In order to avoid the #0/0# indeterminate form, which you would get if you tried to evaluate this limit for #x->1#, you can a little algebraic manipulation to rewrite your initial function.
The numerator of the fraction can be factored using the formula for the difference of two cubes
#color(blue)(a^3 - b^3 = (a-b)^3 + 3ab(a-b)#
This means that you can write
#lim_(x->1)((x^3-1)/(x^2 + 2x - 3)) = lim_(x->1)( (x-1)^3 + 3 * x * 1 * (x-1))/(x^2 + 2x - 3)#
Now focus on the denominator, which can be factored to get
#x^2 + 2x - 3 = x^2 -x + 3x - 3 = (x-1)(x+3)#
Your limit will now be
#lim_(x->1)(color(red)(cancel(color(black)((x-1)))) * [(x-1)^2 + 3x])/(color(red)(cancel(color(black)( (x-1)))) (x+3)) = lim_(x->1)((x-1)^2 + 3x)/(x+3)#
As #x->1#, this limit will be equal to
#lim_(x->1) ( (1-1)^2 + 3 * 1)/(1+3) = color(green)(3/4)#
