# Find the limit: `lim_(x->1) sin(x-1)/(x^2 + x - 2)`

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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**