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Question Number 28151 by tawa tawa last updated on 21/Jan/18

lim_(x→1) ((1/(log_e x)) − (x/(x − 1)))

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\left(\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{e}} \mathrm{x}}\:−\:\frac{\mathrm{x}}{\mathrm{x}\:−\:\mathrm{1}}\right) \\ $$

Commented by çhëý böý last updated on 21/Jan/18

lim_(x→1) ((((x−1)−xlnx)/((x−1)lnx)))=(0/0) lim_(x→1) (((x−((x/x)+lnx))/(lnx+(((x−1))/( x ))))) lim_(x→1 ) (((x(x−1+1nx))/(xlnx+(x−1))))=(0/0) lim_(x→1) ((((x−1+lnx)+x(1−0+(1/x)))/(1+lnx+1))) lim_(x→1) (((0+0+2)/(1+0+1)))=(2/2)=1

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\left({x}−\mathrm{1}\right)−{xlnx}}{\left({x}−\mathrm{1}\right){lnx}}\right)=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{{x}−\left(\frac{{x}}{{x}}+{lnx}\right)}{{lnx}+\frac{\left({x}−\mathrm{1}\right)}{\:{x}\:}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}\:} {\mathrm{lim}}\:\:\left(\frac{{x}\left({x}−\mathrm{1}+\mathrm{1}{nx}\right)}{{xlnx}+\left({x}−\mathrm{1}\right)}\right)=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\left({x}−\mathrm{1}+{lnx}\right)+{x}\left(\mathrm{1}−\mathrm{0}+\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+{lnx}+\mathrm{1}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{0}+\mathrm{0}+\mathrm{2}}{\mathrm{1}+\mathrm{0}+\mathrm{1}}\right)=\frac{\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$

Commented by tawa tawa last updated on 21/Jan/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by abdo imad last updated on 21/Jan/18

=lim_(x→1) ( (1/(lnx)) −(x/(x−1))) let use the ch.x−1=t x→1⇔ t→o lim(...)= lim_(t→0) ((1/(ln(1+t))) −((1+t)/t)) =lim_(t→0) ((t −(1+t)ln(1+t))/(tln(1+t)))let take f(t)=t −(1+t)ln(1+t) and g(t)=tln(1+t) we have f^′ (t)= 1−(ln(1+t)+1)=−ln(1+t) and f^(′′) (t) =((−1)/(1+t)) g^′ (t)=ln(1+t) +(t/(1+t))=ln(1+t) +1−(1/(1+t)) and g^(′′) (t)= (1/(1+t)) +(1/((1+t)^2 )) so lim_(x→1) ((1/(lnx)) −(x/(x−1)))=lim_(t→0) ((f^(′′) (t))/(g^(′′) (t))) = ((−1)/2) .

$$={lim}_{{x}\rightarrow\mathrm{1}} \:\left(\:\frac{\mathrm{1}}{{lnx}}\:−\frac{{x}}{{x}−\mathrm{1}}\right)\:{let}\:{use}\:{the}\:{ch}.{x}−\mathrm{1}={t} \\ $$$${x}\rightarrow\mathrm{1}\Leftrightarrow\:{t}\rightarrow{o}\:\:\:{lim}\left(...\right)=\:{lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+{t}\right)}\:−\frac{\mathrm{1}+{t}}{{t}}\right) \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\frac{{t}\:−\left(\mathrm{1}+{t}\right){ln}\left(\mathrm{1}+{t}\right)}{{tln}\left(\mathrm{1}+{t}\right)}{let}\:{take}\:{f}\left({t}\right)={t}\:−\left(\mathrm{1}+{t}\right){ln}\left(\mathrm{1}+{t}\right) \\ $$$${and}\:{g}\left({t}\right)={tln}\left(\mathrm{1}+{t}\right)\:{we}\:{have}\: \\ $$$${f}^{'} \left({t}\right)=\:\mathrm{1}−\left({ln}\left(\mathrm{1}+{t}\right)+\mathrm{1}\right)=−{ln}\left(\mathrm{1}+{t}\right)\:{and}\:{f}^{''} \left({t}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}+{t}} \\ $$$${g}^{'} \left({t}\right)={ln}\left(\mathrm{1}+{t}\right)\:+\frac{{t}}{\mathrm{1}+{t}}={ln}\left(\mathrm{1}+{t}\right)\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{t}}\:{and} \\ $$$${g}^{''} \left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:+\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\:\:{so} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \left(\frac{\mathrm{1}}{{lnx}}\:−\frac{{x}}{{x}−\mathrm{1}}\right)={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{{f}^{''} \left({t}\right)}{{g}^{''} \left({t}\right)}\:=\:\:\frac{−\mathrm{1}}{\mathrm{2}}\:. \\ $$

Commented by abdo imad last updated on 21/Jan/18

we have used hospital theorem.

$${we}\:{have}\:{used}\:{hospital}\:{theorem}. \\ $$

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