Question Number 97577 by Rio Michael last updated on 08/Jun/20
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{4}{x}\:\mathrm{ln}{x}}{{x}−\mathrm{1}}\:=?? \\ $$
Commented by Dwaipayan Shikari last updated on 23/Jun/20
$$\mathrm{4}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{xlog}\left(\mathrm{1}+{x}−\mathrm{1}\right)}{{x}−\mathrm{1}}=\mathrm{4} \\ $$
Answered by smridha last updated on 08/Jun/20
![(0/0)form using L′Ho^ pital rule 4lim_(x→1) (([1+ln(x)])/1)=4](https://www.tinkutara.com/question/Q97579.png)
$$\frac{\mathrm{0}}{\mathrm{0}}{form}\:\boldsymbol{{using}}\:\boldsymbol{{L}}'\boldsymbol{{H}}\hat {\boldsymbol{{o}pital}}\:\boldsymbol{{rule}} \\ $$$$\mathrm{4}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left[\mathrm{1}+\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\right]}{\mathrm{1}}=\mathrm{4} \\ $$
Commented by Rio Michael last updated on 08/Jun/20
$$\mathrm{thanks}!\:\mathrm{please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{use}\:\mathrm{another}\:\mathrm{method}\: \\ $$$$\mathrm{other}\:\mathrm{thank}\:\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}} \\ $$
Commented by smridha last updated on 08/Jun/20
$${okk} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{4}\boldsymbol{{xlnx}}}{\boldsymbol{{x}}−\mathrm{1}}=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\boldsymbol{{h}}+\mathrm{1}\right)\boldsymbol{{ln}}\left(\boldsymbol{{h}}+\mathrm{1}\right)}{\boldsymbol{{h}}} \\ $$$$=\mathrm{4}\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{ln}}\left(\boldsymbol{{h}}+\mathrm{1}\right)+\mathrm{4}\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{h}}\right)}{\boldsymbol{{h}}} \\ $$$$=\mathrm{0}+\mathrm{4}×\mathrm{1}=\mathrm{4} \\ $$$$ \\ $$
Commented by Rio Michael last updated on 08/Jun/20
$$\mathrm{thank}\:\mathrm{you}\: \\ $$
Answered by abdomathmax last updated on 08/Jun/20
$$\mathrm{changement}\:\mathrm{x}−\mathrm{1}=\mathrm{t}\:\mathrm{give}\:\frac{\mathrm{4xlnx}}{\mathrm{x}−\mathrm{1}}\:=\frac{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{t}} \\ $$$$\mathrm{t}\rightarrow\mathrm{1}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\:\mathrm{so}\:\frac{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{t}}\:\sim\frac{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\mathrm{t}}{\mathrm{t}} \\ $$$$=\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\rightarrow\mathrm{4}\:\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \:\frac{\mathrm{4xlnx}}{\mathrm{x}−\mathrm{1}}\:=\mathrm{4} \\ $$