To find the limit, we can factor the numerator and cancel the common factor of $(x - 2)$ in the numerator and denominator. This gives $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x + 2)(x - 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 2 + 2 = 4$.

A function is continuous at a point if its limit at that point is equal to the value of the function at that point. Of the given functions, only $f(x) = \sin(x)$ is continuous at $x = 0$ because $\lim_{x \to 0} \sin(x) = \sin(0) = 0$.

To find the derivative, we can use the power rule of differentiation. The derivative of $x^n$ is $n x^{n-1}$. Applying this rule, we get $f'(x) = \frac{d}{dx} (x^3 - 2x^2 + 3x - 4) = 3x^2 - 4x + 3$.

To evaluate the integral, we can use the power rule of integration. The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$. Applying this rule, we get $\int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$.

To determine the convergence of the sequence, we can use the limit comparison test. We compare $a_n$ with the sequence $b_n = \frac{n^2}{n^2} = 1$. Since $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^2 + 3n + 2}{n^2 + 1} = \lim_{n \to \infty} \frac{n^2/n^2 + 3n/n^2 + 2/n^2}{n^2/n^2 + 1/n^2} = \lim_{n \to \infty} \frac{1 + 3/n + 2/n^2}{1 + 1/n^2} = 1$, and $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} 1$ is a convergent harmonic series, therefore $\sum_{n=1}^{\infty} a_n$ also converges by the limit comparison test. Furthermore, $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^2 + 3n + 2}{n^2 + 1} = \lim_{n \to \infty} \frac{n^2/n^2 + 3n/n^2 + 2/n^2}{n^2/n^2 + 1/n^2} = \lim_{n \to \infty} \frac{1 + 3/n + 2/n^2}{1 + 1/n^2} = 1$, so the sequence converges to 1.