Find the value of `lim_(xto0^(+)) (sinx)^((1)/(x))`.

Explanation

To solve the limit \( \lim_{x \to 0^+} (\sin x)^{\frac{1}{x}} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ L = \lim_{x \to 0^+} (\sin x)^{\frac{1}{x}} \]

### Step 2: Take the natural logarithm
To simplify the limit, we can take the natural logarithm of both sides:
\[ \ln L = \lim_{x \to 0^+} \frac{1}{x} \ln(\sin x) \]

### Step 3: Analyze \(\ln(\sin x)\) as \(x \to 0^+\)
As \(x\) approaches \(0\), \(\sin x\) approaches \(0\). Therefore, \(\ln(\sin x)\) approaches \(-\infty\). We need to analyze the limit:
\[ \ln L = \lim_{x \to 0^+} \frac{\ln(\sin x)}{x} \]

### Step 4: Use L'Hôpital's Rule
Since both the numerator and denominator approach \(0\) as \(x \to 0^+\), we can apply L'Hôpital's Rule:
\[ \ln L = \lim_{x \to 0^+} \frac{\ln(\sin x)}{x} = \lim_{x \to 0^+} \frac{\frac{d}{dx}(\ln(\sin x))}{\frac{d}{dx}(x)} \]

### Step 5: Differentiate the numerator and denominator
The derivative of \(\ln(\sin x)\) is:
\[ \frac{d}{dx}(\ln(\sin x)) = \frac{\cos x}{\sin x} = \cot x \]
The derivative of \(x\) is \(1\). Therefore, we have:
\[ \ln L = \lim_{x \to 0^+} \cot x \]

### Step 6: Evaluate the limit of \(\cot x\)
As \(x\) approaches \(0\), \(\cot x\) approaches \(\infty\). Thus:
\[ \ln L = \infty \]

### Step 7: Exponentiate to find \(L\)
Since \(\ln L = \infty\), we have:
\[ L = e^{\infty} = 0 \]

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