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lim-x-1-1-ln-x-1-x-1-




Question Number 131962 by EDWIN88 last updated on 10/Feb/21
 lim_(x→1^+ ) ((1/(ln x)) − (1/(x−1)))=?
$$\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{x}}\:−\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\right)=? \\ $$
Answered by liberty last updated on 10/Feb/21
 let u=ln x ⇒x=e^u    lim_(u→0) ((1/u)−(1/(e^u −1)))   = lim_(u→0) (((e^u −1−u)/(u(e^u −1))))  = lim_(u→0) (((e^u −1)/(e^u −1+ue^u )))  = lim_(u→0) ((e^u /(e^u +e^u +ue^u )))=(1/2)
$$\:\mathrm{let}\:\mathrm{u}=\mathrm{ln}\:\mathrm{x}\:\Rightarrow\mathrm{x}=\mathrm{e}^{\mathrm{u}} \\ $$$$\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{u}}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{u}} −\mathrm{1}}\right) \\ $$$$\:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{u}} −\mathrm{1}−\mathrm{u}}{\mathrm{u}\left(\mathrm{e}^{\mathrm{u}} −\mathrm{1}\right)}\right) \\ $$$$=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{u}} −\mathrm{1}}{\mathrm{e}^{\mathrm{u}} −\mathrm{1}+\mathrm{ue}^{\mathrm{u}} }\right) \\ $$$$=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{e}^{\mathrm{u}} }{\mathrm{e}^{\mathrm{u}} +\mathrm{e}^{\mathrm{u}} +\mathrm{ue}^{\mathrm{u}} }\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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