This is a problem involving the computation of a limit involving square roots. The given expression is:

$\lim_{x \to 3} \frac{\sqrt{2x + 3} - \sqrt{x + 6}}{\sqrt{x - 1} - \sqrt{8 - 2x}}$

Step-by-step Solution:

Step 1: Analyze the Expression

At $x = 3$, directly substituting leads to an indeterminate form $\frac{0}{0}$. To resolve this, we rationalize both the numerator and denominator.

Step 2: Rationalize the Numerator

The numerator is $\sqrt{2x + 3} - \sqrt{x + 6}$. Multiply and divide by the conjugate: $\sqrt{2x + 3} - \sqrt{x + 6} = \frac{(\sqrt{2x + 3} - \sqrt{x + 6})(\sqrt{2x + 3} + \sqrt{x + 6})}{\sqrt{2x + 3} + \sqrt{x + 6}}$ $= \frac{(2x + 3) - (x + 6)}{\sqrt{2x + 3} + \sqrt{x + 6}}$ $= \frac{x - 3}{\sqrt{2x + 3} + \sqrt{x + 6}}$

Step 3: Rationalize the Denominator

The denominator is $\sqrt{x - 1} - \sqrt{8 - 2x}$. Similarly, multiply and divide by the conjugate: $\sqrt{x - 1} - \sqrt{8 - 2x} = \frac{(\sqrt{x - 1} - \sqrt{8 - 2x})(\sqrt{x - 1} + \sqrt{8 - 2x})}{\sqrt{x - 1} + \sqrt{8 - 2x}}$ $= \frac{(x - 1) - (8 - 2x)}{\sqrt{x - 1} + \sqrt{8 - 2x}}$ $= \frac{3x - 9}{\sqrt{x - 1} + \sqrt{8 - 2x}}$

Step 4: Simplify the Entire Expression

Now substitute these results into the original limit: $\lim_{x \to 3} \frac{\sqrt{2x + 3} - \sqrt{x + 6}}{\sqrt{x - 1} - \sqrt{8 - 2x}} = \lim_{x \to 3} \frac{\frac{x - 3}{\sqrt{2x + 3} + \sqrt{x + 6}}}{\frac{3x - 9}{\sqrt{x - 1} + \sqrt{8 - 2x}}}$ Simplify the complex fraction: $= \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 1} + \sqrt{8 - 2x})}{(3x - 9)(\sqrt{2x + 3} + \sqrt{x + 6})}$ Factor $3x - 9$ as $3(x - 3)$: $= \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 1} + \sqrt{8 - 2x})}{3(x - 3)(\sqrt{2x + 3} + \sqrt{x + 6})}$ Cancel $x - 3$ (valid as $x \to 3, x - 3 \neq 0$): $= \lim_{x \to 3} \frac{\sqrt{x - 1} + \sqrt{8 - 2x}}{3(\sqrt{2x + 3} + \sqrt{x + 6})}$

Step 5: Substitute $x = 3$

Now substitute $x = 3$ into the simplified expression: $\frac{\sqrt{3 - 1} + \sqrt{8 - 2(3)}}{3(\sqrt{2(3) + 3} + \sqrt{3 + 6})}$ $= \frac{\sqrt{2} + \sqrt{2}}{3(\sqrt{9} + \sqrt{9})}$ $= \frac{2\sqrt{2}}{3(3 + 3)}$ $= \frac{2\sqrt{2}}{18} = \frac{\sqrt{2}}{9}$

Final Answer:

$\lim_{x \to 3} \frac{\sqrt{2x + 3} - \sqrt{x + 6}}{\sqrt{x - 1} - \sqrt{8 - 2x}} = \frac{\sqrt{2}}{9}$

Would you like further clarification or more details on any step?

Related Questions:

1. What are common strategies for solving indeterminate forms in limits?
2. How do you recognize when to rationalize in a limit problem?
3. Why is the cancellation of terms like $x - 3$ valid in this context?
4. Can L’Hôpital’s Rule be applied to this problem?
5. How would this problem change if the limit approached a different value (e.g., $x \to 2$)?

Tip:

Always check for indeterminate forms before proceeding with algebraic manipulations or special techniques like rationalization or L’Hôpital’s Rule.