This is a limit problem involving radicals:

$\lim_{x \to a} \frac{\sqrt{9+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}$

To solve this, rationalize both the numerator and denominator separately by multiplying and dividing by their respective conjugates. Here's the process:

Step 1: Rationalize the numerator

The numerator is: $\sqrt{9+2x} - \sqrt{3x}$ Multiply and divide by its conjugate: $\frac{\sqrt{9+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}} \cdot \frac{\sqrt{9+2x} + \sqrt{3x}}{\sqrt{9+2x} + \sqrt{3x}}$

After simplification: $\text{Numerator becomes: } \frac{(9+2x) - (3x)}{\sqrt{9+2x} + \sqrt{3x}} = \frac{9 - x}{\sqrt{9+2x} + \sqrt{3x}}$

Step 2: Rationalize the denominator

The denominator is: $\sqrt{3a+x} - 2\sqrt{x}$ Multiply and divide by its conjugate: $\frac{\sqrt{3a+x} - 2\sqrt{x}}{\sqrt{3a+x} - 2\sqrt{x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{3a+x} + 2\sqrt{x}}$

After simplification: $\text{Denominator becomes: } \frac{(3a+x) - 4x}{\sqrt{3a+x} + 2\sqrt{x}} = \frac{3a - 3x}{\sqrt{3a+x} + 2\sqrt{x}}$

Step 3: Substitute into the limit

Now the fraction becomes: $\lim_{x \to a} \frac{\frac{9 - x}{\sqrt{9+2x} + \sqrt{3x}}}{\frac{3a - 3x}{\sqrt{3a+x} + 2\sqrt{x}}}$

This simplifies further to: $\lim_{x \to a} \frac{(9-x)(\sqrt{3a+x} + 2\sqrt{x})}{(3a-3x)(\sqrt{9+2x} + \sqrt{3x})}$

Step 4: Substitute $x = a$ (if no indeterminate form arises)

After substituting $x = a$: $\frac{(9-a)(\sqrt{3a+a} + 2\sqrt{a})}{(3a-3a)(\sqrt{9+2a} + \sqrt{3a})}$

The full expansion simplifies numerical simpl ax$$\lim_{x \to a}\ .Let's simplify this carefully:

We are solving: [ \lim_{x \to a} \frac{\sqrt{9+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}$$

Step 1: Analyze numerator

The numerator is: $\sqrt{9+2x} - \sqrt{3x}$ Multiply and divide by its conjugate: $\frac{\sqrt{9+2x} - \sqrt{3x}}{\sqrt{9+2x} + \sqrt{3x}}$ This simplifies to: $\frac{(9+2x) - (3x)}{\sqrt{9+2x} + \sqrt{3x}} = \frac{9 - x}{\sqrt{9+2x} + \sqrt{3x}}$

Step 2: Analyze denominator

The denominator is: $\sqrt{3a+x} - 2\sqrt{x}$ Multiply and divide by its conjugate: $\frac{\sqrt{3a+x} - 2\sqrt{x}}{\sqrt{3a+x} + 2\sqrt{x}}$ This simplifies to: $\frac{(3a+x) - 4x}{\sqrt{3a+x} + 2\sqrt{x}} = \frac{3a - 3x}{\sqrt{3a+x} + 2\sqrt{x}}$

Step 3: Combine numerator and denominator

The limit becomes: $\lim_{x \to a} \frac{\frac{9-x}{\sqrt{9+2x} + \sqrt{3x}}}{\frac{3a-3x}{\sqrt{3a+x} + 2\sqrt{x}}}$

This simplifies further to: $\lim_{x \to a} \frac{(9-x)(\sqrt{3a+x} + 2\sqrt{x})}{(3a-3x)(\sqrt{9+2x} + \sqrt{3x})}$

Step 4: Factorize and evaluate the limit

Observe that as $x \to a$, the numerators and denominators approach non-zero values. Substitute $x = a$ to find the final result:

Substitute $x = a$ directly: $\frac{(9-a)(\sqrt{3a+a} + 2\sqrt{a})}{(3a-3a)(\sqrt{9+2a} + \sqrt{3a})}$

If indeterminate forms appear, further algebraic adjustments are necessary. Would you like more help refining this solution? Let me know!