The limit `lim_(x->1)(x^2+2x-3)/(x^2-x)` has to be determined.

If x is substituted with 1, `(x^2+2x-3)/(x^2-x) = (1+2-3)/(1-1) = 0/0` which is an indeterminate form. l'Hopital's rule can be used here and the numerator and denominator substituted by their derivatives.

This gives: `lim_(x->1)(2x+2)/(2x-1)`

Substituting x = 1 gives `(2 + 2)/(2 - 1) = 4/1 = 4`

**The limit **`lim_(x->1)(x^2+2x-3)/(x^2-x) = 4`