`lim_(xrarr0)(sinx)/(sqrt(x+1)-sqrt(1-x))` is equal to

Options

A. 2
B. 0
C. 1
D. `-1`

Explanation

To solve the limit \( \lim_{x \to 0} \frac{\sin x}{\sqrt{x+1} - \sqrt{1-x}} \), we can follow these steps:

### Step 1: Rationalize the Denominator
We start by rationalizing the denominator. We multiply the numerator and the denominator by the conjugate of the denominator, which is \( \sqrt{x+1} + \sqrt{1-x} \).

\[ \lim_{x \to 0} \frac{\sin x}{\sqrt{x+1} - \sqrt{1-x}} \cdot \frac{\sqrt{x+1} + \sqrt{1-x}}{\sqrt{x+1} + \sqrt{1-x}} \]

### Step 2: Simplify the Expression
This gives us:

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

The denominator simplifies as follows:

\[ (\sqrt{x+1})^2 - (\sqrt{1-x})^2 = (x + 1) - (1 - x) = x + 1 - 1 + x = 2x \]

So, we have:

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

### Step 3: Split the Limit
Now we can separate the limit:

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

### Step 4: Evaluate the First Limit
We know from calculus that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

### Step 5: Evaluate the Second Limit
Now we need to evaluate:

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

Substituting \( x = 0 \):

\[ \frac{\sqrt{0+1} + \sqrt{1-0}}{2} = \frac{\sqrt{1} + \sqrt{1}}{2} = \frac{1 + 1}{2} = 1 \]

### Step 6: Combine the Results
Now we can combine our results:

\[ 1 \cdot 1 = 1 \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to 0} \frac{\sin x}{\sqrt{x+1} - \sqrt{1-x}} = 1 \]

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