How do you find the limit #lnx/(x-1)# as #x->1#?

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Answer 1 · 2016-10-19T17:47:34

The answer#=1#

Use l'Hôpital Rule
Limit#= 0/0#
So the derivative of numerator and denominator gives
#=(lnx')/((x-1)')=1/x# and the limit is #1/1=1# as #x->1#

Answer 2 · 2016-10-19T19:54:48

#1#

#lim_(x->0)log_e x/(x-1) = lim_(x->1)log_e(x^(1/(x-1)))#

Making #y = x-1# we have #x = 1+y# and #lim_(x->1) equiv lim_(y->0)#

then

# lim_(x->1)log_e(x^(1/(x-1)))equiv lim_(y->0)log_e((1+y)^(1/y)) = log_e e = 1#