How do you find the limit #(xlnx)/(x^2-1)# as #x->1#?

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Answer 1 · 2018-05-19T16:50:16

The answer is #=1/2#

The limit is

#lim_(x->1)(xlnx)/(x^2-1)=0/0#

This result is an indeterminate form.

So, apply l'Hospital's rule

#lim_(x->1)(xlnx)/(x^2-1)=lim_(x->1)((xlnx)')/((x^2-1)')#

#=lim_(x->1)((x*1/x+lnx))/((2x)#

#=lim_(x->1)((1+lnx))/((2x)#

#=(1+ln1)/(2*1)#

#=1/2#
graph{(xlnx)/(x^2-1) [-2.735, 2.74, -1.368, 1.368]}