what is the limit of : ln(x) / (x^(2)-1) as {x --> 1}?

Answer 1

if you say y = x -1, you have to find
limit of ln(y+1)/y/(y+2) as y--->0
ln(y+1) = y -y^2/2 +y^3/3 -y^4/4....
simplify by y and you are looking for the limit of:
(1-y/2 +y^2/3....0)/(y+2) as y ---> o
the limit is 1/2

Answer 2

Make a substitution.

Limit as x→1 of ln(x) / (x² - 1)

Let u = x - 1
x² - 1 = (x - 1)(x + 1) = u(u + 2) = u² + 2u

Limit as x→1 of ln(x) / (x² - 1)
= Limit as u→0 of ln(u + 1) / (u² + 2u)

This is now the limit of a quotient in the indeterminant form of 0/0 so we can use L'Hospital's rule. It says that the limit of the quotient is equal to the derivative of the numerator divided by the derivative of the denominator.

= Limit as u→0 of ln(u + 1) / (u² + 2u)

= Limit as u→0 of [1/(u + 1)] / (2u + 2)

= Limit as u→0 of 1/[2(u + 1)²] = 1/2