What is #lim_(xrarroo)(ln(6x+10)-ln(7+3x))#?

Question

2294 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-31T15:22:37

I found: #ln2#

Using a property of logs we can write:

#lim_(x->oo)[ln(6x+10)-ln(7+3x)]=#

#=lim_(x->oo)[ln((6x+10)/(7+3x))]=#

collect #x#:

#=lim_(x->oo)[ln((cancel(x)(6+10/x))/(cancel(x)(7/x+3)))]=#

as #x->oo# then #a/x->0# so:

#=lim_(x->oo)[ln(((6+10/x))/((7/x+3)))]=ln(6/3)=ln2#