How do you find the limit of #(ln x^2) / (x^2-1)# as x approaches 1?

1 Answer
Cesareo R.
Jul 1, 2016

#lim_{x->1}(log_e x^2) / (x^2-1)=1#

Explanation:

#(log_e x^2) / (x^2-1) = log_e x/(x-1)-log_e x/(x+1)#
but the #log_e x# series expansion in the #x=1# vicinity is
#log_e x = sum_{k=1}^{oo}((x-1)^k)/k# so

#log_e x/(x-1) = 1 + sum_{k=2}^{oo}((x-1)^{k-1})/k#

so

#lim_{x->1}log_e x/(x-1) = 1#

Finally

#lim_{x->1}(log_e x^2) / (x^2-1) = lim_{x->1}log_e x/(x-1)-lim_{x->1}log_e x/(x+1) = 1 + 0=1 #

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