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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)) )
limx1lnx(0xdtlnt)
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..
=limx1(x1)lnx×0x1lntdtx1wehavelimx1(x1)lnx=0limx10x1lntdtx1=limx11lnx1(hospital)=limx11lnx=.perhapsthislimitdontexist..

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