Show that the `lim_(xto2) ((sqrt(1-cos{2(x-2)}))/(x-2))` doesnot exist.

Explanation

To show that the limit

\[ \lim_{x \to 2} \frac{\sqrt{1 - \cos(2(x - 2))}}{x - 2} \]

does not exist, we will analyze the left-hand limit and the right-hand limit separately.

### Step 1: Rewrite the expression using trigonometric identity
We know the identity:

\[ 1 - \cos(2\theta) = 2 \sin^2(\theta) \]

Letting \(\theta = x - 2\), we can rewrite the limit as:

\[ 1 - \cos(2(x - 2)) = 2 \sin^2(x - 2) \]

Thus, our limit becomes:

\[ \lim_{x \to 2} \frac{\sqrt{2 \sin^2(x - 2)}}{x - 2} \]

### Step 2: Simplify the expression
This simplifies to:

\[ \lim_{x \to 2} \frac{\sqrt{2} \cdot |\sin(x - 2)|}{x - 2} \]

### Step 3: Analyze the left-hand limit
For \(x\) approaching \(2\) from the left (i.e., \(x \to 2^{-}\)), \(x - 2\) is negative, so \(|\sin(x - 2)| = -\sin(x - 2)\). The left-hand limit is:

\[ \lim_{x \to 2^{-}} \frac{\sqrt{2} \cdot (-\sin(x - 2))}{x - 2} \]

Substituting \(t = x - 2\) (where \(t \to 0^{-}\)), we have:

\[ \lim_{t \to 0^{-}} \frac{\sqrt{2} \cdot (-\sin(t))}{t} \]

Using the limit property \(\lim_{t \to 0} \frac{\sin(t)}{t} = 1\), this becomes:

\[ -\sqrt{2} \cdot 1 = -\sqrt{2} \]

### Step 4: Analyze the right-hand limit
For \(x\) approaching \(2\) from the right (i.e., \(x \to 2^{+}\)), \(x - 2\) is positive, so \(|\sin(x - 2)| = \sin(x - 2)\). The right-hand limit is:

\[ \lim_{x \to 2^{+}} \frac{\sqrt{2} \cdot \sin(x - 2)}{x - 2} \]

Again substituting \(t = x - 2\) (where \(t \to 0^{+}\)), we have:

\[ \lim_{t \to 0^{+}} \frac{\sqrt{2} \cdot \sin(t)}{t} \]

Using the same limit property, this becomes:

\[ \sqrt{2} \cdot 1 = \sqrt{2} \]

### Step 5: Compare the left-hand and right-hand limits
Now we have:

- Left-hand limit: \(-\sqrt{2}\)
- Right-hand limit: \(\sqrt{2}\)

Since the left-hand limit and the right-hand limit are not equal:

\[ -\sqrt{2} \neq \sqrt{2} \]

### Conclusion
Thus, we conclude that:

\[ \lim_{x \to 2} \frac{\sqrt{1 - \cos(2(x - 2))}}{x - 2} \text{ does not exist.} \]