`lim_(x->0)(x+2sinx)/(sqrt(x^2+2sinx+1)-sqrt(sin^2x-x+1))`

Options

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

Answer

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 will follow these steps:

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

- In the numerator:
\[ 0 + 2\sin(0) = 0 + 0 = 0 \]

- In the denominator:
\[ \sqrt{0^2 + 2\sin(0) + 1} - \sqrt{\sin^2(0) - 0 + 1} = \sqrt{0 + 0 + 1} - \sqrt{0 - 0 + 1} = 1 - 1 = 0 \]

Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Rationalize the denominator
To resolve the indeterminate form, we will multiply the numerator and denominator by the conjugate of 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:
\[ (\sqrt{x^2 + 2\sin x + 1})^2 - (\sqrt{\sin^2 x - x + 1})^2 = (x^2 + 2\sin x + 1) - (\sin^2 x - x + 1) \]

### Step 3: Simplify the denominator
Now, simplifying the denominator:
\[ x^2 + 2\sin x + 1 - \sin^2 x + x - 1 = x^2 + x + 2\sin x - \sin^2 x \]

### Step 4: Rewrite the limit
Now we rewrite the limit:
\[ \lim_{x \to 0} \frac{(x + 2\sin x)(\sqrt{x^2 + 2\sin x + 1} + \sqrt{\sin^2 x - x + 1})}{x^2 + x + 2\sin x - \sin^2 x} \]

### Step 5: Evaluate the limit again
Substituting \( x = 0 \) again:
- The numerator becomes:
\[ (0 + 2\sin(0))(\sqrt{0 + 0 + 1} + \sqrt{0 - 0 + 1}) = 0 \cdot (1 + 1) = 0 \]

- The denominator becomes:
\[ {0^2 + 0 + 2\sin(0) - \sin^2(0)} = 0 + 0 + 0 - 0 = 0 \]

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

### Step 6: Apply L'Hôpital's Rule
Differentiate the numerator and 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:
\[ 2x + 1 + 2\cos x - 2\sin x \cos x \]

### Step 7: Evaluate the limit again
Now we evaluate:
\[ \lim_{x \to 0} \frac{1 + 2\cos x}{2x + 1 + 2\cos x - 2\sin x \cos x} \]

Substituting \( x = 0 \):
- The numerator becomes:
\[ 1 + 2\cos(0) = 1 + 2 \cdot 1 = 3 \]

- The denominator becomes:
\[ 2 \cdot 0 + 1 + 2\cos(0) - 2\sin(0)\cos(0) = 0 + 1 + 2 \cdot 1 - 0 = 3 \]

Thus, the limit simplifies to:
\[ \frac{3}{3} = 1 \]

### Final Answer
So the limit is:
\[ \lim_{x \to 0} \frac{x + 2\sin x}{\sqrt{x^2 + 2\sin x + 1} - \sqrt{\sin^2 x - x + 1}} = 1 \]