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lim-x-log-x-2021x-log-x-pi-iberty-




Question Number 129655 by liberty last updated on 17/Jan/21
  lim_(x→∞) (log _x (2021x))^(log x)  =?                                                   ⌈∗⌉((√π^(ℓiberty) ) )
limx(logx(2021x))logx=?(πiberty)
Answered by bemath last updated on 17/Jan/21
 lim_(x→∞) (((ln (2021x))/(ln x)))^(ln x) = lim_(x→∞) (1+((ln 2021)/(ln x)))^(ln x)     lim_(x→∞) [(1+(1/((((ln x)/(ln 2021))))))((ln x)/(ln 2021)) ]^(ln 2021) = e^(ln 2021)  = 2021
limx(ln(2021x)lnx)lnx=limx(1+ln2021lnx)lnxlimx[(1+1(lnxln2021))lnxln2021]ln2021=eln2021=2021

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