`lim_(x to 0) (x +2 sin x)/(sqrt(x^(2)+2 sin x + 1)-sqrt(sin^(2) x - x+ 1))`is

Options

A. 6
B. 2
C. 3
D. 1

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{x + 2 \sin x}{\sqrt{x^2 + 2 \sin x + 1} - \sqrt{\sin^2 x - x + 1}}, \]

we'll follow these steps:

### Step 1: Evaluate the limit directly
First, substitute \(x = 0\) into the expression:

\[ \frac{0 + 2 \sin(0)}{\sqrt{0^2 + 2 \sin(0) + 1} - \sqrt{\sin^2(0) - 0 + 1}} = \frac{0}{\sqrt{1} - \sqrt{1}} = \frac{0}{0}. \]

Since we get the indeterminate form \( \frac{0}{0} \), we need to manipulate the expression.

**Hint:** If you encounter \( \frac{0}{0} \), consider rationalizing the denominator or using L'Hôpital's rule.

### Step 2: Rationalize the denominator
To simplify the expression, we can rationalize the denominator:

\[ \lim_{x \to 0} \frac{(x + 2 \sin x)(\sqrt{x^2 + 2 \sin x + 1} + \sqrt{\sin^2 x - x + 1})}{( \sqrt{x^2 + 2 \sin x + 1} - \sqrt{\sin^2 x - x + 1})(\sqrt{x^2 + 2 \sin x + 1} + \sqrt{\sin^2 x - x + 1})}. \]

The denominator simplifies to:

\[ (x^2 + 2 \sin x + 1) - (\sin^2 x - x + 1) = x^2 + 2 \sin x + 1 - \sin^2 x + x - 1. \]

This simplifies to:

\[ x^2 + x + 2 \sin x - \sin^2 x. \]

**Hint:** Rationalizing the denominator can help eliminate the indeterminate form.

### Step 3: Simplify the expression
Now, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{(x + 2 \sin x)(\sqrt{x^2 + 2 \sin x + 1} + \sqrt{\sin^2 x - x + 1})}{x^2 + x + 2 \sin x - \sin^2 x}. \]

### Step 4: Evaluate the limit again
Now substitute \(x = 0\) again into the simplified expression.

The numerator becomes:

\[ (0 + 2 \sin(0))(\sqrt{0^2 + 2 \sin(0) + 1} + \sqrt{\sin^2(0) - 0 + 1}) = 0. \]

The denominator becomes:

\[ 0^2 + 0 + 2 \sin(0) - \sin^2(0) = 0. \]

We still have \( \frac{0}{0} \). So we apply L'Hôpital's Rule.

**Hint:** If you still get \( \frac{0}{0} \) after simplification, apply L'Hôpital's Rule.

### Step 5: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator:

- The derivative of the numerator \(x + 2 \sin x\) is \(1 + 2 \cos x\).
- The derivative of the denominator \(x^2 + x + 2 \sin x - \sin^2 x\) is \(2x + 1 + 2 \cos x - 2 \sin x \cos x\).

Now, we have:

\[ \lim_{x \to 0} \frac{1 + 2 \cos x}{2x + 1 + 2 \cos x - 2 \sin x \cos x}. \]

### Step 6: Substitute \(x = 0\) again
Now substitute \(x = 0\):

\[ \frac{1 + 2 \cdot 1}{2 \cdot 0 + 1 + 2 \cdot 1 - 2 \cdot 0 \cdot 1} = \frac{1 + 2}{0 + 1 + 2 - 0} = \frac{3}{3} = 1. \]

### Final Answer
Thus, the limit is:

\[ \boxed{1}. \]