Question

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

Answer 1

`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`