The value of $$\lim_{x \to 1} \frac{(x^2-1)\sin^2(\pi x)}{x^4-2x^3+2x-1}$$ is equal to:

We need to find $$\lim_{x \to 1} \frac{(x^2-1)\sin^2(\pi x)}{x^4-2x^3+2x-1}$$. Numerator: $$(x^2-1)\sin^2(\pi x) = (x-1)(x+1)\sin^2(\pi x)$$. For the denominator, since $$x = 1$$ is a root, we factor out $$(x-1)$$: $$x^4 - 2x^3 + 2x - 1 = (x-1)(x^3 - x^2 - x + 1)$$, and as $$x = 1$$ is again a root of $$x^3 - x^2 - x + 1$$: $$x^3 - x^2 - x + 1 = (x-1)(x^2 - 1) = (x-1)^2(x+1)$$, so finally $$x^4 - 2x^3 + 2x - 1 = (x-1)^3(x+1)$$.

Hence the expression simplifies to $$\frac{(x-1)(x+1)\sin^2(\pi x)}{(x-1)^3(x+1)} = \frac{\sin^2(\pi x)}{(x-1)^2}$$.

Let $$t = x - 1$$, so as $$x \to 1$$, $$t \to 0$$. Then $$\sin(\pi x) = \sin(\pi + \pi t) = -\sin(\pi t)$$ and hence $$\sin^2(\pi x) = \sin^2(\pi t)$$. Therefore, $$\lim_{t \to 0} \frac{\sin^2(\pi t)}{t^2} = \left(\lim_{t \to 0} \frac{\sin(\pi t)}{t}\right)^2 = \pi^2$$ (using the standard limit $$\lim_{u \to 0} \frac{\sin u}{u} = 1$$ with $$u = \pi t$$).

Therefore, the answer is Option D: $$\pi^2$$.