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lim-x-1-lnx-0-x-dt-lnt-




Question Number 91228 by ~blr237~ last updated on 28/Apr/20
lim_(x→1)   lnx(∫_0 ^x  (dt/(lnt)) )
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:{lnx}\left(\int_{\mathrm{0}} ^{{x}} \:\frac{{dt}}{{lnt}}\:\right)\: \\ $$
Commented by abdomathmax last updated on 29/Apr/20
=lim_(x→1)  (x−1)lnx ×((∫_0 ^x  (1/(lnt))dt)/(x−1))  we have lim_(x→1) (x−1)lnx =0  lim_(x→1)    ((∫_0 ^x  (1/(lnt))dt)/(x−1)) =lim_(x→1)   ((1/(lnx))/1)  (hospital)  =lim_(x→1) (1/(lnx)) =∞  .perhaps this limit dont exist..
$$={lim}_{{x}\rightarrow\mathrm{1}} \:\left({x}−\mathrm{1}\right){lnx}\:×\frac{\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{1}}{{lnt}}{dt}}{{x}−\mathrm{1}} \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){lnx}\:=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \:\:\:\frac{\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{1}}{{lnt}}{dt}}{{x}−\mathrm{1}}\:={lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{\frac{\mathrm{1}}{{lnx}}}{\mathrm{1}}\:\:\left({hospital}\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}} \frac{\mathrm{1}}{{lnx}}\:=\infty\:\:.{perhaps}\:{this}\:{limit}\:{dont}\:{exist}.. \\ $$

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