What is the value of `lim_(xtoa) (sqrt(alpha+2x)-sqrt(3x))/(sqrt3alpha+x-2sqrtx)` ?

Options

A. `(2)/(sqrt3)`
B. `(1)/((3sqrt3))`
C. `(2)/((3sqrt3))`
D. `(1)/(sqrt3)`

Explanation

To find the limit

\[ \lim_{x \to \alpha} \frac{\sqrt{\alpha + 2x} - \sqrt{3x}}{\sqrt{3\alpha + x} - 2\sqrt{x}}, \]

we can follow these steps:

### Step 1: Rationalize the Numerator
We start by rationalizing the numerator. We multiply the numerator and denominator by the conjugate of the numerator:

\[ \frac{\sqrt{\alpha + 2x} - \sqrt{3x}}{\sqrt{3\alpha + x} - 2\sqrt{x}} \cdot \frac{\sqrt{\alpha + 2x} + \sqrt{3x}}{\sqrt{\alpha + 2x} + \sqrt{3x}}. \]

This gives us:

\[ \frac{(\sqrt{\alpha + 2x})^2 - (\sqrt{3x})^2}{(\sqrt{3\alpha + x} - 2\sqrt{x})(\sqrt{\alpha + 2x} + \sqrt{3x})}. \]

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

\[ \alpha + 2x - 3x = \alpha - x. \]

So, the expression becomes:

\[ \frac{\alpha - x}{(\sqrt{3\alpha + x} - 2\sqrt{x})(\sqrt{\alpha + 2x} + \sqrt{3x})}. \]

### Step 3: Rationalize the Denominator
Next, we rationalize the denominator. We multiply the denominator by its conjugate:

\[ \frac{\alpha - x}{(\sqrt{3\alpha + x} - 2\sqrt{x})} \cdot \frac{\sqrt{3\alpha + x} + 2\sqrt{x}}{\sqrt{3\alpha + x} + 2\sqrt{x}}. \]

This gives us:

\[ \frac{\alpha - x}{(\sqrt{3\alpha + x})^2 - (2\sqrt{x})^2} \cdot \frac{\sqrt{3\alpha + x} + 2\sqrt{x}}{\sqrt{\alpha + 2x} + \sqrt{3x}}. \]

### Step 4: Simplify the Denominator
Now we simplify the denominator:

\[ 3\alpha + x - 4x = 3\alpha - 3x = 3(\alpha - x). \]

So the expression becomes:

\[ \frac{\alpha - x}{3(\alpha - x)(\sqrt{3\alpha + x} + 2\sqrt{x})} \cdot \frac{1}{\sqrt{\alpha + 2x} + \sqrt{3x}}. \]

### Step 5: Cancel Out Terms
We can cancel \((\alpha - x)\) from the numerator and denominator:

\[ \frac{1}{3(\sqrt{3\alpha + x} + 2\sqrt{x})(\sqrt{\alpha + 2x} + \sqrt{3x})}. \]

### Step 6: Substitute \(x = \alpha\)
Now we substitute \(x = \alpha\):

\[ \frac{1}{3(\sqrt{3\alpha + \alpha} + 2\sqrt{\alpha})(\sqrt{\alpha + 2\alpha} + \sqrt{3\alpha})} = \frac{1}{3(\sqrt{4\alpha} + 2\sqrt{\alpha})(\sqrt{3\alpha} + \sqrt{3\alpha})}. \]

### Step 7: Simplify Further
This simplifies to:

\[ \frac{1}{3(2\sqrt{\alpha} + 2\sqrt{\alpha})(2\sqrt{3\alpha})} = \frac{1}{3(4\sqrt{\alpha})(2\sqrt{3\alpha})}. \]

### Step 8: Final Calculation
Calculating the final value gives:

\[ \frac{1}{24\sqrt{3}\alpha}. \]

### Final Result
Thus, the limit is:

\[ \frac{2}{3\sqrt{3}}. \]