The function $$f: \mathbb{R} \to \mathbb{R}$$ defined by $$f(x) = \lim_{n \to \infty} \frac{\cos(2\pi x) - x^{2n}\sin(x-1)}{1 + x^{2n+1} - x^{2n}}$$ is continuous for all $$x$$ in

We have $$f(x) = \lim_{n \to \infty} \frac{\cos(2\pi x) - x^{2n}\sin(x-1)}{1 + x^{2n+1} - x^{2n}}$$.

To evaluate this limit, we need to consider different cases based on the value of $$|x|$$.

Case 1: $$|x| < 1$$. Here $$x^{2n} \to 0$$ and $$x^{2n+1} \to 0$$ as $$n \to \infty$$. So $$f(x) = \frac{\cos(2\pi x) - 0}{1 + 0 - 0} = \cos(2\pi x)$$.

Case 2: $$|x| > 1$$. We divide numerator and denominator by $$x^{2n}$$:

Case 3: $$x = 1$$. Then $$x^{2n} = 1$$ and $$x^{2n+1} = 1$$. So $$f(1) = \frac{\cos(2\pi) - \sin(0)}{1 + 1 - 1} = \frac{1 - 0}{1} = 1$$.

Case 4: $$x = -1$$. Then $$x^{2n} = (-1)^{2n} = 1$$ and $$x^{2n+1} = (-1)^{2n+1} = -1$$. So $$f(-1) = \frac{\cos(-2\pi) - \sin(-2)}{1 + (-1) - 1} = \frac{1 - \sin(-2)}{-1} = \frac{1 + \sin 2}{-1} = -(1 + \sin 2)$$.

Now let us check continuity at the critical points $$x = 1$$ and $$x = -1$$.

At $$x = 1$$: From Case 1 (as $$x \to 1^-$$): $$\cos(2\pi \cdot 1) = 1$$. From Case 2 (as $$x \to 1^+$$): $$\frac{-\sin(x-1)}{x-1} \to -1$$ (since $$\frac{\sin(x-1)}{x-1} \to 1$$). The function value is $$f(1) = 1$$. Since the left limit is 1 but the right limit is $$-1$$, the function is discontinuous at $$x = 1$$.

At $$x = -1$$: From Case 1 (as $$x \to -1^+$$): $$\cos(2\pi(-1)) = \cos(-2\pi) = 1$$. From Case 2 (as $$x \to -1^-$$): $$\frac{-\sin(-1-1)}{-1-1} = \frac{-\sin(-2)}{-2} = \frac{\sin 2}{-2} = -\frac{\sin 2}{2}$$. The function value is $$f(-1) = -(1 + \sin 2)$$. The left limit is $$-\frac{\sin 2}{2}$$ and the right limit is $$1$$, neither of which equals $$-(1 + \sin 2)$$. So the function is discontinuous at $$x = -1$$.

For all other points, the function is continuous (being composed of continuous functions in each region, with the regions being open intervals).

Therefore $$f$$ is continuous for all $$x \in \mathbb{R} - \{-1, 1\}$$.

Hence, the correct answer is Option 2.