lim-x-1-4x-lnx-x-1-

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

$$\frac{\mathrm{0}}{\mathrm{0}}{form}\:\boldsymbol{{using}}\:\boldsymbol{{L}}'\boldsymbol{{H}}\hat {\boldsymbol{{o}pital}}\:\boldsymbol{{rule}} \\ $$$$\mathrm{4}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left[\mathrm{1}+\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\right]}{\mathrm{1}}=\mathrm{4} \\ $$

Commented by Rio Michael last updated on 08/Jun/20

thanks! please can you use another method other thank L′Hopital′s rule

$$\mathrm{thanks}!\:\mathrm{please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{use}\:\mathrm{another}\:\mathrm{method}\: \\ $$$$\mathrm{other}\:\mathrm{thank}\:\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}} \\ $$

Commented by smridha last updated on 08/Jun/20

okk lim_(x→1) ((4xlnx)/(x−1))=lim_(h→0) ((4(h+1)ln(h+1))/h) =4lim_(h→0) ln(h+1)+4lim_(h→0) ((ln(1+h))/h) =0+4×1=4

$${okk} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{4}\boldsymbol{{xlnx}}}{\boldsymbol{{x}}−\mathrm{1}}=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\boldsymbol{{h}}+\mathrm{1}\right)\boldsymbol{{ln}}\left(\boldsymbol{{h}}+\mathrm{1}\right)}{\boldsymbol{{h}}} \\ $$$$=\mathrm{4}\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{ln}}\left(\boldsymbol{{h}}+\mathrm{1}\right)+\mathrm{4}\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{h}}\right)}{\boldsymbol{{h}}} \\ $$$$=\mathrm{0}+\mathrm{4}×\mathrm{1}=\mathrm{4} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 08/Jun/20

$$\mathrm{thank}\:\mathrm{you}\: \\ $$

Answered by abdomathmax last updated on 08/Jun/20

changement x−1=t give ((4xlnx)/(x−1)) =((4(t+1)ln(1+t))/t) t→1 ⇒t→0 so ((4(t+1)ln(1+t))/t) ∼((4(t+1)t)/t) =4(t+1)→4 ⇒lim_(x→1) ((4xlnx)/(x−1)) =4

$$\mathrm{changement}\:\mathrm{x}−\mathrm{1}=\mathrm{t}\:\mathrm{give}\:\frac{\mathrm{4xlnx}}{\mathrm{x}−\mathrm{1}}\:=\frac{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{t}} \\ $$$$\mathrm{t}\rightarrow\mathrm{1}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\:\mathrm{so}\:\frac{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{t}}\:\sim\frac{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\mathrm{t}}{\mathrm{t}} \\ $$$$=\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)\rightarrow\mathrm{4}\:\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \:\frac{\mathrm{4xlnx}}{\mathrm{x}−\mathrm{1}}\:=\mathrm{4} \\ $$

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