The value of `lim_(x to 0) ("sinx"/x)^("sin x"/"x-sinx")` is

Options

A. e
B. 1
C. `e^2`
D. 1/e

Answer

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

### Step 1: Evaluate the base \( \frac{\sin x}{x} \)
As \( x \) approaches 0, we know from the standard limit that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

### Step 2: Evaluate the exponent \( \frac{\sin x}{x - \sin x} \)
We need to analyze the limit of the exponent as \( x \) approaches 0. We can rewrite the expression \( x - \sin x \) using the Taylor series expansion for \( \sin x \):
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
Thus,
\[ x - \sin x = x - \left( x - \frac{x^3}{6} + O(x^5) \right) = \frac{x^3}{6} - O(x^5) \]
As \( x \to 0 \), the dominant term is \( \frac{x^3}{6} \).

Now substituting this back into the exponent:
\[ \frac{\sin x}{x - \sin x} = \frac{\sin x}{\frac{x^3}{6}} = \frac{6 \sin x}{x^3} \]
As \( x \to 0 \), we can again use the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\[ \lim_{x \to 0} \frac{6 \sin x}{x^3} = \lim_{x \to 0} \frac{6 \cdot 1}{x^2} = \infty \]

### Step 3: Combine the results
Now we have:
\[ \left( \frac{\sin x}{x} \right)^{\frac{\sin x}{x - \sin x}} = 1^{\infty} \]
This is an indeterminate form. To resolve this, we can use the logarithm:
Let \( L = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{\sin x}{x - \sin x}} \).
Taking the natural logarithm:
\[ \ln L = \lim_{x \to 0} \frac{\sin x}{x - \sin x} \ln\left( \frac{\sin x}{x} \right) \]

### Step 4: Evaluate \( \ln\left( \frac{\sin x}{x} \right) \)
Using the approximation \( \frac{\sin x}{x} \approx 1 - \frac{x^2}{6} \) for small \( x \):
\[ \ln\left( \frac{\sin x}{x} \right) \approx \ln\left( 1 - \frac{x^2}{6} \right) \approx -\frac{x^2}{6} \]

### Step 5: Substitute back into the limit
Now substituting back:
\[ \ln L = \lim_{x \to 0} \frac{6 \sin x}{x^3} \left(-\frac{x^2}{6}\right) = -\lim_{x \to 0} \frac{\sin x}{x} = -1 \]
Thus,
\[ L = e^{-1} = \frac{1}{e} \]

### Final Answer
Therefore, the value of the limit is:
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{\sin x}{x - \sin x}} = \frac{1}{e} \]