Question Number 34160 by candre last updated on 01/May/18
$${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}ln}\:{x}\centerdot\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}ln}\:{x}\centerdot\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18
Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18
$${see}\:{the}\:{graph}\:{LHS}\:{limit}\:{is}\:{zero}\:{also}\:{RHS}\:{limit} \\ $$$${also}\:{zero} \\ $$
Commented by abdo mathsup 649 cc last updated on 02/May/18
$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\left({xlnx}\right)\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}=\mathrm{0}\:{because}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:{xln}\left({x}\right)=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:=\:\mathrm{1}\:\:{from}\:{another}\:{side} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} {ln}\left({x}\right).{ln}\left(\mathrm{1}+{x}\right)\:={ln}\left(\mathrm{1}\right){ln}\left(\mathrm{2}\right)=\mathrm{0}\:{because} \\ $$$${the}\:{function}\:{ln}\left({x}\right)\:{is}\:{continue}\:{at}\:\mathrm{1}\:{and}\:\mathrm{2}. \\ $$$$ \\ $$
Answered by MJS last updated on 02/May/18
$$\mathrm{right}\:\mathrm{side}: \\ $$$$\mathrm{ln1}=\mathrm{0} \\ $$$$\mathrm{0ln2}=\mathrm{0} \\ $$$$\mathrm{left}\:\mathrm{side}: \\ $$$$\mathrm{e}^{\mathrm{ln}{x}\:\mathrm{ln}\left({x}+\mathrm{1}\right)} =\left({x}+\mathrm{1}\right)^{\mathrm{ln}{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left({x}+\mathrm{1}\right)^{\mathrm{ln}{x}} =\mathrm{1}^{−\infty} =\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}ln}{x}\:\mathrm{ln}\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$