Step-by-step Solution:

We need to evaluate the limit: \[ \lim_{x \to a} \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}} \] Since both numerator and denominator contain square roots, we use rationalization to simplify. Numerator: \[\] \[ \sqrt{a + 2x} - \sqrt{3x} \] Multiply and divide by the conjugate: \[ \frac{(\sqrt{a + 2x} - \sqrt{3x})(\sqrt{a + 2x} + \sqrt{3x})}{\sqrt{a + 2x} + \sqrt{3x}} \] Using the identity \((\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B\): \[ a + 2x - 3x = a - x \] Thus, the numerator becomes: \[ \frac{a - x}{\sqrt{a + 2x} + \sqrt{3x}} \] Denominator: \[ \sqrt{3a + x} - 2\sqrt{x} \] Multiply and divide by the conjugate: \[ \frac{(\sqrt{3a + x} - 2\sqrt{x})(\sqrt{3a + x} + 2\sqrt{x})}{\sqrt{3a + x} + 2\sqrt{x}} \] Using the identity: \[ 3a + x - 4x = 3a - 3x = 3(a - x) \] Thus, the denominator becomes: \[ \frac{3(a - x)}{\sqrt{3a + x} + 2\sqrt{x}} \] Step 2: Compute the Limit, Now the fraction simplifies to: \[ \frac{\frac{a - x}{\sqrt{a + 2x} + \sqrt{3x}}}{\frac{3(a - x)}{\sqrt{3a + x} + 2\sqrt{x}}} \] Cancel \( a - x \): \[ \frac{1}{\sqrt{a + 2x} + \sqrt{3x}} \times \frac{\sqrt{3a + x} + 2\sqrt{x}}{3} \] Taking the limit as \( x \to a \), substitute \( x = a \): \[ \frac{1}{\sqrt{a + 2a} + \sqrt{3a}} \times \frac{\sqrt{3a + a} + 2\sqrt{a}}{3} \] \[ \frac{1}{\sqrt{3a} + \sqrt{3a}} \times \frac{\sqrt{4a} + 2\sqrt{a}}{3} \] \[ \frac{1}{2\sqrt{3a}} \times \frac{2\sqrt{a} + 2\sqrt{a}}{3} \] \[ \frac{1}{2\sqrt{3a}} \times \frac{4\sqrt{a}}{3} \] \[ \frac{4\sqrt{a}}{6\sqrt{3a}} \] \[ \frac{2}{3\sqrt{3}} \] Thus, the correct answer is: \( D. \frac{2}{3\sqrt{3}} \)