Question #7515d

1 Answer
Jim H
Apr 2, 2015

The limit is #0#

#ln(2+x) - ln (1+x) = ln((2+x)/(1+x))#

#lim_(xrarroo)(2+x)/(1+x) = 1#

#ln# is continuous on its domain so #lim_(urarr1)lnu = ln1 = 0#

So,
#lim_(xrarroo)[ln(2+x) - ln (1+x)] = lim_(xrarroo)ln((2+x)/(1+x)) = 0#

(You colud insert #=ln1=# before the #0#

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