`lim_(xto3)sqrt(1-cos2(x-3))/(x-3)`

Options

A. `=sqrt2`
B. does not exist
C. `=1`
D. `=-sqrt2`

Answer

To solve the limit \( \lim_{x \to 3} \frac{\sqrt{1 - \cos(2(x - 3))}}{x - 3} \), we will follow these steps:

### Step 1: Substitute the limit
First, we substitute \( x = 3 \) into the limit expression to check if it results in an indeterminate form:
\[ \cos(2(3 - 3)) = \cos(0) = 1 \quad \Rightarrow \quad 1 - \cos(0) = 0 \]
Thus, the expression becomes:
\[ \frac{\sqrt{0}}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can proceed with simplification.

### Step 2: Use the identity for cosine
We can use the trigonometric identity \( 1 - \cos(2\theta) = 2\sin^2(\theta) \). Here, we replace \( \theta \) with \( x - 3 \):
\[ 1 - \cos(2(x - 3)) = 2\sin^2(x - 3) \]
Thus, the limit becomes:
\[ \lim_{x \to 3} \frac{\sqrt{2\sin^2(x - 3)}}{x - 3} \]

### Step 3: Simplify the expression
We can simplify the square root:
\[ \sqrt{2\sin^2(x - 3)} = \sqrt{2} \cdot |\sin(x - 3)| \]
So the limit now is:
\[ \lim_{x \to 3} \frac{\sqrt{2} |\sin(x - 3)|}{x - 3} \]

### Step 4: Analyze the limit from both sides
Next, we need to consider the left-hand limit and the right-hand limit as \( x \) approaches \( 3 \).

#### Left-hand limit (\( x \to 3^- \)):
As \( x \) approaches \( 3 \) from the left, \( x - 3 \) is negative, so \( |\sin(x - 3)| = -\sin(x - 3) \):
\[ \lim_{x \to 3^-} \frac{\sqrt{2} (-\sin(x - 3))}{x - 3} = -\sqrt{2} \lim_{x \to 3^-} \frac{\sin(x - 3)}{x - 3} \]
Using the limit property \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \) where \( u = x - 3 \):
\[ = -\sqrt{2} \cdot 1 = -\sqrt{2} \]

#### Right-hand limit (\( x \to 3^+ \)):
As \( x \) approaches \( 3 \) from the right, \( x - 3 \) is positive, so \( |\sin(x - 3)| = \sin(x - 3) \):
\[ \lim_{x \to 3^+} \frac{\sqrt{2} \sin(x - 3)}{x - 3} = \sqrt{2} \lim_{x \to 3^+} \frac{\sin(x - 3)}{x - 3} \]
Again using the same limit property:
\[ = \sqrt{2} \cdot 1 = \sqrt{2} \]

### Step 5: Conclusion
Since the left-hand limit and the right-hand limit are not equal:
\[ \lim_{x \to 3^-} = -\sqrt{2} \quad \text{and} \quad \lim_{x \to 3^+} = \sqrt{2} \]
Thus, the limit does not exist.

### Final Answer:
\[ \text{Limit does not exist.} \]