Let's evaluate each limit step by step.

(i) $\lim_{x \to 1^-} 2^{\frac{1}{x-1}}$

As $x \to 1^-$, the exponent $\frac{1}{x-1}$ tends towards $-\infty$, since $x - 1$ approaches a small negative value, meaning the denominator approaches 0 from the negative side.

Thus, the expression becomes $2^{-\infty}$, which tends to 0.

$\boxed{0}$

(ii) $\lim_{x \to 0} \frac{\sqrt{4+x} - 4}{x}$

To solve this, we rationalize the numerator by multiplying and dividing by the conjugate:

$\frac{\sqrt{4+x} - 4}{x} \cdot \frac{\sqrt{4+x} + 4}{\sqrt{4+x} + 4} = \frac{(4+x) - 16}{x(\sqrt{4+x} + 4)} = \frac{x - 12}{x(\sqrt{4+x} + 4)}$

Simplifying further:

$= \frac{-12}{\sqrt{4+x} + 4}$

Now, as $x \to 0$:

$\frac{-12}{\sqrt{4+0} + 4} = \frac{-12}{4+4} = \frac{-12}{8} = -\frac{3}{2}$

Thus, the limit is:

$\boxed{-\frac{3}{2}}$

(iii) $\lim_{x \to 2} \frac{x^2 - 4}{x + 2}$

First, factor the numerator:

$\frac{x^2 - 4}{x + 2} = \frac{(x - 2)(x + 2)}{x + 2}$

For $x \neq -2$, cancel the $x + 2$ terms:

$= x - 2$

Now, as $x \to 2$:

$2 - 2 = 0$

Thus, the limit is:

$\boxed{0}$

(iv) $\lim_{x \to a} \frac{\sin x - \sin a}{x - a}$

This limit is in the standard form of the derivative of the sine function. The derivative of $\sin x$ at $x = a$ is:

$\cos a$

Thus, the limit is:

$\boxed{\cos a}$

Do you want further details or have any questions?

Here are five related questions you might explore:

1. How do you evaluate limits involving logarithms?
2. What is the limit of exponential functions as the base approaches 0?
3. How does rationalizing the numerator help in solving limits?
4. What is L'Hopital's rule and when can it be applied?
5. How do you find the derivative of trigonometric functions?

Tip: Always check if a limit problem involves an indeterminate form (like $\frac{0}{0}$), which may suggest factoring, conjugation, or applying L'Hopital's rule for simplification.