Limits, Continuity & Differentiability
Rationalisation in Limits
Grade 12
Question:
<p>Evaluate:
\[\lim_{x \to 3} \frac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt{2}}\]</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
<p>\(\sqrt{2}\)</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(2\)</p>
Step-by-Step Solution
Key Concept: Rationalize both numerator and denominator separately by multiplying by conjugates to eliminate indeterminate form 0/0, then simplify algebraically.
<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='text-decoration:overline;'>√3x + 3</span>/<span style='text-decoration:overline;'>√3x + 3</span>:</p><p>Numerator becomes: (3x - 9)/(√3x + 3) = 3(x - 3)/(√3x + 3)</p><p><strong>Step 3:</strong> Rationalize the denominator by multiplying by (√(2x-4) + √2)/(√(2x-4) + √2):</p><p>Denominator becomes: (2x - 4 - 2)/(√(2x-4) + √2) = 2(x - 3)/(√(2x-4) + √2)</p><p><strong>Step 4:</strong> The limit becomes:</p><p>lim_{x→3} [3(x-3)/(√3x + 3)] · [(√(2x-4) + √2)/(2(x-3))]</p><p><strong>Step 5:</strong> Cancel (x - 3) terms:</p><p>lim_{x→3} [3(√(2x-4) + √2)]/(2(√3x + 3))</p><p><strong>Step 6:</strong> Substitute x = 3:</p><p>[3(√2 + √2)]/(2(√9 + 3)) = [3(2√2)]/(2(6)) = 6√2/12 = <strong>√2/2</strong></p><p>∴ Answer: A</p>
Correct Answer: A