Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>\(\lim_{x \to 3} \dfrac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt{2}}\) is equal to</p>
<p>\(\sqrt{3}\)</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>\(\dfrac{1}{2\sqrt{2}}\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: Rationalize both numerator and denominator separately by multiplying by their conjugates to eliminate the indeterminate form 0/0, then simplify the resulting algebraic expressions.
<p><strong>Step 1:</strong> Check the form at x = 3: numerator = √9 - 3 = 0, denominator = √2 - √2 = 0. This is 0/0 form.</p><p><strong>Step 2:</strong> Rationalize the numerator by multiplying by <span style='color:blue'>√(3x) + 3</span>:<br/>Numerator becomes: (3x - 9)/(√(3x) + 3) = 3(x - 3)/(√(3x) + 3)</p><p><strong>Step 3:</strong> Rationalize the denominator by multiplying by <span style='color:blue'>√(2x - 4) + √2</span>:<br/>Denominator becomes: (2x - 4 - 2)/(√(2x - 4) + √2) = 2(x - 3)/(√(2x - 4) + √2)</p><p><strong>Step 4:</strong> The limit becomes:<br/>lim(x→3) [3(x-3)/(√(3x)+3)] × [√(2x-4)+√2]/[2(x-3)]</p><p><strong>Step 5:</strong> Cancel (x - 3) terms:<br/>= lim(x→3) [3(√(2x-4)+√2)] / [2(√(3x)+3)]</p><p><strong>Step 6:</strong> Substitute x = 3:<br/>= 3(√2 + √2) / [2(√9 + 3)] = 3(2√2) / [2(3 + 3)] = 6√2/12 = <span style='color:green'>√2/2</span></p><p>∴ Answer: A</p>
Correct Answer: A