`"lt"_(x to 0) ((sin x)/x)^((sin x)/(x- sinx))` is:

Options

A. `e^(-1)`
B. e
C. 1
D. none

Answer

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

### Step-by-Step Solution:

1. **Identify the components of the limit:**
We have \( f(x) = \frac{\sin x}{x} \) and \( g(x) = \frac{\sin x}{x - \sin x} \).

2. **Evaluate the limit of \( f(x) \) as \( x \to 0 \):**
We know that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

3. **Evaluate the limit of \( g(x) \) as \( x \to 0 \):**
First, we need to simplify \( g(x) \):
\[ g(x) = \frac{\sin x}{x - \sin x} \]
As \( x \to 0 \), both the numerator and denominator approach 0. We can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{\sin x}{x - \sin x} = \lim_{x \to 0} \frac{\cos x}{1 - \cos x} = \frac{1}{0} \text{ (undefined)} \]
However, we can rewrite \( x - \sin x \) using the Taylor series expansion:
\[ x - \sin x \approx x - \left(x - \frac{x^3}{6} + O(x^5)\right) = \frac{x^3}{6} + O(x^5) \]
Thus,
\[ g(x) \approx \frac{\sin x}{\frac{x^3}{6}} = \frac{6 \sin x}{x^3} \]
As \( x \to 0 \), we know \( \frac{\sin x}{x} \to 1 \), so:
\[ g(x) \to \frac{6 \cdot 1}{0} \text{ (undefined)} \]

4. **Combine the results using the exponential limit form:**
Since we have \( f(x) \to 1 \) and \( g(x) \to \infty \) (as \( g(x) \) approaches infinity), we can rewrite the limit in the form \( f(x)^{g(x)} \):
\[ \lim_{x \to 0} f(x)^{g(x)} = e^{\lim_{x \to 0} (f(x) - 1) g(x)} \]

5. **Calculate \( (f(x) - 1) g(x) \):**
We know:
\[ f(x) - 1 = \frac{\sin x - x}{x} \]
Thus,
\[ (f(x) - 1) g(x) = \frac{\sin x - x}{x} \cdot \frac{\sin x}{x - \sin x} \]
We can substitute \( \sin x - x \approx -\frac{x^3}{6} \) and \( x - \sin x \approx \frac{x^3}{6} \):
\[ \lim_{x \to 0} \frac{-\frac{x^3}{6}}{x} \cdot \frac{\sin x}{\frac{x^3}{6}} = \lim_{x \to 0} -1 \cdot 1 = -1 \]

6. **Final result:**
Therefore, we have:
\[ \lim_{x \to 0} f(x)^{g(x)} = e^{-1} = \frac{1}{e} \]

### Conclusion:
The limit is:
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{\sin x}{x - \sin x}} = \frac{1}{e} \]