Question

How do you evaluate `[ ( ln x ) / (x^2 + x - 2 )]` as x approaches 1?

Answer 1

`lim_(x-> 1) (Ln(x))/(x^2+x-2) = 1/3`

Explanation:

If we evaluate the expression at the limit, we get `Ln(1) / (1^2 +1 -2 ) = 0/0`

This `0/0` result is a L'Hopital indeterminate form, it does not mean the limit does not exist, it means the limit does exist but needs to be derived using L'Hopitals Rule.

To find the limit using L'Hopitals rule simple evaluate the same limit for the derivative of the numerator over the derivative of the denominator.

So `lim_(x-> 1) ((Ln(x))')/(((x^2+x-2))') = (1/x)/(2x+1) = (1/1)/(2(1)+1) = 1/3`