`lim_(x to 0) (sqrt(1+ sin x) - sqrt(1- sinx))/x`=

Options

A. 1
B. 2
C. 0
D. none

Answer

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

### Step 1: Multiply by the Conjugate
We start by multiplying the numerator and denominator by the conjugate of the numerator:

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

This gives us:

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

### Step 2: Simplify the Numerator
Now simplify the numerator:

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

So we have:

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

### Step 3: Factor Out \(x\)
Now we can factor out \(x\) from the limit:

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

### Step 4: Evaluate the Limit
We know that \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). Therefore, we can evaluate the limit:

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

### Final Answer
Thus, the limit is:

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

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