`lim_(x to pi//2) (sin x - (sin x)^(sin x))/(1- sin x + log sin x)`=

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the limit

\[ \lim_{x \to \frac{\pi}{2}} \frac{\sin x - (\sin x)^{\sin x}}{1 - \sin x + \log(\sin x)}, \]

we will follow these steps:

### Step 1: Substitute \(\sin x\) with \(t\)

Let \(t = \sin x\). As \(x\) approaches \(\frac{\pi}{2}\), \(t\) approaches \(1\). Therefore, we can rewrite the limit as:

\[ \lim_{t \to 1} \frac{t - t^t}{1 - t + \log t}. \]

### Step 2: Evaluate the limit

Substituting \(t = 1\) directly into the expression gives us:

\[ \frac{1 - 1^1}{1 - 1 + \log 1} = \frac{0}{0}, \]

which is an indeterminate form. We can apply L'Hôpital's Rule.

### Step 3: Differentiate the numerator and denominator

Using L'Hôpital's Rule, we differentiate the numerator and denominator:

1. **Numerator**:
- The derivative of \(t - t^t\) can be computed using the product and chain rule.
- Let \(y = t^t\), then \(\log y = t \log t\).
- Differentiating both sides gives us:
\[ \frac{1}{y} \frac{dy}{dt} = \log t + 1. \]
Thus,
\[ \frac{dy}{dt} = t^t (\log t + 1). \]
- Therefore, the derivative of the numerator is:
\[ 1 - \frac{dy}{dt} = 1 - t^t (\log t + 1). \]

2. **Denominator**:
- The derivative of \(1 - t + \log t\) is:
\[ -1 + \frac{1}{t}. \]

### Step 4: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{t \to 1} \frac{1 - t^t (\log t + 1)}{-1 + \frac{1}{t}}. \]

### Step 5: Substitute \(t = 1\) again

Substituting \(t = 1\) into the new limit gives us:

- For the numerator:
\[ 1 - 1^1(0 + 1) = 1 - 1 = 0. \]
- For the denominator:
\[ -1 + 1 = 0. \]

This is again an indeterminate form, so we apply L'Hôpital's Rule a second time.

### Step 6: Differentiate again

1. **Numerator**:
- Differentiate \(1 - t^t (\log t + 1)\):
- The derivative is:
\[ -\left( t^t (\log t + 1) + t^t \cdot \frac{1}{t} \cdot (1 + \log t) \right). \]

2. **Denominator**:
- The derivative of \(-1 + \frac{1}{t}\) is:
\[ -\frac{1}{t^2}. \]

### Step 7: Apply L'Hôpital's Rule again

Now we can apply L'Hôpital's Rule again:

\[ \lim_{t \to 1} \frac{-\left( t^t (\log t + 1) + t^t \cdot \frac{1}{t} \cdot (1 + \log t) \right)}{-\frac{1}{t^2}}. \]

### Step 8: Substitute \(t = 1\) again

Substituting \(t = 1\):

- The numerator becomes:
\[ -\left(1(0 + 1) + 1 \cdot (1)(1 + 0)\right) = -2. \]
- The denominator becomes:
\[ -(-1) = 1. \]

Thus, the limit evaluates to:

\[ \lim_{t \to 1} \frac{-2}{1} = 2. \]

### Final Answer

Therefore, the limit is

\[ \boxed{2}. \]