What is the limit of #ln(1+5x) - ln(3+4x)# as x approaches #oo#?

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Answer 1 · 2016-09-05T13:12:36

#= ln (5/4)#

#lim_(x to oo) ln(1+5x) - ln(3+4x)#

#= lim_(x to oo) ln( (1+5x)/(3+4x))#

#= lim_(x to oo) ln( (1/x+5)/(3/x+4))#

#= ln( (lim_(x to oo) 1/x+5)/(lim_(x to oo)3/x+4))#

#= ln (5/4)#

Answer 2 · 2016-09-05T13:13:17

#:." The Reqd. Limit"=ln(5/4)#.

The Reqd. Limit#=lim_(xrarroo) {ln(1+5x)-ln(3+4x)}#

#=lim_(xrarroo) ln{(1+5x)/(3+4x)}#

#=lim_(xrarroo) ln{(cancelx(5+1/x))/(cancelx(4+3/x))}#

#=lim_(xrarroo) ln{(5+1/x)/(4+3/x)}#

As #ln# is a continuous fun., The Reqd. Limit

#=ln{lim_(xrarroo) (5+1/x)/(4+3/x)}#

#=ln((5+0)/(4+0))=ln(5/4)#

#:." The Reqd. Limit"=ln(5/4)#.