`Lt_(xto0)(sqrt(4+x)-2)/(sinx)` is equal to
(i) 4
(ii) 1
(iii) `1/4`
(iv) 0

Options

A. 4
B. 1
C. `1/4`
D. 0

Answer

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

### Step 1: Rationalize the Numerator
We start by rationalizing the numerator. We multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{4+x} + 2 \):

\[ \lim_{x \to 0} \frac{\sqrt{4+x} - 2}{\sin x} \cdot \frac{\sqrt{4+x} + 2}{\sqrt{4+x} + 2} \]

### Step 2: Simplify the Expression
This gives us:

\[ \lim_{x \to 0} \frac{(\sqrt{4+x} - 2)(\sqrt{4+x} + 2)}{\sin x (\sqrt{4+x} + 2)} \]

Using the difference of squares, we can simplify the numerator:

\[ \sqrt{4+x}^2 - 2^2 = (4+x) - 4 = x \]

So our limit now looks like:

\[ \lim_{x \to 0} \frac{x}{\sin x (\sqrt{4+x} + 2)} \]

### Step 3: Split the Limit
Now we can separate the limit:

\[ \lim_{x \to 0} \frac{x}{\sin x} \cdot \lim_{x \to 0} \frac{1}{\sqrt{4+x} + 2} \]

### Step 4: Evaluate the Limits
We know from the standard limit that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \Rightarrow \quad \lim_{x \to 0} \frac{x}{\sin x} = 1 \]

Now, we evaluate the second limit:

\[ \lim_{x \to 0} \frac{1}{\sqrt{4+x} + 2} = \frac{1}{\sqrt{4+0} + 2} = \frac{1}{2 + 2} = \frac{1}{4} \]

### Step 5: Combine the Results
Now we combine both results:

\[ \lim_{x \to 0} \frac{x}{\sin x} \cdot \lim_{x \to 0} \frac{1}{\sqrt{4+x} + 2} = 1 \cdot \frac{1}{4} = \frac{1}{4} \]

Thus, the final answer is:

\[ \boxed{\frac{1}{4}} \]