x}−{ln}\left({x}\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{{x}}\:=\frac{{x}−\mathrm{1}}{{x}}\:{and}\:{f}^{'} \left({x}\right)\geqslant\mathrm{0}\:\Leftrightarrow{x}\geqslant\mathrm{1} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{f}\left({x}\right)=+\infty\:\:{and}\:{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)= \\ $$$${lim}_{{x}\rightarrow+\infty} \:{x}\left(\mathrm{1}−\frac{{lnx}}{{x}}\right)={lim}_{+\infty} {x}=+\infty \\ $$$${variation}\:{of}\:{f}\left({x}\right) \\ $$$${x}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${f}\left({x}\right)\:\:\:\:+\infty\:{dec}\:\:\mathrm{1}\:\:\:\:\:{inc}\:\:\:\:+\infty \\ $$$${we}\:{see}\:{that}\:{f}\left({x}\right)\geqslant\mathrm{0}\:\:\forall\:{x}>\mathrm{0}\:\Rightarrow{ln}\left({x}\right)\leqslant{x}\:\:\forall{x}>\mathrm{0} \\ $$
Commented by yorrick23ralph@gmail.com last updated on 09/Feb/19
thanks sir
$${thanks}\:{sir} \\ $$
Commented by mr W last updated on 10/Feb/19
x<e^x   ⇒ln x<x  (not ln x≤x)
$${x}<{e}^{{x}} \\ $$$$\Rightarrow\mathrm{ln}\:{x}<{x}\:\:\left({not}\:\mathrm{ln}\:{x}\leqslant{x}\right) \\ $$

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