$$\lim_{x \to 2} \left( \dfrac{\sqrt{1 - \cos\{2(x-2)\}}}{x - 2} \right)$$

$$L = \lim_{x\to2} \left( \frac{\sqrt{1 - \cos\{2(x - 2)\}}}{x - 2} \right)$$

Let $$x - 2 = h$$. As $$x \to 2$$, $$h \to 0$$:

$$L = \lim_{h\to0} \frac{\sqrt{1 - \cos 2h}}{h} = \lim_{h\to0} \frac{\sqrt{2\sin^2 h}}{h} = \lim_{h\to0} \frac{\sqrt{2}\vert{}\sin h\vert{}}{h}$$

Evaluating Left-Hand Limit (LHL), as $$h \to 0^-$$:

$$\text{LHL} = \lim_{h\to0^-} \frac{\sqrt{2}(-\sin h)}{h} = -\sqrt{2} \lim_{h\to0^-} \frac{\sin h}{h} = -\sqrt{2}$$

Evaluating Right-Hand Limit (RHL), as $$h \to 0^+$$:

$$\text{RHL} = \lim_{h\to0^+} \frac{\sqrt{2}(\sin h)}{h} = \sqrt{2} \lim_{h\to0^+} \frac{\sin h}{h} = \sqrt{2}$$

$$\text{LHL} \neq \text{RHL}$$

Answer: Option (D): does not exist