For `xgt0, lim_(xrarr0) {(sinx)^(1//x)+((1)/(x))^sinx}`, is

Options

A. 0
B. -1
C. 1
D. 2

Explanation

To solve the limit \( \lim_{x \to 0^+} \left( \sin x^{\frac{1}{x}} + \left( \frac{1}{x} \right)^{\sin x} \right) \), we will break it down step by step.

### Step 1: Rewrite the Limit
We can express the limit as:
\[ \lim_{x \to 0^+} \left( \sin x^{\frac{1}{x}} + \left( \frac{1}{x} \right)^{\sin x} \right) \]

### Step 2: Evaluate Each Term Separately
We will evaluate each term separately.

1. **First Term: \( \lim_{x \to 0^+} \sin x^{\frac{1}{x}} \)**

As \( x \to 0^+ \), \( \sin x \to 0 \) and \( \frac{1}{x} \to \infty \). Therefore, we have \( \sin x^{\frac{1}{x}} \to 0^{\infty} \), which is an indeterminate form.

Using the fact that \( \sin x \approx x \) for small \( x \):
\[ \sin x^{\frac{1}{x}} \approx \left( x \right)^{\frac{1}{x}} \to 0 \]

2. **Second Term: \( \lim_{x \to 0^+} \left( \frac{1}{x} \right)^{\sin x} \)**

Here, \( \sin x \to 0 \) as \( x \to 0^+ \). Thus, we can rewrite this term:
\[ \left( \frac{1}{x} \right)^{\sin x} = e^{\sin x \log \left( \frac{1}{x} \right)} = e^{-\sin x \log x} \]
As \( x \to 0^+ \), \( \sin x \approx x \), hence:
\[ -\sin x \log x \approx -x \log x \]
Now, we need to evaluate \( \lim_{x \to 0^+} -x \log x \).

### Step 3: Evaluate \( \lim_{x \to 0^+} -x \log x \)
To evaluate this limit, we can use L'Hôpital's Rule since it is of the form \( 0 \cdot (-\infty) \):
\[ \lim_{x \to 0^+} -x \log x = \lim_{x \to 0^+} \frac{-\log x}{\frac{1}{x}} \]
Applying L'Hôpital's Rule:
\[ = \lim_{x \to 0^+} \frac{-\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} x = 0 \]

### Step 4: Combine the Results
Now we have:
1. \( \lim_{x \to 0^+} \sin x^{\frac{1}{x}} = 0 \)
2. \( \lim_{x \to 0^+} \left( \frac{1}{x} \right)^{\sin x} = e^0 = 1 \)

Thus, combining both results:
\[ \lim_{x \to 0^+} \left( \sin x^{\frac{1}{x}} + \left( \frac{1}{x} \right)^{\sin x} \right) = 0 + 1 = 1 \]

### Final Answer
\[ \lim_{x \to 0^+} \left( \sin x^{\frac{1}{x}} + \left( \frac{1}{x} \right)^{\sin x} \right) = 1 \]