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Question Number 103654 by Study last updated on 16/Jul/20
if a,b>1 lim_(x→0^+ ) ((ln(b−x))/(ax))=???
ifa,b>1limx→0+ln(b−x)ax=???
Answered by Worm_Tail last updated on 16/Jul/20
lim_(x→0^+ ) ((ln(b−x))/(ax))=lim((1/a)ln(b−x)^(1/x) ) lim_(x→0^+ ) ((ln(b−x))/(ax))=lim((1/a)ln(b(1−(x/b)))^(1/x) lim_(x→0^+ ) ((ln(b−x))/(ax))=lim(1/a)lnb^(1/x) (1−(x/b))^(1/x) lim_(x→0^+ ) ((ln(b−x))/(ax))=lim(1/a)lnb^(1/x) +lim(1/a)ln(1+(x/(−b))) lim_(x→0^+ ) ((ln(b−x))/(ax))=lim(1/a)lnb^(1/x) +lim(1/(a(−b)))ln(1+(x/(−b)))^((−b)/x) lim_(x→0^+ ) ((ln(b−x))/(ax))=lim(1/a)lnb^(1/x) −(1/(ab)) since b>1 lim_(x→0^+ ) ((ln(b−x))/(ax))=infinity−(1/(ab)) =infinity
limx→0+ln(b−x)ax=lim(1aln(b−x)1x)limx→0+ln(b−x)ax=lim(1aln(b(1−xb))1xlimx→0+ln(b−x)ax=lim1alnb1x(1−xb)1xlimx→0+ln(b−x)ax=lim1alnb1x+lim1aln(1+x−b)limx→0+ln(b−x)ax=lim1alnb1x+lim1a(−b)ln(1+x−b)−bxlimx→0+ln(b−x)ax=lim1alnb1x−1absinceb>1limx→0+ln(b−x)ax=infinity−1ab=infinity

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