`lim_(x to 0) ("sin"2X)/(2 - sqrt(4 - x))` is

Options

A. 2
B. 4
C. 8
D. 0

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin(2x)}{2 - \sqrt{4 - x}} \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \frac{\sin(2x)}{2 - \sqrt{4 - x}} \]

### Step 2: Rationalize the denominator
To simplify the expression, we can multiply the numerator and the denominator by the conjugate of the denominator:
\[ \lim_{x \to 0} \frac{\sin(2x)}{2 - \sqrt{4 - x}} \cdot \frac{2 + \sqrt{4 - x}}{2 + \sqrt{4 - x}} \]
This gives us:
\[ \lim_{x \to 0} \frac{\sin(2x)(2 + \sqrt{4 - x})}{(2 - \sqrt{4 - x})(2 + \sqrt{4 - x})} \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:
\[ (2 - \sqrt{4 - x})(2 + \sqrt{4 - x}) = 2^2 - (\sqrt{4 - x})^2 = 4 - (4 - x) = x \]
So, we can rewrite our limit:
\[ \lim_{x \to 0} \frac{\sin(2x)(2 + \sqrt{4 - x})}{x} \]

### Step 4: Separate the limit
Now we can separate the limit into two parts:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} \cdot \lim_{x \to 0} (2 + \sqrt{4 - x}) \]

### Step 5: Evaluate the first limit
Using the standard limit \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \) where \( k = 2 \):
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \]

### Step 6: Evaluate the second limit
Now evaluate the second limit:
\[ \lim_{x \to 0} (2 + \sqrt{4 - x}) = 2 + \sqrt{4} = 2 + 2 = 4 \]

### Step 7: Combine the results
Now we can combine the results from the two limits:
\[ 2 \cdot 4 = 8 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin(2x)}{2 - \sqrt{4 - x}} = 8 \]

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