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

1 Answer
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