`lim_(xrarr0+) (sinsqrt(x))/(sqrt(sinx))` is equal to

Options

A. 0
B. 1
C. -1
D. none of these

Explanation

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

### Step-by-step Solution:

1. **Rewrite the Limit**:
We start with the limit:
\[ L = \lim_{x \to 0^+} \frac{\sin(\sqrt{x})}{\sqrt{\sin(x)}} \]

2. **Divide Numerator and Denominator**:
We can divide both the numerator and the denominator by \( \sqrt{x} \):
\[ L = \lim_{x \to 0^+} \frac{\sin(\sqrt{x})}{\sqrt{x}} \cdot \frac{1}{\sqrt{\sin(x)} / \sqrt{x}} \]

3. **Simplify the Expression**:
This can be rewritten as:
\[ L = \lim_{x \to 0^+} \frac{\sin(\sqrt{x})}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{\sin(x)}} \]

4. **Evaluate Each Limit**:
- The first part \( \lim_{x \to 0^+} \frac{\sin(\sqrt{x})}{\sqrt{x}} \) can be evaluated using the substitution \( y = \sqrt{x} \), which gives \( x = y^2 \) and as \( x \to 0^+ \), \( y \to 0^+ \):
\[ \lim_{y \to 0^+} \frac{\sin(y)}{y} = 1 \]
- The second part \( \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{\sin(x)}} \) can be evaluated using the fact that as \( x \to 0 \), \( \sin(x) \approx x \):
\[ \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{\sin(x)}} = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{x}} = 1 \]

5. **Combine the Results**:
Now we combine both results:
\[ L = 1 \cdot 1 = 1 \]

### Final Result:
Thus, the limit is:
\[ \lim_{x \to 0^+} \frac{\sin(\sqrt{x})}{\sqrt{\sin(x)}} = 1 \]