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Question-120127




Question Number 120127 by huotpat last updated on 29/Oct/20
Answered by Dwaipayan Shikari last updated on 29/Oct/20
lim_(x→1) ((1−x+logx)/((x−1)^2 ))  =((1−x+log(1+x−1))/((x−1)^2 ))=((1−x+x−1−(((x−1)^2 )/2)+(((x−1)^3 )/3)−...)/((x−1)^2 ))  =lim_(x→0) ((−(x−1)^2 )/(2(x−1)^2 ))+O(x−1)^3   =−(1/2)         (As   O(x−1)^3 →0   )
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}−{x}+{logx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−{x}+{log}\left(\mathrm{1}+{x}−\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−{x}+{x}−\mathrm{1}−\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{3}}−…}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+{O}\left({x}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\left({As}\:\:\:{O}\left({x}−\mathrm{1}\right)^{\mathrm{3}} \rightarrow\mathrm{0}\:\:\:\right) \\ $$
Answered by Bird last updated on 29/Oct/20
g(x)=((1−x+lnx)/((x−1)^2 )) we do the ch.  x−1=t ⇒g(x)=g(1+t)  =((−t+ln(1+t))/t^2 )  we have ln^′ (1+t)=(1/(1+t)) =1−t +o(t^2 )  ⇒ln(1+t) =t−(t^2 /2)+o(t^3 ) ⇒  ln(1+t)−t =−(t^2 /2)+o(t^3 ) ⇒  g(1+t)=−(1/2)+o(t) ⇒  lim_(t→0) g(1+t)=−(1/2)=lim_(x→0) g(x)
$${g}\left({x}\right)=\frac{\mathrm{1}−{x}+{lnx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:{we}\:{do}\:{the}\:{ch}. \\ $$$${x}−\mathrm{1}={t}\:\Rightarrow{g}\left({x}\right)={g}\left(\mathrm{1}+{t}\right) \\ $$$$=\frac{−{t}+{ln}\left(\mathrm{1}+{t}\right)}{{t}^{\mathrm{2}} } \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{t}\right)=\frac{\mathrm{1}}{\mathrm{1}+{t}}\:=\mathrm{1}−{t}\:+{o}\left({t}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{t}\right)\:={t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({t}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{t}\right)−{t}\:=−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({t}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${g}\left(\mathrm{1}+{t}\right)=−\frac{\mathrm{1}}{\mathrm{2}}+{o}\left({t}\right)\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} {g}\left(\mathrm{1}+{t}\right)=−\frac{\mathrm{1}}{\mathrm{2}}={lim}_{{x}\rightarrow\mathrm{0}} {g}\left({x}\right) \\ $$

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