Find the limit: `lim_(x->1) sin(x-1)/(x^2 + x - 2)`
We have to determine `lim_(x->1) sin(x - 1)/(x^2 + x - 2)`
If we substitute x = 1, we get the the result as 0/0 which is not determinate. This allows the use of l'Hopital's rule and the numerator and the denominator can be substituted with their derivatives.
=> `lim_(x->1) cos(x - 1)/(2x +1)`
substituting x = 1, gives (cos 0)/3 = 1/3
The required value of the limit is 1/3