Evaluate the algebraic limit: $\lim_{x \to 0} \frac{\sqrt{1 + 2x} - \sqrt{1 - 3x}}{x}$.

Determine the value of the constant $k$ such that the function $f(x)$ is continuous at $x = 2$, where: $$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x \ne 2 \\ k & \text{if } x = 2 \end{cases}$$

Evaluate the following algebraic limits: (a) $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$ (b) $\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$

State and geometrically prove the fundamental trigonometric limit: $$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \quad (\theta \text{ in radians})$$ Hence evaluate $\lim_{x \to 0} \frac{1 - \cos(4x)}{x^2}$.