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Question Number 91228 by ~blr237~ last updated on 28/Apr/20

$$\underset{x\to 1}{\lim }lnx\left({\int }_0^x\frac{dt}{lnt}\right)$$

Commented by abdomathmax last updated on 29/Apr/20

$$={lim}_{x\to 1}(x-1)lnx\times \frac{{\int }_0^x\frac{1}{lnt}dt}{x-1}$$$$wehave{lim}_{x\to 1}(x-1)lnx=0$$$${lim}_{x\to 1}\frac{{\int }_0^x\frac{1}{lnt}dt}{x-1}={lim}_{x\to 1}\frac{\frac{1}{lnx}}{1}\left(hospital\right)$$$$={lim}_{x\to 1}\frac{1}{lnx}=\mathrm{\infty }.perhapsthislimitdontexist..$$

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