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lim-x-ln-ax-b-ln-bx-a-




Question Number 1368 by prakash jain last updated on 25/Jul/15
lim_(x→∞)   ((ln(ax+b))/(ln (bx+a))) =?
limxln(ax+b)ln(bx+a)=?
Commented by 123456 last updated on 25/Jul/15
f(x)=((ln (ax+b))/(ln (bx+a)))  (((d/dx)[ln (ax+b)])/((d/dx)[ln (bx+a)]))=((a/(ax+b))/(b/(bx+a)))=((a(bx+a))/(b(ax+b)))→((ab)/(ba))=1
f(x)=ln(ax+b)ln(bx+a)ddx[ln(ax+b)]ddx[ln(bx+a)]=aax+bbbx+a=a(bx+a)b(ax+b)abba=1
Answered by 112358 last updated on 25/Jul/15
L=lim_(x→∞) ((ln(ax+b))/(ln(bx+a)))=(∞/∞) which is indeterminate.  Perhaps L′Ho^� ptial′s rule can be used.
L=limxln(ax+b)ln(bx+a)=whichisindeterminate.PerhapsLHoptial^srulecanbeused.
Commented by 123456 last updated on 26/Jul/15
a>0,b>0
a>0,b>0

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