Question

lim(x趋于0+)(ln(xln a)ln(ln ax/ln(x/a))),其中a>1

高数题目

Answer 1

原极限=lim ln( ln(ax)/ln(x/a) )/[1/ln(xlna)] (这部化为无穷比无穷型,然后用洛必达法则)
=lim { [ ln(x/a)/ln(ax) ][ ( (1/x)ln(x/a)-(1/x)ln(ax) )/(ln(x/a)²) ] }/{ (-1/x)/[ ln(xlna) ]² }
=lim { ln(x/a)ln(1/a²)/[ln(ax)(ln(x/a))²] }/{ (-1)/[ ln(xlna) ]² }
=lim -ln(1/a²) [ ln(x/a)/ln(ax) ] { [ ln(xlna) / ln(x/a)]² }
=-ln(1/a²)·1·1
=-ln(1/a²)