In #28:
ln(x² - 1): ln(x) is defined only for positive values of x, so:
x² - 1 > 0 ---> x² > 1 ---> x > 1 or x < -1
In the denominator: x² - 2x must be positive:
x² - 2x > 0 ---> x(x - 2) > 0
either x > 0 and x - 2 > 0 or x < 0 and x - 2 < 0
---> x > 2 or x < 0
Now put the first two answers with the second two answers and take the most restrictive cases: x > 1 and x > 2 means x >2 X < -1 and x < 0 means x < -1
Therefore, it will be continuous in the region where x < -1 and again in the region where x > 2 (but not between these two regions).
