Key Concepts and Formulas

• Indeterminate Forms: Recognizing when a limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Differentiation of Square Roots: The derivative of $\sqrt{u}$ with respect to $x$ is $\frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$.

Step-by-Step Solution

Step 1: Evaluate the limit by direct substitution. We are asked to find the limit: $L = \mathop {\lim }\limits_{x \to 3} \frac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt 2}$ Substitute $x = 3$ into the expression: Numerator: $\sqrt{3(3)} - 3 = \sqrt{9} - 3 = 3 - 3 = 0$ Denominator: $\sqrt{2(3) - 4} - \sqrt 2 = \sqrt{6 - 4} - \sqrt 2 = \sqrt{2} - \sqrt 2 = 0$ Since we obtain the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule. L'Hôpital's Rule states that if $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$. Let $f(x) = \sqrt{3x} - 3$ and $g(x) = \sqrt{2x - 4} - \sqrt 2$. We need to find the derivatives of $f(x)$ and $g(x)$.

Step 3: Calculate the derivative of the numerator, $f'(x)$. $f(x) = \sqrt{3x} - 3 = (3x)^{1/2} - 3$ Using the power rule and chain rule for differentiation: $\frac{d}{dx}(\sqrt{3x}) = \frac{1}{2\sqrt{3x}} \cdot \frac{d}{dx}(3x) = \frac{1}{2\sqrt{3x}} \cdot 3 = \frac{3}{2\sqrt{3x}}$ The derivative of a constant (-3) is 0. So, $f'(x) = \frac{3}{2\sqrt{3x}}$.

Step 4: Calculate the derivative of the denominator, $g'(x)$. $g(x) = \sqrt{2x - 4} - \sqrt 2 = (2x - 4)^{1/2} - \sqrt 2$ Using the power rule and chain rule for differentiation: $\frac{d}{dx}(\sqrt{2x - 4}) = \frac{1}{2\sqrt{2x - 4}} \cdot \frac{d}{dx}(2x - 4) = \frac{1}{2\sqrt{2x - 4}} \cdot 2 = \frac{1}{\sqrt{2x - 4}}$ The derivative of a constant ($-\sqrt 2$) is 0. So, $g'(x) = \frac{1}{\sqrt{2x - 4}}$.

Step 5: Apply L'Hôpital's Rule by taking the limit of the ratio of the derivatives. $L = \mathop {\lim }\limits_{x \to 3} \frac{f'(x)}{g'(x)} = \mathop {\lim }\limits_{x \to 3} \frac{\frac{3}{2\sqrt{3x}}}{\frac{1}{\sqrt{2x - 4}}}$ Simplify the complex fraction: $L = \mathop {\lim }\limits_{x \to 3} \frac{3}{2\sqrt{3x}} \cdot \sqrt{2x - 4}$ $L = \mathop {\lim }\limits_{x \to 3} \frac{3\sqrt{2x - 4}}{2\sqrt{3x}}$

Step 6: Evaluate the limit by direct substitution after applying L'Hôpital's Rule. Now, substitute $x = 3$ into the simplified expression: $L = \frac{3\sqrt{2(3) - 4}}{2\sqrt{3(3)}}$ $L = \frac{3\sqrt{6 - 4}}{2\sqrt{9}}$ $L = \frac{3\sqrt{2}}{2 \cdot 3}$ $L = \frac{3\sqrt{2}}{6}$ $L = \frac{\sqrt{2}}{2}$ To express this in a form that matches the options, we can write $\frac{\sqrt{2}}{2}$ as $\frac{\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{\sqrt{2}}$.

Step 7: (Alternative Method - Rationalization) Alternatively, we could rationalize the numerator and the denominator without using L'Hôpital's Rule. Multiply the numerator and denominator by the conjugate of the numerator $(\sqrt{3x} + 3)$ and the conjugate of the denominator $(\sqrt{2x - 4} + \sqrt 2)$: $L = \mathop {\lim }\limits_{x \to 3} \frac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt 2} \times \frac{\sqrt{3x} + 3}{\sqrt{3x} + 3} \times \frac{\sqrt{2x - 4} + \sqrt 2}{\sqrt{2x - 4} + \sqrt 2}$ $L = \mathop {\lim }\limits_{x \to 3} \frac{(3x - 9)}{(\sqrt{2x - 4} - \sqrt 2)(\sqrt{3x} + 3)} \times \frac{\sqrt{2x - 4} + \sqrt 2}{(2x - 4 - 2)}$ $L = \mathop {\lim }\limits_{x \to 3} \frac{(3x - 9)}{(2x - 6)} \times \frac{(\sqrt{2x - 4} + \sqrt 2)}{(\sqrt{3x} + 3)}$ Factor out constants: $L = \mathop {\lim }\limits_{x \to 3} \frac{3(x - 3)}{2(x - 3)} \times \frac{(\sqrt{2x - 4} + \sqrt 2)}{(\sqrt{3x} + 3)}$ Cancel out $(x - 3)$: $L = \mathop {\lim }\limits_{x \to 3} \frac{3}{2} \times \frac{(\sqrt{2x - 4} + \sqrt 2)}{(\sqrt{3x} + 3)}$ Now, substitute $x = 3$: $L = \frac{3}{2} \times \frac{(\sqrt{2(3) - 4} + \sqrt 2)}{(\sqrt{3(3)} + 3)}$ $L = \frac{3}{2} \times \frac{(\sqrt{6 - 4} + \sqrt 2)}{(\sqrt{9} + 3)}$ $L = \frac{3}{2} \times \frac{(\sqrt{2} + \sqrt 2)}{(3 + 3)}$ $L = \frac{3}{2} \times \frac{2\sqrt{2}}{6}$ $L = \frac{3}{2} \times \frac{\sqrt{2}}{3}$ $L = \frac{\sqrt{2}}{2}$ $L = \frac{1}{\sqrt{2}}$

Common Mistakes & Tips

• Incorrect Differentiation: Ensure accurate application of the chain rule when differentiating square roots. For $\sqrt{ax+b}$, the derivative is $\frac{a}{2\sqrt{ax+b}}$.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when simplifying fractions and rationalizing terms.
• Forgetting the Indeterminate Form: Always check for the indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule. Applying it otherwise can lead to incorrect results.
• Simplification of Radicals: Simplify radical expressions to match the format of the given options if necessary. For example, $\frac{\sqrt{2}}{2}$ can be written as $\frac{1}{\sqrt{2}}$.

Summary

The problem requires evaluating a limit that results in the indeterminate form $\frac{0}{0}$. We can solve this using L'Hôpital's Rule by differentiating the numerator and the denominator separately and then evaluating the limit of the resulting ratio. Alternatively, rationalizing the numerator and denominator also leads to the same result. Both methods confirm that the limit evaluates to $\frac{\sqrt{2}}{2}$, which is equivalent to $\frac{1}{\sqrt{2}}$.

The final answer is $\boxed{\frac{1}{\sqrt{2}}}$ which corresponds to option (B).