Evaluate the following limits:
`lim_(xrarr0)((sqrt(1+2x)-sqrt(1-2x)))/(sinx)`

Explanation

To evaluate the limit

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

we can follow these steps:

### Step 1: Multiply by the Conjugate
To simplify the expression, we will multiply the numerator and the denominator by the conjugate of the numerator:

\[ \frac{\sqrt{1 + 2x} - \sqrt{1 - 2x}}{\sin x} \cdot \frac{\sqrt{1 + 2x} + \sqrt{1 - 2x}}{\sqrt{1 + 2x} + \sqrt{1 - 2x}}. \]

This gives us:

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

### Step 2: Simplify the Numerator
Using the difference of squares, we can simplify the numerator:

\[ (1 + 2x) - (1 - 2x) = 2x + 2x = 4x. \]

So, we have:

\[ \frac{4x}{\sin x \cdot (\sqrt{1 + 2x} + \sqrt{1 - 2x})}. \]

### Step 3: Rewrite the Limit
Now we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{4x}{\sin x \cdot (\sqrt{1 + 2x} + \sqrt{1 - 2x})}. \]

### Step 4: Use the Limit Property
We know that

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

which implies

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

Thus, we can express our limit as:

\[ 4 \cdot \lim_{x \to 0} \frac{x}{\sin x} \cdot \frac{1}{\sqrt{1 + 2x} + \sqrt{1 - 2x}}. \]

### Step 5: Evaluate the Limit
Now we evaluate the limit of the remaining expression as \(x\) approaches 0:

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

So we have:

\[ \lim_{x \to 0} \frac{4x}{\sin x \cdot (\sqrt{1 + 2x} + \sqrt{1 - 2x})} = 4 \cdot 1 \cdot \frac{1}{2} = 2. \]

### Final Answer
Thus, the limit evaluates to:

\[ \boxed{2}. \]