(ii) $$ \lim _{x \rightarrow 3} \frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}=1 $$

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Evaluate the limit by following steps:
- step0: Evaluate using transformations:
  $$\lim _{x\rightarrow 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)$$
- step1: Multiply by the Conjugate:
  $$\lim _{x\rightarrow 3}\left(\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\right)$$
- step2: Rewrite the expression:
  $$\frac{\lim _{x\rightarrow 3}\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\lim _{x\rightarrow 3}\left(2\right)}$$
- step3: Evaluate:
  $$\frac{2}{\lim _{x\rightarrow 3}\left(2\right)}$$
- step4: Evaluate:
  $$\frac{2}{2}$$
- step5: Reduce the fraction:
  $$1$$
To evaluate the limit

$$
\lim_{x \rightarrow 3} \frac{x-3}{\sqrt{x-2}-\sqrt{4-x}},
$$

we can follow these steps:

1. **Substituting the limit value**: First, we substitute $$x = 3$$ into the expression:

$$
   \frac{3-3}{\sqrt{3-2}-\sqrt{4-3}} = \frac{0}{\sqrt{1}-\sqrt{1}} = \frac{0}{0}.
   $$

This results in an indeterminate form $$ \frac{0}{0} $$, so we need to simplify the expression.

2. **Rationalizing the denominator**: To eliminate the square roots in the denominator, we can multiply the numerator and the denominator by the conjugate of the denominator:

$$
   \frac{x-3}{\sqrt{x-2}-\sqrt{4-x}} \cdot \frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}} = \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(\sqrt{x-2})^2 - (\sqrt{4-x})^2}.
   $$

3. **Simplifying the denominator**: The denominator simplifies as follows:

$$
   (\sqrt{x-2})^2 - (\sqrt{4-x})^2 = (x-2) - (4-x) = x - 2 - 4 + x = 2x - 6.
   $$

4. **Rewriting the limit**: Now we can rewrite the limit:

$$
   \lim_{x \rightarrow 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2x-6}.
   $$

5. **Factoring the denominator**: Notice that $$2x - 6 = 2(x - 3)$$. Thus, we can rewrite the limit as:

$$
   \lim_{x \rightarrow 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}.
   $$

6. **Canceling the common factor**: We can cancel $$x - 3$$ from the numerator and denominator (as long as $$x \neq 3$$):

$$
   \lim_{x \rightarrow 3} \frac{\sqrt{x-2}+\sqrt{4-x}}{2}.
   $$

7. **Substituting again**: Now we substitute $$x = 3$$:

$$
   \frac{\sqrt{3-2}+\sqrt{4-3}}{2} = \frac{\sqrt{1}+\sqrt{1}}{2} = \frac{1+1}{2} = \frac{2}{2} = 1.
   $$

Thus, the limit is

$$
\lim_{x \rightarrow 3} \frac{x-3}{\sqrt{x-2}-\sqrt{4-x}} = 1.
$$

This confirms the original statement.
The limit is 1.

(ii)

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